Rd Sharma 2022 for Class 10 Maths Chapter 6 - Co Ordinate Geometry

Answer:

According to the Rectangular Cartesian Co-ordinate system of representing a point (x, y),

If then the point lies in the 1^st quadrant

If then the point lies in the 2^nd quadrant

If then the point lies in the 3^rd quadrant

If then the point lies in the 4^th quadrant

But in case

If then the point lies on the y-axis

If then the point lies on the x-axis

(i) Here the point is given to be P (5, 0). Comparing this with the standard form of

(x, y) we have

Here we see that

Hence the given point lies on the

(ii) Here the point is given to be Q (0, -−2). Comparing this with the standard form of (x, y) we have

Here we see that

Hence the given point lies on the

(iii) Here the point is given to be R (-4, 0). Comparing this with the standard form of (x, y) we have

Here we see that

Hence the given point lies on the

(iv) Here the point is given to be S (0, 5). Comparing this with the standard form of (x, y) we have

Here we see that

Hence the given point lies on the

Answer:

The distance between any two adjacent vertices of a square will always be equal. This distance is nothing but the side of the square.

Here, the side of the square ‘ABCD’ is given to be ‘2a’.

(i) Since it is given that the vertex ‘A’ coincides with the origin we know that the co-ordinates of this point is (0, 0).

We also understand that the side ‘AB’ is along the x-axis. So, the vertex ‘B’ has got to be at a distance of ‘2a’ from ‘A’.

Hence the vertex ‘B’ has the co-ordinates (2a, 0).

Also it is said that the side ‘AD’ is along the y-axis. So, the vertex ‘D’ it has got to be at a distance of ‘2a’ from ‘A’.

Hence the vertex ‘D’ has the co-ordinates (0, 2a)

Finally we have vertex ‘C’ at a distance of ‘2a’ both from vertex ‘B’ as well as ‘D’.

Hence the vertex of ‘C’ has the co-ordinates (2a, 2a)

So, the co-ordinates of the different vertices of the square are

(ii) Here it is said that the centre of the square is at the origin and that the sides of the square are parallel to the axes.

Moving a distance of half the side of the square in either the ‘upward’ or ‘downward’ direction and also along either the ‘right’ or ‘left’ direction will give us all the four vertices of the square.

Half the side of the given square is ‘a’.

The centre of the square is the origin and its vertices are (0, 0). Moving a distance of ‘a’ to the right as well as up will lead us to the vertex ‘A’ and it will have vertices (a, a).

Moving a distance of ‘a’ to the left as well as up will lead us to the vertex ‘B’ and it will have vertices (-(−a, a).

Moving a distance of ‘a’ to the left as well as down will lead us to the vertex ‘C’ and it will have vertices (-(−a, -−a).

Moving a distance of ‘a’ to the right as well as down will lead us to the vertex ‘D’ and it will have vertices (a,-,−a).

So, the co-ordinates of the different vertices of the square are

Answer:

In an equilateral triangle, the height ‘h’ is given by

Here it is given that ‘PQ’ forms the base of two equilateral triangles whose side measures ‘2a’ units.

The height of these two equilateral triangles has got to be

In an equilateral triangle the height drawn from one vertex meets the midpoint of the side opposite this vertex.

So here we have ‘PQ’ being the base lying along the y-axis with its midpoint at the origin, that is at (0, 0).

So the vertices ‘R’ and ‘R’’ will lie perpendicularly to the y-axis on either sides of the origin at a distance of ‘’ units.

Hence the co-ordinates of ‘R’ and ‘R’’ are

Answer:

The distance d between two points and is given by the formula

(i) The two given points are (−6, 7) and (−1, −5)

The distance between these two points is

Hence the distance is.

(ii) The two given points are

The distance between these two points is

Hence the distance is.

(iii) The two given points are and

The distance between these two points is

Hence the distance is.

(iv) The two given points are (a, 0) and (0, b)

The distance between these two points is

Hence the distance is.

Answer:

The distance d between two points and is given by the formula

The distance between two points (3, a) and (4, 1) is given as. Substituting these values in the formula for distance between two points we have,

Now, squaring the above equation on both sides of the equals sign

Thus we arrive at a quadratic equation. Let us solve this now,

The roots of the above quadratic equation are thus 4 and −2.

Thus the value of ‘a’ could either be.

Answer:

The distance d between two points and is given by the formula

Here it is given that one end of a line segment has co−ordinates (2,−3). The abscissa of the other end of the line segment is given to be 10. Let the ordinate of this point be ‘y’.

So, the co−ordinates of the other end of the line segment is (10, y).

The distance between these two points is given to be 10 units.

Substituting these values in the formula for distance between two points we have,

Squaring on both sides of the equation we have,

We have a quadratic equation for ‘y’. Solving for the roots of this equation we have,

The roots of the above equation are ‘−9’ and ‘3’

Thus the ordinates of the other end of the line segment could be.

Answer:

The distance d between two points and is given by the formula

In a rectangle, the opposite sides are equal in length. The diagonals of a rectangle are also equal in length.

Here the four points are A(−4,−1), B(−2,−4), C(4,0) and D(2,3).

First let us check the length of the opposite sides of the quadrilateral that is formed by these points.

We have one pair of opposite sides equal.

Now, let us check the other pair of opposite sides.

The other pair of opposite sides are also equal. So, the quadrilateral formed by these four points is definitely a parallelogram.

For a parallelogram to be a rectangle we need to check if the diagonals are also equal in length.

Now since the diagonals are also equal we can say that the parallelogram is definitely a rectangle.

Hence we have proved that the quadrilateral formed by the four given points is a.

Answer:

The distance d between two points and is given by the formula

In a parallelogram the opposite sides are equal in length.

Here the four points are A(1, −2), B(3, 6), C(5, 10) and D(3, 2).

Let us check the length of the opposite sides of the quadrilateral that is formed by these points.

We have one pair of opposite sides equal.

Now, let us check the other pair of opposite sides.

The other pair of opposite sides is also equal. So, the quadrilateral formed by these four points is definitely a parallelogram.

Hence we have proved that the quadrilateral formed by the given four points is a .

Answer:

In ΔABC, the coordinates of the vertices are A(–2, 0), B(2, 0), C(0, 2). 

$$\begin{gathered} AB=\sqrt{{\left(2+2\right)}^2+{\left(0-0\right)}^2}=4 \\ BC=\sqrt{{\left(0-2\right)}^2+{\left(2-0\right)}^2}=\sqrt{8}=2\sqrt{2} \\ CA=\sqrt{{\left(0+2\right)}^2+{\left(2-0\right)}^2}=\sqrt{8}=2\sqrt{2} \end{gathered}$$

In ΔPQR, the coordinates of the vertices are P(–4, 0), Q(4, 0), R(0, 4).

$$\begin{gathered} PQ=\sqrt{{\left(4+4\right)}^2+{\left(0-0\right)}^2}=8 \\ QR=\sqrt{{\left(0-4\right)}^2+{\left(4-0\right)}^2}=4\sqrt{2} \\ PR=\sqrt{{\left(0+4\right)}^2+{\left(4-0\right)}^2}=4\sqrt{2} \end{gathered}$$

Now, for ΔABC and ΔPQR to be similar, the corresponding sides should be proportional. 

$$\begin{gathered} \mathrm{So}, \frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{PR} \\ \Rightarrow \frac{4}{8}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{1}{2} \end{gathered}$$
Thus, ΔABC is similar to ΔPQR. 

Answer:

The distance d between two points and is given by the formula

In an isosceles triangle there are two sides which are equal in length.

Here the three points are A(3, 0), B(6, 4) and C(−1, 3).

Let us check the length of the three sides of the triangle.

Here, we see that two sides of the triangle are equal. So the triangle formed should be an isosceles triangle.

We can also observe that

Hence proved that the triangle formed by the three given points is an.

Answer:

The distance d between two points and is given by the formula

In a right-angled triangle, by Pythagoras theorem, the square of the longest side is equal to the sum of squares of the other two sides in the triangle.

Here the three points are A(2,−2), B(−2,1) and C(5,2).

Let us find out the lengths of all the sides of the triangle.

Here we have,

Since the square of the longest side is equal to the sum of squares of the other two sides the given triangle is a .

In a right angled triangle the area of the triangle ‘A’ is given by,

In a right angled triangle the sides containing the right angle will not be the longest side.

Hence the area of the given right angled triangle is,

Hence the area of the given right-angled triangle is.

In a right-angled triangle the hypotenuse will be the longest side. Here the longest side is ‘BC’.

Hence the hypotenuse of the given right-angled triangle is

Answer:

The distance d between two points and is given by the formula

In any triangle the sum of lengths of any two sides need to be greater than the third side.

Here the three points are, and

Let us now find out the lengths of all the three sides of the given triangle.

Here we see that,

This is in violation of the basic property of any triangle to exist. Therefore these points cannot give rise to a triangle.

Hence we have proved that the given three points do not form a triangle.

Answer:

Consider the figure. 
Using distance formula, 
$$\begin{gathered} \mathrm{AC}=\sqrt{{\left(2-5\right)}^2+{\left(9-5\right)}^2} \\ \mathrm{AC}=\sqrt{{\left(-3\right)}^2+{\left(4\right)}^2} \\ \mathrm{AC}=\sqrt{25} \\ {\mathrm{AC}}^2=25 \mathrm{units} \end{gathered}$$
$$\begin{gathered} \mathrm{AB}=\sqrt{{\left(2-a\right)}^2+{\left(9-5\right)}^2} \\ \mathrm{AB}=\sqrt{\left(4+a^2-4a\right)+\left(16\right)} \\ \mathrm{AB}=\sqrt{4+a^2-4a+16}=\sqrt{a^2-4a+20} \\ {\mathrm{AB}}^2=(a^2-4a+20) \mathrm{units} \end{gathered}$$
$$\begin{gathered} \mathrm{BC}=\sqrt{{\left(5-a\right)}^2+{\left(5-5\right)}^2} \\ \mathrm{BC}=\sqrt{\left(25+a^2-10a\right)+0} \\ \mathrm{BC}=\sqrt{a^2-10a+25} \\ {\mathrm{BC}}^2=(a^2-10a+25) \mathrm{units} \end{gathered}$$
We are given that ABC is a right triangle right angled at B. 
By Pythagoras theorem, we have;
$$\begin{gathered} {\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{BC}}^2 \\ \Rightarrow 25=(a^2-4a+20)+(a^2-10a+25) \\ \Rightarrow 25=2a^2-14a+45 \\ \Rightarrow 2a^2-14a+20=0 \\ \Rightarrow a^2-7a+10=0 \\ \Rightarrow a^2-5a-2a+10=0 \\ \Rightarrow a(a-5)-2(a-5)=0 \\ \Rightarrow (a-2)(a-5)=0 \\ \Rightarrow a=2 or a=5 \end{gathered}$$
We cannot put a = 5 as it will make BC = 0. So, we ignore a = 5 and accept a = 2.
$$\begin{gathered} \Rightarrow {\mathrm{AB}}^2=4-8+20=16 \\ \Rightarrow \mathrm{AB}=4 \mathrm{units} \\ \Rightarrow {\mathrm{BC}}^2=4+25-20=9 \\ \Rightarrow \mathrm{BC}=3 \mathrm{units} \\ \mathrm{And}, \mathrm{AC}=5 \mathrm{units} \\ \mathrm{Area}=\frac{1}{2}\times \mathrm{base}\times \mathrm{height} \\ \mathrm{Area}=\frac{1}{2}\times 3\times 4=6 \mathrm{square} \mathrm{units} \end{gathered}$$
 

Answer:

It is given that P(2, 2) is equidistant from the points A(−2, k) and B(−2k, −3).

∴ AP = BP

$\Rightarrow \sqrt{{\left(-2-2\right)}^2+{\left(k-2\right)}^2}=\sqrt{{\left(-2k-2\right)}^2+{\left(-3-2\right)}^2}$                (Distance formula)

Squaring on both sides, we get

$$\begin{gathered} 16+k^2-4k+4=4k^2+8k+4+25 \\ \Rightarrow 3k^2+12k+9=0 \\ \Rightarrow k^2+4k+3=0 \\ \Rightarrow \left(k+3\right)\left(k+1\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow k+3=0 \mathrm{or} k+1=0 \\ \Rightarrow k=-3 \mathrm{or} k=-1 \end{gathered}$$

Thus, the value of k is −1 or −3.

When k = −1,

$\mathrm{AP}=\sqrt{{\left(-2-2\right)}^2+{\left(-1-2\right)}^2}=\sqrt{{\left(-4\right)}^2+{\left(-3\right)}^2}=\sqrt{16+9}=\sqrt{25}=5$ units

When k = −3,

$\mathrm{AP}=\sqrt{{\left(-2-2\right)}^2+{\left(-3-2\right)}^2}=\sqrt{{\left(-4\right)}^2+{\left(-5\right)}^2}=\sqrt{16+25}=\sqrt{41}$ units

Answer:

The distance d between two points and is given by the formula

Let the three given points be P(x, y), A(3,6) and B(−3,4).

Now let us find the distance between ‘P’ and ‘A’.

Now, let us find the distance between ‘P’ and ‘B’.

It is given that both these distances are equal. So, let us equate both the above equations,

Squaring on both sides of the equation we get,

Hence the relationship between ‘x’ and ‘y’ based on the given condition is.

Answer:

The distance d between two points and is given by the formula

For three points to be collinear the sum of distances between two pairs of points should be equal to the third pair of points.

The given points are A (−2, 5), B (0, 1) and C (2, −3)

Let us find the distances between the possible pairs of points.

We see that

Since sum of distances between two pairs of points equals the distance between the third pair of points the three points must be collinear.

Hence we have proved that the three given points are.

Answer:

The distance between two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given as:
$d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Now, the point A(2, –4) is equidistant from P(3, 8) and Q(–10, y).
Thus,
$$\begin{gathered} \sqrt{{\left(3-2\right)}^2+{\left(8+4\right)}^2}=\sqrt{{\left(2+10\right)}^2+{\left(-4-y\right)}^2} \\ \Rightarrow \sqrt{1+144}=\sqrt{144+16+y^2+8y} \\ \Rightarrow \sqrt{145}=\sqrt{y^2+8y+160} \end{gathered}$$

Squaring both sides,
$$\begin{gathered} 145=y^2+8y+160 \\ \Rightarrow y^2+8y+15=0 \\ \Rightarrow y^2+5y+3y+15=0 \\ \Rightarrow y\left(y+5\right)+3\left(y+5\right)=0 \\ \Rightarrow \left(y+3\right)\left(y+5\right)=0 \\ \Rightarrow y=-3 \mathrm{or} y=-5 \end{gathered}$$

Thus, the value of y is either $-3 \mathrm{or} -5$. Therefore, Q(–10, –3) or Q(–10, –5). 

Now, the distance PQ if Q(–10, –3) is given as:
$\begin{array}{rcl}PQ& =& \sqrt{{\left(3+10\right)}^2+{\left(8+3\right)}^2}\\ & =& \sqrt{169+121}\\ & =& \sqrt{290}\end{array}$

Similarly, the distance PQ if Q(–10, –5) is given as:
$\begin{array}{rcl}PQ& =& \sqrt{{\left(3+10\right)}^2+{\left(8+5\right)}^2}\\ & =& \sqrt{169+169}\\ & =& \sqrt{238}\\ & =& 13\sqrt{2}\end{array}$

Therefore, the distance PQ is either $\sqrt{290}$ units or $13\sqrt{2}$ units.

Answer:

We are given three vertices of a parallelogram A(3, 4),  B(3, 8) and C(9, 8). 

We know that diagonals of a parallelogram bisect each other. Let the fourth vertex be D(x, y).
Mid point of BD = $\left(\frac{x+3}{2}, \frac{y+8}{2}\right)$
Mid point of AC = $\left(\frac{9+3}{2}, \frac{4+8}{2}\right)$ = $\left(6, 6\right)$
Since the mid point of BD = mid point of AC
So, $\left(\frac{x+3}{2}, \frac{y+8}{2}\right)$ = (6, 6)
Thus, x = 9, y = 4.
So, the fourth vertex is (9, 4).

 

Answer:

Disclaimer: In (i), the coordinates of point C should be (11, –1).

(i) A(2, –2), B(7, 3), C(11, –1), D(6, –6)

Now, using the distance formula we can find the sides of the quadrilateral.

AB$=\sqrt{{\left(7-2\right)}^2+{\left(3+2\right)}^2}=\sqrt{25+25}=\sqrt{50}=5\sqrt{2}$
BC$=\sqrt{{\left(11-7\right)}^2+{\left(-1-3\right)}^2}=\sqrt{4^2+{\left(-4\right)}^2}=\sqrt{32}=4\sqrt{2}$
CD$=\sqrt{{\left(6-11\right)}^2+{\left(-6+1\right)}^2}=\sqrt{50}=5\sqrt{2}$
AD$=\sqrt{{\left(2-6\right)}^2+{\left(-2+6\right)}^2}=\sqrt{32}=4\sqrt{2}$

Similarly, we can find the length of diagonals.
AC$=\sqrt{{\left(11-2\right)}^2+{\left(-1+2\right)}^2}=\sqrt{81+1}=\sqrt{82}$
BD$=\sqrt{{\left(6-7\right)}^2+{\left(-6-3\right)}^2}=\sqrt{1+81}=\sqrt{82}$

Thus, we conclude AB = CD and BC = AD.
Also, AC = BD, diagonals are equal.
Hence, the required quadrilateral is a rectangle.

(ii) A(4, 5), B(7, 6), C(4, 3), D(1, 2)

The distance between two points Pand Qis given by distance formula:

Thus,
$\begin{array}{rcl}AB& =& \sqrt{{\left(7-4\right)}^2+{\left(6-5\right)}^2}\\ & =& \sqrt{9+1}\\ & =& \sqrt{10}\end{array}$

$\begin{array}{rcl}BC& =& \sqrt{{\left(4-7\right)}^2+{\left(3-6\right)}^2}\\ & =& \sqrt{9+9}\\ & =& \sqrt{18}\end{array}$

$\begin{array}{rcl}CD& =& \sqrt{{\left(1-4\right)}^2+{\left(2-3\right)}^2}\\ & =& \sqrt{9+1}\\ & =& \sqrt{10}\end{array}$

$\begin{array}{rcl}AD& =& \sqrt{{\left(4-1\right)}^2+{\left(5-2\right)}^2}\\ & =& \sqrt{9+9}\\ & =& \sqrt{18}\end{array}$

Hence,

AB = CD and BC = DA

Here, opposite sides of the quadrilateral is equal.

Now, 
$\begin{array}{rcl}AC& =& \sqrt{{\left(4-4\right)}^2+{\left(3-5\right)}^2}\\ & =& \sqrt{4}\\ & =& 2\end{array}$

$\begin{array}{rcl}BD& =& \sqrt{{\left(7-1\right)}^2+{\left(6-2\right)}^2}\\ & =& \sqrt{36+16}\\ & =& \sqrt{52}\end{array}$

Since the diagonals are not equal, it is a parallelogram.

Answer:

The distance d between two points and is given by the formula

In a rhombus all the sides are equal in length. And the area ‘A’ of a rhombus is given as

Here the four points are A(3,0), B(4,5), C(−1,4) and D(−2,−1).

First let us check if all the four sides are equal.

Here, we see that all the sides are equal, so it has to be a rhombus.

Hence we have proved that the quadrilateral formed by the given four vertices is a.

Now let us find out the lengths of the diagonals of the rhombus.

Now using these values in the formula for the area of a rhombus we have,

Thus the area of the given rhombus is.

Answer:

The distance d between two points and is given by the formula

For three points to be collinear the sum of distances between any two pairs of points should be equal to the third pair of points.

The given points are A(3,1), B(6,4) and C(8,6).

Let us find the distances between the possible pairs of points.

We see that.

Since sum of distances between two pairs of points equals the distance between the third pair of points the three points must be collinear.

Hence, the three given points are.

Answer:

The distance d between two points and is given by the formula

Here we are to find out a point on the y−axis which is equidistant from both the points A(5,−2) and B(−3,2).

Let this point be denoted as C(x, y).

Since the point lies on the y-axis the value of its ordinate will be 0. Or in other words we have.

Now let us find out the distances from ‘A’ and ‘B’ to ‘C’

We know that both these distances are the same. So equating both these we get,

Squaring on both sides we have,

Hence the point on the y-axis which lies at equal distances from the mentioned points is.

Answer:

The distance d between two points and is given by the formula

Here we are to find out a point on the x−axis which is equidistant from both the points A(7,6) and B(−3,4).

Let this point be denoted as C(x, y).

Since the point lies on the x-axis the value of its ordinate will be 0. Or in other words we have.

Now let us find out the distances from ‘A’ and ‘B’ to ‘C’

We know that both these distances are the same. So equating both these we get,

Squaring on both sides we have,

Hence the point on the x-axis which lies at equal distances from the mentioned points is.

Answer:

(i) A (2, 3), B(−2, 2), C(−1,−2) and D(3, −1)

Now, using the distance formula we can find the sides of the quadrilateral.

AB$=\sqrt{{\left(-2-2\right)}^2+{\left(2-3\right)}^2}=\sqrt{16+1}=\sqrt{17}$
BC$=\sqrt{{\left(-1+2\right)}^2+{\left(-2-2\right)}^2}=\sqrt{1^2+{\left(-4\right)}^2}=\sqrt{17}$
CD$=\sqrt{{\left(3+1\right)}^2+{\left(-1+2\right)}^2}=\sqrt{17}$
AD$=\sqrt{{\left(3-2\right)}^2+{\left(-1-3\right)}^2}=\sqrt{1+16}=\sqrt{17}$

Similarly, we can find the length of diagonals.
AC$=\sqrt{{\left(-1-2\right)}^2+{\left(-2-3\right)}^2}=\sqrt{9+25}=\sqrt{34}$
BD$=\sqrt{{\left(3+2\right)}^2+{\left(-1-2\right)}^2}=\sqrt{25+9}=\sqrt{34}$

Thus, we conclude AB = BC = CD = AD.
Also, AC = BD, diagonals are equal.
Hence, the required quadrilateral is a square.

(ii) $P\left(\sqrt{2}, \sqrt{2}\right), Q\left(-\sqrt{2}, –\sqrt{2}\right) \mathrm{and} R\left(-\sqrt{6}, \sqrt{6}\right)$
Now, using the distance formula we can find the sides of the triangle.

$\begin{array}{rcl}PQ& =& \sqrt{{\left(-\sqrt{2}-\sqrt{2}\right)}^2+{\left(-\sqrt{2}-\sqrt{2}\right)}^2}\\ & =& \sqrt{8+8}\\ & =& \sqrt{16}\\ & =& 4\end{array}$

$\begin{array}{rcl}QR& =& \sqrt{{\left(-\sqrt{6}+\sqrt{2}\right)}^2+{\left(\sqrt{6}+\sqrt{2}\right)}^2}\\ & =& \sqrt{6+2-2\sqrt{12}+6+2+2\sqrt{12}}\\ & =& \sqrt{16}\\ & =& 4\end{array}$

$\begin{array}{rcl}PR& =& \sqrt{{\left(-\sqrt{6}-\sqrt{2}\right)}^2+{\left(\sqrt{6}-\sqrt{2}\right)}^2}\\ & =& \sqrt{6+2+2\sqrt{12}+6+2-2\sqrt{12}}\\ & =& \sqrt{16}\\ & =& 4\end{array}$

We conclude PQ = QR = PR.
Hence, the required triangle is equilateral.

Answer:

The distance d between two points and is given by the formula

The three given points are P(6,−1), Q(1,3) and R(x,8).

Now let us find the distance between ‘P’ and ‘Q’.

Now, let us find the distance between ‘Q’ and ‘R’.

It is given that both these distances are equal. So, let us equate both the above equations,

Squaring on both sides of the equation we get,

Now we have a quadratic equation. Solving for the roots of the equation we have,

Thus the roots of the above equation are 5 and −3.

Hence the values of ‘x’ are.

Answer:

The distance d between two points and is given by the formula

The three given points are Q(0,1), P(5,−3) and R(x,6).

Now let us find the distance between ‘P’ and ‘Q’.

Now, let us find the distance between ‘Q’ and ‘R’.

It is given that both these distances are equal. So, let us equate both the above equations,

Squaring on both sides of the equation we get,

Hence the values of ‘x’ are.

Now, the required individual distances,

Hence the length of ‘QR’ is.

For ‘PR’ there are two cases. First when the value of ‘x’ is 4,

Then when the value of ‘x’ is −4,

Hence the length of ‘PR’ can beunits

Answer:

The distance d between two points and is given by the formula

The distance between two points P(2,−3) and Q(10,y) is given as 10 units. Substituting these values in the formula for distance between two points we have,

Now, squaring the above equation on both sides of the equals sign

Thus we arrive at a quadratic equation. Let us solve this now,

The roots of the above quadratic equation are thus 3 and −9.

Thus the value of ‘y’ could either be.

Answer:

It is given that A(3, y) is equidistant from points P(8, −3) and Q(7, 6).

∴ AP = AQ

$\Rightarrow \sqrt{{\left(3-8\right)}^2+{\left[y-\left(-3\right)\right]}^2}=\sqrt{{\left(3-7\right)}^2+{\left(y-6\right)}^2}$                 (Distance formula)

Squaring on both sides, we get

$$\begin{gathered} 25+{\left(y+3\right)}^2=16+{\left(y-6\right)}^2 \\ \Rightarrow 25+y^2+6y+9=16+y^2-12y+36 \\ \Rightarrow 12y+6y=52-34 \\ \Rightarrow 18y=18 \\ \Rightarrow y=1 \end{gathered}$$

Thus, the value of y is 1.

$\therefore \mathrm{AQ}=\sqrt{{\left(3-7\right)}^2+{\left(1-6\right)}^2}=\sqrt{{\left(-4\right)}^2+{\left(-5\right)}^2}=\sqrt{16+25}=\sqrt{41}$ units

Answer:

TO FIND: The equation of perpendicular bisector of line segment joining points (7, 1) and (3, 5)

Let P(x, y) be any point on the perpendicular bisector of AB. Then,

PA=PB

Hence the equation of perpendicular bisector of line segment joining points (7, 1) and (3, 5) is

Answer:

The length of the diameter is $10\sqrt{2} units$.
So, the radius is $5\sqrt{2} units$.
The centre of the circle be C(2a, a−7).
Suppose it passes through the point P(11, −9).
Therefore, PC = r
 $$\begin{gathered} \Rightarrow P{C}^2=r^2 \\ \Rightarrow {\left(11-2a\right)}^2+{\left(-9-a+7\right)}^2={\left(5\sqrt{2}\right)}^2 \\ \Rightarrow 121+4a^2-44a+a^2+4+4a=50 \\ \Rightarrow 5a^2-40a+75=0 \\ \Rightarrow \left(a-3\right)\left(a-5\right)=0 \\ \Rightarrow a=3 or a=5 \end{gathered}$$
Hence the values of a are 3 or 5.
 

Answer:

The position of the ayush's house is (2, 4) and that of the bank is (5, 8).
The distance between the house and the bank is
$$\begin{gathered} d_1=\sqrt{{\left(5-2\right)}^2+{\left(8-4\right)}^2} \\ =\sqrt{3^2+4^2} \\ =\sqrt{9+16} \\ =5 units \end{gathered}$$
The position of the the bank is (5, 8) and that of the school is (13, 14).
The distance between the bank and the school is
$$\begin{gathered} d_2=\sqrt{{\left(13-5\right)}^2+{\left(14-8\right)}^2} \\ =\sqrt{8^2+6^2} \\ =\sqrt{64+36} \\ =10 units \end{gathered}$$
The position of the school is (13, 14) and that of the office is (13, 6)
The distance between the bank and the school is
$$\begin{gathered} d_3=\sqrt{{\left(13-13\right)}^2+{\left(26-14\right)}^2} \\ =\sqrt{0^2+{12}^2} \\ =12 units \end{gathered}$$
Suppose d be the total distance travelled by ayush
$$\begin{gathered} d=d_1+d_2+d_3 \\ d=5+10+12 \\ d=27 km \end{gathered}$$
Now, let D be the shortest distance between ayush house and the office,
$$\begin{gathered} D=\sqrt{{\left(13-2\right)}^2+{\left(26-4\right)}^2} \\ =\sqrt{{11}^2+{22}^2} \\ =\sqrt{605} \\ =24.6 km \end{gathered}$$

Thus the extra distance covered by ayush is d − D = 27 − 24.6 = 2.4 km.

Answer:

Let the given points be A(0, −3) and B(0, 3). Suppose the coordinates of the third vertex be C(x, y).

Now, ∆ABC is an equilateral triangle.

∴ AB = BC = CA

$\sqrt{{\left(0-0\right)}^2+{\left(-3-3\right)}^2}=\sqrt{{\left(x-0\right)}^2+{\left(y-3\right)}^2}=\sqrt{{\left(x-0\right)}^2+{\left[y-\left(-3\right)\right]}^2}$                 (Distance formula)

Squaring on both sides, we get

$36=x^2+{\left(y-3\right)}^2=x^2+{\left(y+3\right)}^2$

⇒ $x^2+{\left(y-3\right)}^2=x^2+{\left(y+3\right)}^2$ and $x^2+{\left(y-3\right)}^2=36$

Now,

$$\begin{gathered} x^2+{\left(y-3\right)}^2=x^2+{\left(y+3\right)}^2 \\ \Rightarrow y^2-6y+9=y^2+6y+9 \\ \Rightarrow -12y=0 \\ \Rightarrow y=0 \end{gathered}$$

Putting y = 0 in $x^2+{\left(y-3\right)}^2=36$, we get

$$\begin{gathered} x^2+{\left(0-3\right)}^2=36 \\ \Rightarrow x^2=36-9=27 \\ \Rightarrow x=\pm \sqrt{27}=\pm 3\sqrt{3} \end{gathered}$$

Thus, the coordinates of the third vertex are $\left(3\sqrt{3}, 0\right)$ or $\left(-3\sqrt{3}, 0\right)$.

Answer:

The distance d between two points and is given by the formula

In an equilateral triangle all the sides are of equal length.

Here we are given that A (3, 4) and B (−2, 3) are two vertices of an equilateral triangle. Let C(x, y) be the third vertex of the equilateral triangle.

First let us find out the length of the side of the equilateral triangle.

Hence the side of the equilateral triangle measures units.

Now, since it is an equilateral triangle, all the sides need to measure the same length.

Hence we have

Equating both these equations we have,

Squaring on both sides we have,

From the above equation we have,

Substituting this and the value of the side of the triangle in the equation for one of the sides we have,

Squaring on both sides,

Now we have a quadratic equation for ‘x’. Solving for the roots of this equation,

We know that. Substituting the value of ‘x’ we have,

Hence the two possible values of the third vertex are.

Answer:

The distance d between two points and is given by the formula

The circumcentre of a triangle is the point which is equidistant from each of the three vertices of the triangle.

Here the three vertices of the triangle are given to be A(−2.−3), B(−1,0) and C(7,−6).

Let the circumcentre of the triangle be represented by the point R(x, y).

So we have

Equating the first pair of these equations we have,

Squaring on both sides of the equation we have,

Equating another pair of the equations we have,

Squaring on both sides of the equation we have,

Now we have two equations for ‘x’ and ‘y’, which are

From the second equation we have. Substituting this value of ‘y’ in the first equation we have,

Therefore the value of ‘y’ is,

Hence the co−ordinates of the circumcentre of the triangle with the given vertices are.

Answer:

The distance d between two points and is given by the formula

In a right angled triangle the angle opposite the hypotenuse subtends an angle of.

Here let the given points be A(0,100), B(10,0). Let the origin be denoted by O(0,0).

Let us find the distance between all the pairs of points

Here we can see that.

So, is a right angled triangle with ‘AB’ being the hypotenuse. So the angle opposite it has to be. This angle is nothing but the angle subtended by the line segment ‘AB’ at the origin.

Hence the angle subtended at the origin by the given line segment is.

Answer:

The distance d between two points and is given by the formula

In a square all the sides are of equal length. The diagonals are also equal to each other. Also in a square the diagonal is equal to times the side of the square.

Here let the two points which are said to be the opposite vertices of a diagonal of a square be A(−1,2) and C(3,2).

Let us find the distance between them which is the length of the diagonal of the square.

Now we know that in a square,

Substituting the value of the diagonal we found out earlier in this equation we have,

Now, a vertex of a square has to be at equal distances from each of its adjacent vertices.

Let P(x, y) represent another vertex of the same square adjacent to both ‘A’ and ‘C’.

But these two are nothing but the sides of the square and need to be equal to each other.

Squaring on both sides we have,

Substituting this value of ‘x’ and the length of the side in the equation for ‘AP’ we have,

Squaring on both sides,

We have a quadratic equation. Solving for the roots of the equation we have,

The roots of this equation are 0 and 4.

Therefore the other two vertices of the square are.

Answer:

We have A (−1, 3) and B (4,−7) be two points. Let a pointdivide the line segment joining the points A and B in the ratio 3:4 internally.

Now according to the section formula if point a point P divides a line segment joining andin the ratio m: n internally than,

Now we will use section formula to find the co-ordinates of unknown point P as,

Therefore, co-ordinates of point P is

Answer:

The co-ordinates of the midpoint between two points and is given by,

In a parallelogram the diagonals bisect each other. That is the point of intersection of the diagonals is the midpoint of either of the diagonals.

Here, it is given that the vertices of a parallelogram are A(−2,−1), B(1,0) and C(4,3) and D(1,2).

We see that ‘AC’ and ‘BD’ are the diagonals of the parallelogram.

The midpoint of either one of these diagonals will give us the point of intersection of the diagonals.

Let this point be M(x, y).

Let us find the midpoint of the diagonal ‘AC’.

Hence the co-ordinates of the point of intersection of the diagonals of the given parallelogram are.

Answer:

Given: Point A(3, 1), B(5, 1), C(a, b) and D(4, 3) are vertices of a parallelogram ABCD.

Diagonals of a parallelogram bisect each other.

∴ Mid point of AC = Mid point of BD

$$\begin{gathered} \mathrm{Mid} \mathrm{point} \mathrm{of} \left(x_1, y_1\right) \mathrm{and} \left(x_2, y_2\right) \mathrm{is} \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right). \\ \mathrm{Mid} \mathrm{point} \mathrm{of} AC=\left(\frac{3+a}{2}, \frac{1+b}{2}\right) \\ \mathrm{Mid} \mathrm{point} \mathrm{of} BD=\left(\frac{5+4}{2}, \frac{1+3}{2}\right) \\ =\left(\frac{9}{2}, \frac{4}{2}\right)=\left(\frac{9}{2}, 2\right) \\ \therefore \left(\frac{3+a}{2}, \frac{1+b}{2}\right)=\left(\frac{9}{2}, 2\right) \\ \Rightarrow \frac{3+a}{2}=\frac{9}{2} \mathrm{and} \frac{1+b}{2}=2 \\ \Rightarrow 3+a=9 \mathrm{and} 1+b=4 \\ \Rightarrow a=6 \mathrm{and} b=3 \end{gathered}$$

Hence,  the values of a and b is 6 and 3, respectively.

Answer:

The co-ordinates of a point which divided two points and internally in the ratio is given by the formula,

Here we are given that the point P(2,y) divides the line joining the points A(−2,2) and B(3,7) in some ratio.

Let us substitute these values in the earlier mentioned formula.

Equating the individual components we have

We see that the ratio in which the given point divides the line segment is.

Let us now use this ratio to find out the value of ‘y’.

Equating the individual components we have

Thus the value of ‘y’ is.

Answer:

The distance d between two points and is given by the formula

The co-ordinates of the midpoint between two points and is given by,

Here, it is given that the three vertices of a triangle are A(−1,3), B(1,−1) and C(5,1).

The median of a triangle is the line joining a vertex of a triangle to the mid-point of the side opposite this vertex.

Let ‘D’ be the mid-point of the side ‘BC’.

Let us now find its co-ordinates.

Thus we have the co-ordinates of the point as D(3,0).

Now, let us find the length of the median ‘AD’.

Thus the length of the median through the vertex ‘A’ of the given triangle is.

Answer:

Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (3,−4); B (−1,−3) and C (−6, 2). We have to find the co-ordinates of the forth vertex.

Let the forth vertex be

Since ABCD is a parallelogram, the diagonals bisect each other. Therefore the mid-point of the diagonals of the parallelogram will coincide.

Now to find the mid-point of two pointsand we use section formula as,

The mid-point of the diagonals of the parallelogram will coincide.

So,

Therefore,

Now equate the individual terms to get the unknown value. So,

So the forth vertex is

Answer:

The co-ordinates of the midpoint between two points and is given by,

Here we are supposed to find the points which divide the line joining A(−2,2) and B(2,8) into 4 equal parts.

We shall first find the midpoint M(x, y) of these two points since this point will divide the line into two equal parts.

So the point M(0,5) splits this line into two equal parts.

Now, we need to find the midpoint of A(−2,2) and M(0,5) separately and the midpoint of B(2,8) and M(0,5). These two points along with M(0,5) split the line joining the original two points into four equal parts.

Let be the midpoint of A(−2,2) and M(0,5).

Now let bet the midpoint of B(2,8) and M(0,5).

Hence the co-ordinates of the points which divide the line joining the two given points are.

Answer:

It is given that P, Q, R and S divides the line segment joining A(1, 2) and B(6, 7) in 5 equal parts.

∴ AP = PQ = QR = RS = SB          .....(1)

Now,

AP + PQ + QR + RS + SB = AB

⇒ AP + AP + AP + AP + AP = AB             [From (1)]

⇒ 5AP = AB

⇒ AP = $\frac{1}{5}$AB                  .....(2)   

Now,

PB = PQ + QR + RS + SB = $\frac{1}{5}$AB +$\frac{1}{5}$AB + $\frac{1}{5}$AB + $\frac{1}{5}$AB = $\frac{4}{5}$AB         .....(3)

From (2) and (3), we get

AP : PB = $\frac{1}{5}$AB : $\frac{4}{5}$AB = 1 : 4

Similarly,

AQ : QB = 2 : 3 and AR : RB = 3 : 2

Using section formula, we get

Coordinates of P = $\left(\frac{1\times 6+4\times 1}{1+4},\frac{1\times 7+4\times 2}{1+4}\right)=\left(\frac{10}{5},\frac{15}{5}\right)=\left(2,3\right)$

Coordinates of Q = $\left(\frac{2\times 6+3\times 1}{2+3},\frac{2\times 7+3\times 2}{2+3}\right)=\left(\frac{15}{5},\frac{20}{5}\right)=\left(3,4\right)$

Coordinates of R = $\left(\frac{3\times 6+2\times 1}{3+2},\frac{3\times 7+2\times 2}{3+2}\right)=\left(\frac{20}{5},\frac{25}{5}\right)=\left(4,5\right)$

Answer:

The ratio in which the x−axis divides two points  and  is $\lambda :1$

The ratio in which the y-axis divides two points  and  is $\mu :1$

The co-ordinates of the point dividing two points  and  in the ratio  is given as,

 Where 

Here the two given points are A(−2,−3) and B(5,6).

1. The ratio in which the x-axis divides these points is

$$\begin{gathered} \frac{6\lambda -3}{3}=0 \\ \lambda =\frac{1}{2} \end{gathered}$$

Let point P(x, y) divide the line joining ‘AB’ in the ratio 

Substituting these values in the earlier mentioned formula we have,

Thus the ratio in which the x−axis divides the two given points and the co-ordinates of the point is.

2. The ratio in which the y-axis divides these points is

$$\begin{gathered} \frac{5\mu -2}{3}=0 \\ \Rightarrow \mu =\frac{2}{5} \end{gathered}$$

Let point P(x, y) divide the line joining ‘AB’ in the ratio 

Substituting these values in the earlier mentioned formula we have,

Thus the ratio in which the x-axis divides the two given points and the co-ordinates of the point is.

Answer:

Let A (4, 3); B (6, 4); C (5, 6) and D (3, 5) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a square.

So we should find the lengths of sides of quadrilateral ABCD.

All the sides of quadrilateral are equal.

So now we will check the lengths of the diagonals.

All the sides as well as the diagonals are equal. Hence ABCD is a square.

Answer:

Let A (−4,−1); B (−2,−4); C (4, 0) and D (2, 3) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a rectangle.

So we should find the lengths of opposite sides of quadrilateral ABCD.

Opposite sides are equal. So now we will check the lengths of the diagonals.

Opposite sides are equal as well as the diagonals are equal. Hence ABCD is a rectangle.

Answer:

We have a rectangle ABCD formed by joining the points A (−1,−1); B (−1, 4); C (5, 4) and D (5,−1). The mid-points of the sides AB, BC, CD and DA are P, Q, R, S respectively.

We have to find that whether PQRS is a square, rectangle or rhombus.

In general to find the mid-point of two pointsand we use section formula as,

Therefore mid-point P of side AB can be written as,

Now equate the individual terms to get,

So co-ordinates of P is

Similarly mid-point Q of side BC can be written as,

Now equate the individual terms to get,

So co-ordinates of Q is (2, 4)

Similarly mid-point R of side CD can be written as,

Now equate the individual terms to get,

So co-ordinates of R is

Similarly mid-point S of side DA can be written as,

Now equate the individual terms to get,

So co-ordinates of S is (2,−1)

So we should find the lengths of sides of quadrilateral PQRS.

All the sides of quadrilateral are equal.

So now we will check the lengths of the diagonals.

All the sides are equal but the diagonals are unequal. Hence ABCD is a rhombus.

Answer:

Suppose P(−1, y) divides the line segment joining A(−3, 10) and B(6 −8) in the ratio k : 1.

Using section formula, we get

Coordinates of P = $\left(\frac{6k-3}{k+1},\frac{-8k+10}{k+1}\right)$

$\therefore \left(\frac{6k-3}{k+1},\frac{-8k+10}{k+1}\right)=\left(-1,y\right)$

$\Rightarrow \frac{6k-3}{k+1}=-1$ and $y=\frac{-8k+10}{k+1}$

Now,

$$\begin{gathered} \frac{6k-3}{k+1}=-1 \\ \Rightarrow 6k-3=-k-1 \\ \Rightarrow 7k=2 \\ \Rightarrow k=\frac{2}{7} \end{gathered}$$

So, P divides the line segment AB in the ratio 2 : 7.

Putting k = $\frac{2}{7}$ in $y=\frac{-8k+10}{k+1}$, we get

$y=\frac{-8\times {\displaystyle \frac{2}{7}}+10}{{\displaystyle \frac{2}{7}}+1}=\frac{-16+70}{2+7}=\frac{54}{9}=6$

Hence, the value of y is 6.

Answer:

Let the co-ordinates of point A be.

Centre lies on the mid-point of the diameter. So applying the mid-point formula we get,

Similarly,

So the co-ordinates of A are (3,−10)

Answer:

The co-ordinates of a point which divided two points and internally in the ratio is given by the formula,

Here it is said that the point (−4,6) divides the points A(−6,10) and B(3,−8). Substituting these values in the above formula we have,

Equating the individual components we have,

Therefore the ratio in which the line is divided is

Answer:

The ratio in which the y-axis divides two points and is $\lambda :1$

The co-ordinates of the point dividing two points and in the ratio is given as,

where,

Here the two given points are A(5,−6) and B(−1,−4).

$(x, y)=\left(\frac{-\lambda +5}{\lambda +1},\frac{-4\lambda -6}{\lambda +1}\right)$
Since, the y-axis divided the given line, so the x coordinate will be 0.
$$\begin{gathered} \frac{-\lambda +5}{\lambda +1}=0 \\ \lambda =\frac{5}{1} \end{gathered}$$

Thus the given points are divided by the y-axis in the ratio.

The co-ordinates of this point (x, y) can be found by using the earlier mentioned formula.

Thus the co-ordinates of the point which divides the given points in the required ratio are.

Answer:

The co-ordinates of the point dividing two points and in the ratio is given as,

where,

Here the two given points are A(1,4) and B(5,2). Let point P(x, y) divide the line joining ‘AB’ in the ratio

Substituting these values in the earlier mentioned formula we have,

Thus the co-ordinates of the point which divides the given points in the required ratio are.

Answer:

Let P divides AB in a ratio of λ : 1
Therefore, coordinates of the point P are $\left(\frac{6\lambda +2}{\lambda +1}, \frac{-3\lambda +3}{\lambda +1}\right)$
Given that coordinates of the point P are (4, m).
$$\begin{gathered} \Rightarrow \frac{6\lambda +2}{\lambda +1}=4 \\ \Rightarrow 6\lambda +2=4\lambda +4 \\ \Rightarrow \lambda =1 \end{gathered}$$
Hence, the point P divides AB in a ratio of 1 : 1.
Replacing the value of λ = 1 in y-coordinate of P, we get
$$\begin{gathered} \frac{-3\left(1\right)+3}{1+1}=m \\ \Rightarrow m=0 \end{gathered}$$
Thus, y-coordinate of P is equal to 0. 

Answer:

Given that, point R lies on the line segment PQ where P(–1, 3) and Q(2, 5) and $PR=\frac{3}{5}PQ$. This means that point R divides PQ in the ratio 3 : 2 internally.

If a point X(x, y) divides AB with A(x₁, y₁) and B(x₂, y₂) in a ratio of m : n, then coordinates of the point X are $\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$.

Thus,
Coordinates of the point R = $\left(\frac{3\times 2+2\times \left(-1\right)}{3+2},\frac{3\times 5+2\times 3}{3+2}\right)$
                                           $$\begin{gathered} =\left(\frac{6-2}{5},\frac{15+6}{5}\right) \\ =\left(\frac{4}{5},\frac{21}{5}\right) \end{gathered}$$

Hence, the coordinates of the point R are $\left(\frac{4}{5},\frac{21}{5}\right)$.

Answer:

Let the point X(a, b) be the mid-point of the line segment joining the points A(10, –6) and B(k, 4).

Now, coordinates of the midpoint of a line segment with end points (x₁, y₁) and (x₂, y₂) are given as $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$.

Thus,
 $$\begin{gathered} \left(a,b\right)=\left(\frac{10+k}{2},\frac{-6+4}{2}\right) \\ \Rightarrow b=-1 \mathrm{and} a=\frac{10+k}{2} \end{gathered}$$

Given that, a – 2b = 18.
Putting the respective values in the given equation,
$$\begin{gathered} \frac{10+k}{2}-2\left(-1\right)=18 \\ \Rightarrow \frac{10+k}{2}+2=18 \\ \Rightarrow \frac{10+k}{2}=16 \\ \Rightarrow k=22 \end{gathered}$$

Thus, k = 22.

Now, the distance between the points A(10, –6) and B(22, 4) is given as:
$\begin{array}{rcl}AB& =& \sqrt{{\left(22-10\right)}^2+{\left(4+6\right)}^2} \left(\because d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\right)\\ & =& \sqrt{{\left(12\right)}^2+{\left(10\right)}^2}\\ & =& \sqrt{144+100}\\ & =& \sqrt{244}\\ & =& 2\sqrt{61}\end{array}$

Hence, the distance AB is $\begin{array}{rcl}& & 2\sqrt{61}\end{array}$ units.

Answer:

It is given that P, Q(x, 7), R, S(6, y) divides the line segment joining A(2, p) and B(7, 10) in 5 equal parts.

∴ AP = PQ = QR = RS = SB          .....(1)

Now,

AP + PQ + QR + RS + SB = AB

⇒ SB + SB + SB + SB + SB = AB             [From (1)]

⇒ 5SB = AB

⇒ SB = $\frac{1}{5}$AB                  .....(2)   

Now,

AS = AP + PQ + QR + RS = $\frac{1}{5}$AB +$\frac{1}{5}$AB + $\frac{1}{5}$AB + $\frac{1}{5}$AB = $\frac{4}{5}$AB         .....(3)

From (2) and (3), we get

AS : SB = $\frac{4}{5}$AB : $\frac{1}{5}$AB = 4 : 1

Similarly,

AQ : QB = 2 : 3

Using section formula, we get

Coordinates of Q = $\left(\frac{2\times 7+3\times 2}{2+3},\frac{2\times 10+3\times p}{2+3}\right)=\left(\frac{20}{5},\frac{20+3p}{5}\right)=\left(4,\frac{20+3p}{5}\right)$

$\therefore \left(x,7\right)=\left(4,\frac{20+3p}{5}\right)$

$\Rightarrow x=4$ and $7=\frac{20+3p}{5}$

Now,

$$\begin{gathered} 7=\frac{20+3p}{5} \\ \Rightarrow 20+3p=35 \\ \Rightarrow 3p=15 \\ \Rightarrow p=5 \end{gathered}$$

Coordinates of S = $\left(\frac{4\times 7+1\times 2}{4+1},\frac{4\times 10+1\times p}{4+1}\right)=\left(\frac{30}{5},\frac{40+5}{5}\right)=\left(6,9\right)$

$$\begin{gathered} \therefore \left(6,y\right)=\left(6,9\right) \\ \Rightarrow y=9 \end{gathered}$$

Thus, the values of x, y and p are 4, 9 and 5, respectively.

Answer:

Let a in which P and Q are the mid-points of sides AB and AC respectively. The coordinates are: A (1, 1); P (−2, 3) and Q (5, 2).

We have to find the co-ordinates of and.

In general to find the mid-point of two pointsand we use section formula as,

Therefore mid-point P of side AB can be written as,

Now equate the individual terms to get,

So, co-ordinates of B is (−5, 5)

Similarly, mid-point Q of side AC can be written as,

Now equate the individual terms to get,

So, co-ordinates of C is (9, 3)

Answer:

We have two points A (3, 4) and B (k, 7) such that its mid-point is.

It is also given that point P lies on a line whose equation is

In general to find the mid-point of two pointsand we use section formula as,

Therefore mid-point P of side AB can be written as,

Now equate the individual terms to get,

Since, P lies on the given line. So,

Put the values of co-ordinates of point P in the equation of line to get,

On further simplification we get,

So,

Answer:

We have two points A (−2,−2) and B (2,−4). Let P be any point which divide AB as,

Since,

So,

Now according to the section formula if any point P divides a line segment joining andin the ratio m: n internally than,

Therefore P divides AB in the ratio 3: 4. So,

Answer:

We have a parallelogram ABCD in which A (3, 2) and B (−1, 0) and the co-ordinate of the intersection of diagonals is M (2,−5).

We have to find the co-ordinates of vertices C and D.

So let the co-ordinates be and

In general to find the mid-point of two pointsand we use section formula as,

The mid-point of the diagonals of the parallelogram will coincide.

So,

Therefore,

Now equate the individual terms to get the unknown value. So,

So the co-ordinate of vertex C is (1,−12)

Similarly,

Therefore,

Now equate the individual terms to get the unknown value. So,

So the co-ordinate of vertex C is (5,−10)

Answer:

The co-ordinates of the midpoint between two points and is given by,

Let the three vertices of the triangle be, and.

The three midpoints are given. Let these points be, and.

Let us now equate these points using the earlier mentioned formula,

Equating the individual components we get,

Using the midpoint of another side we have,

Equating the individual components we get,

Using the midpoint of the last side we have,

Equating the individual components we get,

Adding up all the three equations which have variable ‘x’ alone we have,

Substituting in the above equation we have,

Therefore,

And

Adding up all the three equations which have variable ‘y’ alone we have,

Substituting in the above equation we have,

Therefore,

And

Therefore the co-ordinates of the three vertices of the triangle are.

Answer:

We have two points P (3, 3) and Q (6,−6). There are two points A and B which trisect the line segment joining P and Q.

Let the co-ordinate of A be

Now according to the section formula if any point P divides a line segment joining andin the ratio m: n internally than,

The point A is the point of trisection of the line segment PQ. So, A divides PQ in the ratio 1: 2

Now we will use section formula to find the co-ordinates of unknown point A as,

Therefore, co-ordinates of point A is(4, 0)

It is given that point A lies on the line whose equation is

So point A will satisfy this equation.

So,

Answer:

Let the points A(3, −4) and B(1, 2) is trisected at the points P(p, −2) and Q(5/3, q).

Thus, AP = PQ = QB.

Therefore, P divides AB internally in the ratio 1 : 2. 

Section formula: if the point (x, y) divides the line segment joining the points (x₁, y₁) and (x₂, y₂) internally in the ratio m : n, then the coordinates (x, y) = $\left(\frac{m{x}_2+n{x}_1}{m+n}, \frac{m{y}_2+n{y}_1}{m+n}\right)$

Therefore, using section formula, the coordinates of P are:

$$\begin{gathered} \left(p, -2\right)=\left(\frac{1\left(1\right)+2\left(3\right)}{1+2},\frac{1\left(2\right)+2\left(-4\right)}{1+2}\right) \\ \Rightarrow \left(p, -2\right)=\left(\frac{1+6}{3},\frac{2-8}{3}\right) \\ \Rightarrow \left(p, -2\right)=\left(\frac{7}{3},\frac{-6}{3}\right) \\ \Rightarrow \left(p, -2\right)=\left(\frac{7}{3},-2\right) \\ \Rightarrow p=\frac{7}{3} \end{gathered}$$

Also, Q divides AB internally in the ratio 2 : 1.

Therefore, using section formula, the coordinates of Q are:

$$\begin{gathered} \left(\frac{5}{3}, q\right)=\left(\frac{2\left(1\right)+1\left(3\right)}{2+1},\frac{2\left(2\right)+1\left(-4\right)}{2+1}\right) \\ \Rightarrow \left(\frac{5}{3}, q\right)=\left(\frac{2+3}{3},\frac{4-4}{3}\right) \\ \Rightarrow \left(\frac{5}{3}, q\right)=\left(\frac{5}{3},\frac{0}{3}\right) \\ \Rightarrow \left(\frac{5}{3}, q\right)=\left(\frac{5}{3},0\right) \\ \Rightarrow q=0 \end{gathered}$$

Hence,  the values of p and q is $\frac{7}{3}$ and 0, respectively.

Answer:

We have two points A (2, 1) and B (5,−8). There are two points P and Q which trisect the line segment joining A and B.

Now according to the section formula if any point P divides a line segment joining andin the ratio m: n internally than,

The point P is the point of trisection of the line segment AB. So, P divides AB in the ratio 1: 2

Now we will use section formula to find the co-ordinates of unknown point A as,

Therefore, co-ordinates of point P is(3,−2)

It is given that point P lies on the line whose equation is

So point A will satisfy this equation.

So,

Answer:

Given that, a line 2x + 3y – 5 = 0 divides the line segment joining the points A(8, –9) and B(2, 1). Let the point that divides AB in the ratio k : 1 be X(a, b).

Now,
 $\begin{array}{rcl}\mathrm{Coordinates} \mathrm{of} X& =& \left(\frac{k{x}_2+x_1}{k+1},\frac{k{y}_2+y_1}{k+1}\right)\\ & =& \left(\frac{2k+8}{k+1},\frac{k-9}{k+1}\right)\end{array}$

These coordinates satisfy the equation of the line 2x + 3y – 5 = 0.
So,
$$\begin{gathered} 2\left(\frac{2k+8}{k+1}\right)+3\left(\frac{k-9}{k+1}\right)-5=0 \\ \Rightarrow 4k+16+3k-27-5k-5=0 \\ \Rightarrow 2k-16=0 \\ \Rightarrow k=8 \end{gathered}$$

Thus, the point X divides the line in the ratio 8 : 1.

Since $k=8$, the coordinates of point X are:
$\begin{array}{rcl}\mathrm{Coordinates} \mathrm{of} X& =& \left(\frac{2\times {\displaystyle 8}+8}{8+1},\frac{{\displaystyle 8}-9}{{\displaystyle 8}+1}\right)\\ & =& \left(\frac{{\displaystyle 16}+8}{{\displaystyle 9}},\frac{{\displaystyle -1}}{{\displaystyle 9}}\right)\\ & =& \left(\frac{{\displaystyle 24}}{{\displaystyle 9}},\frac{{\displaystyle -1}}{{\displaystyle 9}}\right)\\ & =& \left(\frac{8}{3},\frac{-1}{9}\right)\end{array}$

Hence, the coordinates of the point of division is $\begin{array}{rcl}& & \left(\frac{8}{3},\frac{-1}{9}\right)\end{array}$.

Answer:

It is given that $\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{k}{1}.$

So, P divides the line segment joining the points A(3, −5) and B(−4, 8) in the ratio k : 1.

Using the section formula, we get

Coordinates of P = $\left(\frac{-4k+3}{k+1},\frac{8k-5}{k+1}\right)$

Since P lies on the line x + y = 0, so

$$\begin{gathered} \frac{-4k+3}{k+1}+\frac{8k-5}{k+1}=0 \\ \Rightarrow \frac{-4k+3+8k-5}{k+1}=0 \\ \Rightarrow 4k-2=0 \\ \Rightarrow k=\frac{1}{2} \end{gathered}$$

Hence, the value of k is $\frac{1}{2}$.

Answer:

It is given that P is the mid-point of the line segment joining the points A(−10, 4) and B(−2, 0).

∴ Coordinates of P = $\left(\frac{-10+\left(-2\right)}{2},\frac{4+0}{2}\right)=\left(\frac{-12}{2},\frac{4}{2}\right)=\left(-6,2\right)$

Suppose P divides the line segment joining the points C(−9, −4) and D(−4, y) in the ratio k : 1.

Using section formula, we get

Coordinates of P = $\left(\frac{-4k-9}{k+1},\frac{ky-4}{k+1}\right)$

$$\begin{gathered} \therefore \left(\frac{-4k-9}{k+1},\frac{ky-4}{k+1}\right)=\left(-6,2\right) \\ \Rightarrow \frac{-4k-9}{k+1}=-6 \mathrm{and} \frac{ky-4}{k+1}=2 \end{gathered}$$

Now,

$$\begin{gathered} \frac{-4k-9}{k+1}=-6 \\ \Rightarrow -4k-9=-6k-6 \\ \Rightarrow 2k=3 \\ \Rightarrow k=\frac{3}{2} \end{gathered}$$

So, P divides the line segment CD in the ratio 3 : 2.

Putting k = $\frac{3}{2}$ in $\frac{ky-4}{k+1}=2$, we get
$$\begin{gathered} \frac{{\displaystyle \frac{3y}{2}}-4}{{\displaystyle \frac{3}{2}}+1}=2 \\ \Rightarrow \frac{3y-8}{5}=2 \\ \Rightarrow 3y-8=10 \\ \Rightarrow 3y=18 \\ \Rightarrow y=6 \end{gathered}$$

Hence, the value of y is 6.

Answer:

It is given that the point C(–1, 2) divides the line segment joining the points A(2, 5) and B(x, y) in the ratio 3 : 4 internally.
Using the section formula, we get
$$\begin{gathered} \left(-1, 2\right)=\left(\frac{3\times x+4\times 2}{3+4},\frac{3\times y+4\times 5}{3+4}\right) \\ \Rightarrow \left(-1, 2\right)=\left(\frac{3x+8}{7},\frac{3y+20}{7}\right) \\ \Rightarrow \frac{3x+8}{7}=-1 \mathrm{and} \frac{3y+20}{7}=2 \end{gathered}$$
⇒ 3x + 8 = –7 and 3y + 20 = 14
⇒ 3x = –15 and 3y = –6
⇒ x = –5 and y = –2
∴ x² + y² = 25 + 4 = 29
Hence, the value of x² + y² is 29.