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

Answer:

A point P is given with coordinates (2, 3).

The distance of a point from the x-axis is given by its y-coordinate. Thus, the distance of point P(2, 3) from the x-axis is 3 units.

Hence, the correct answer is option (b).

Answer:

Given that, AOBC is a rectangle with vertices A(0, 3), O(0, 0) and B(5, 0).

Length of the diagonal AB = Distance between the points A(0,3) and B(5,0)

Now,
Distance between the points (x₁, y₁) and (x₂, y₂) = $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Thus,
$\begin{array}{rcl}AB& =& \sqrt{{\left(5-0\right)}^2+{\left(0-3\right)}^2}\\ & =& \sqrt{25+9}\\ & =& \sqrt{34}\end{array}$

Hence, the correct answer is option (c).
 

Answer:

Given that, the three points A(–4, 0), B(4, 0) and C(0, 3) are the vertices of a triangle and we have to determine the type of triangle formed by them.

Distance between the points (x₁, y₁) and (x₂, y₂) = $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

$\begin{array}{rcl}AB& =& \sqrt{{\left[4-\left(-4\right)\right]}^2+{\left(0-0\right)}^2}\\ & =& \sqrt{{\left(8\right)}^2}\\ & =& 8\end{array}$

$\begin{array}{rcl}BC& =& \sqrt{{\left(0-4\right)}^2+{\left(3-0\right)}^2}\\ & =& \sqrt{16+9}\\ & =& \sqrt{25}\\ & =& 5\end{array}$

$\begin{array}{rcl}CA& =& \sqrt{{\left[0-\left(-4\right)\right]}^2+{\left(3-0\right)}^2}\\ & =& \sqrt{16+9}\\ & =& \sqrt{25}\\ & =& 5\end{array}$

Thus, two sides of the triangle are equal, which means it is an isosceles triangle.

Hence, the correct answer is option (b).

Answer:

Given that, a line segment is formed by joining the points A(–2, –5) and B(2, 5). The perpendicular bisector of a line segment passes through its mid-point.

Let C be the point of the line segment AB through which its perpendicular bisector passes. Thus, point C is the mid-point of AB, i.e., it divides the line segment in the ratio 1 : 1.

$\begin{array}{rcl}\mathrm{Coordinates} \mathrm{of} C& =& \left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ & =& \left(\frac{-2+2}{2},\frac{-5+5}{2}\right)\\ & =& \left(0,0\right)\end{array}$

Hence, the correct answer is option (a).
 

Answer:

Given that, a parallelogram ABCD has three vertices as A(–2, 3), B(6, 7) and C(8, 3).

The diagonals of a parallelogram bisect each other. So, the mid-point of AC is same as the mid-point of BD. Let the point of intersection of AC and BD be O.

Therefore, the coordinates of O (or mid-point of AC) are given as:
$\begin{array}{rcl}\mathrm{Coordinates} \mathrm{of} O& =& \left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ & =& \left(\frac{8-2}{2},\frac{3+3}{2}\right)\\ & =& \left(3,3\right)\end{array}$

Let the coordinates of D be (x, y).
Then,
$$\begin{gathered} \mathrm{Mid}-\mathrm{point} \mathrm{of} BD=\left(3,3\right) \\ \Rightarrow \left(3,3\right)=\left(\frac{6+x}{2},\frac{7+y}{2}\right) \\ \Rightarrow x=0 \mathrm{and} y=-1 \end{gathered}$$

Hence, the correct answer is option (b).

Answer:

Given that, O(0, 0), A(2x, 0) and B(0, 2y) are the vertices of triangle OAB. This is a right angled triangle as two vertices are on the coordinate axes and the third one is the origin, thus making a right angle at the origin.

For a right triangle, the circumcentre is the mid-point of the hypotenuse. ​The point which is equidistant from vertices of triangle is termed as circumcentre of that triangle.

Here, the end-points of the hypotenuse are A(2x, 0) and B(0, 2y). Let the mid-point of AB be C.

So,
$\begin{array}{rcl}\mathrm{Coordinates} \mathrm{of} C& =& \left(\frac{2x+0}{2},\frac{0+2y}{2}\right) \left[\because \mathrm{Coordinates} \mathrm{of} \mathrm{the} \mathrm{mid}-\mathrm{point}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\right]\\ & =& \left(\frac{2x}{2},\frac{2y}{2}\right)\\ & =& \left(x,y\right)\end{array}$

Hence, the correct answer is option (a).

Answer:

Given that, a circle is drawn with origin as the centre passes through the point A$\left(\frac{13}{2},0\right)$.

The radius of the circle is equal to the distance between the centre O(0, 0) and A$\left(\frac{13}{2},0\right)$.

Now,
$\begin{array}{rcl}OA& =& \sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2} \left[\because \mathrm{Distance} \mathrm{between} \mathrm{two} \mathrm{points} \left(x_1,y_1\right) \mathrm{and} \left(x_2,y_2\right)= \sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\right]\\ & =& \sqrt{{\left(\frac{13}{2}-0\right)}^2+{\left(0-0\right)}^2}\\ & =& \sqrt{{\left(\frac{13}{2}\right)}^2}\\ & =& \frac{13}{2}\\ & =& 6.5 \mathrm{units}\end{array}$

A point lies inside, on or outside the circle if the distance of the point from the centre of the circle is less than, equal to or greater than the radius of the circle respectively.

The points $\left(\frac{-3}{4},1\right)$, $\left(2,\frac{7}{3}\right)$ and $\left(5,\frac{-1}{2}\right)$ seem to lie inside the circle with radius 6.5 units. So, consider the distance between O(0, 0) and point D$\left(-6,\frac{5}{2}\right)$.

$\begin{array}{rcl}OD& =& \sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2} \left[\because \mathrm{Distance} \mathrm{between} \mathrm{two} \mathrm{points} \left(x_1,y_1\right) \mathrm{and} \left(x_2,y_2\right)= \sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\right]\\ & =& \sqrt{{\left(-6-0\right)}^2+{\left(\frac{5}{2}-0\right)}^2}\\ & =& \sqrt{36+\frac{25}{4}}\\ & =& \sqrt{\frac{169}{4}}\\ & =& \frac{13}{2}\\ & =& 6.5 \mathrm{units}\end{array}$

Thus, point D lies on the circle and not in its interior.

Hence, the correct answer is option (d).

Answer:

It is given that distance between (4, p) and (1, 0) is 5.

In general, the distance between A and B is given by .

So,

On further simplification,

Hence, the correct answer is option (b).

Answer:

Given that, the vertices of a quadrilateral are A(9, 0), B(9, 6), C(–9, 6) and D(–9, 0).

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

Here,
$\begin{array}{rcl}AB& =& \sqrt{{\left(9-9\right)}^2+{\left(6-0\right)}^2}\\ & =& 6 \mathrm{units}\end{array}$

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

$\begin{array}{rcl}CD& =& \sqrt{{\left(-9+9\right)}^2+{\left(0-6\right)}^2}\\ & =& \sqrt{{\left(-6\right)}^2}\\ & =& 6 \mathrm{units}\end{array}$

$\begin{array}{rcl}DA& =& \sqrt{{\left(-9-9\right)}^2+{\left(0-0\right)}^2}\\ & =& \sqrt{{\left(-18\right)}^2}\\ & =& 18 \mathrm{units}\end{array}$

Since the opposite sides are equal, it is a rectangle.

Hence, the correct answer is option (b).

Answer:

Given that, the distance between two points A(2, –2) and B(–1, x) is 5.

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

Here,
$$\begin{gathered} 5=\sqrt{{\left(-1-2\right)}^2+{\left(x+2\right)}^2} \\ \Rightarrow 5=\sqrt{9+x^2+4x+4} \\ \Rightarrow 25=x^2+4x+13 \\ \Rightarrow x^2+4x-12=0 \\ \Rightarrow x^2+6x-2x-12=0 \\ \Rightarrow x\left(x+6\right)-2\left(x+6\right)=0 \\ \Rightarrow \left(x-2\right)\left(x+6\right)=0 \\ \Rightarrow x=2 \mathrm{or} x=-6 \end{gathered}$$

Thus, the values of x are $2 \mathrm{or} -6$.

Therefore, the sum of the values of x = $-4$.

Hence, the correct answer is option (b).

Answer:

We have to find the distance between and.

In general, the distance between A and B is given by,

So,

But according to the trigonometric identity,

Therefore,

So, the answer is (b)

Answer:

We have to find the distance between A(a cos 25°, 0)  and.

In general, the distance between A and B is given by,

So,

 $$\begin{gathered} AB=\sqrt{{\left(0-a\cos 25°\right)}^2+{\left(a\cos 65°-0\right)}^2} \\ =\sqrt{{\left(a\cos 25°\right)}^2+{\left(a\cos 65°\right)}^2} \end{gathered}$$

$\cos 25°=\sin 65° \mathrm{and} \cos 65°=\sin 25°$

But according to the trigonometric identity,

Therefore,

So, the answer is (a)

Answer:

It is given that distance between P (x, 2) and is 10.

In general, the distance between A and B is given by,

So,

On further simplification,

We will neglect the negative value. So,

So the answer is (c)

Answer:

We have to find the distance between and.

In general, the distance between A and B is given by,

So,

But according to the trigonometric identity,

Therefore,

So, the answer is (d)

Answer:

Given that, the distance between two points A(4, p) and B(1, 0) is 5.

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

Here,
$$\begin{gathered} 5=\sqrt{{\left(1-4\right)}^2+{\left(0-p\right)}^2} \\ \Rightarrow 5=\sqrt{9+p^2} \\ \Rightarrow 25=p^2+9 \\ \Rightarrow p^2-16=0 \\ \Rightarrow p=\pm 4 \end{gathered}$$

Hence, the correct answer is option (a).

Answer:

It is given that distance between P (2,−3) and is 10.

In general, the distance between A and B is given by,

So,

On further simplification,

We will neglect the negative value. So,

So the answer is (b)

Answer:

We have a triangle whose co-ordinates are A (0, 0); B (1, 0); C (0, 1). So clearly the triangle is right angled triangle, right angled at A. So,

Now apply Pythagoras theorem to get the hypotenuse,

So the perimeter of the triangle is,

Therefore the answer is (d)

Answer:

We have a triangle in which the co-ordinates of the vertices are A (2, 2) B (−4,−4) and C (5,−8).

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

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

Now equate the individual terms to get,

So co-ordinates of D is (−1,−1)

So the length of median from C to the side AB,

So the answer is (c)

Answer:

Given that, the three points A(x, 2), B(–3, –4) and C(7, –5) are collinear.

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

Here,
$\begin{array}{rcl}AB& =& \sqrt{{\left(-3-x\right)}^2+{\left(-4-2\right)}^2}\\ & =& \sqrt{9+x^2+6x+36}\\ & =& \sqrt{x^2+6x+45}\end{array}$

$\begin{array}{rcl}BC& =& \sqrt{{\left(7+3\right)}^2+{\left(-5+4\right)}^2}\\ & =& \sqrt{100+1}\\ & =& \sqrt{101}\end{array}$

$\begin{array}{rcl}CA& =& \sqrt{{\left(7-x\right)}^2+{\left(-5-2\right)}^2}\\ & =& \sqrt{49+x^2-14x+49}\\ & =& \sqrt{x^2-14x+98}\end{array}$

If they are collinear, then AB + BC = CA.
Thus,
$\sqrt{x^2+6x+45}+\sqrt{101}=\sqrt{x^2-14x+98}$

Squaring both sides,
$x^2+6x+45+101+2\sqrt{101x^2+606x+4545}=x^2-14x+98$
$\Rightarrow 20x+48+2\sqrt{101x^2+606x+4545}=0$
$\Rightarrow 10x+24=-\sqrt{101x^2+606x+4545}$

Squaring both sides,
$$\begin{gathered} \Rightarrow 100x^2+576+480x=101x^2+606x+4545 \\ \Rightarrow x^2+126x+3969=0 \\ \Rightarrow \left(x+63\right)\left(x+63\right)=0 \end{gathered}$$

Thus, the value of x is –63.

Hence, the correct answer is option (a).

Answer:

Let P be the point of intersection of y-axis with the line segment joining A (−3,−4) and B (1,−2) which divides the line segment AB in the ratio.

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

Now we will use section formula as,

Now equate the x component on both the sides,

On further simplification,

So y-axis divides AB in the ratio

So the answer is (c)

Answer:

We have three collinear points.

In general if are collinear then,

So,

So,

So,

Therefore,

So the answer is (b)

Answer:

We have three collinear points A(3, 1) B(5, p) and C(7, −5).

In general, if are collinear then, the area of the triangle = 0. 
i.e., 

Here, $x_1=3, y_1=1, x_2=5, y_2=p, x_3=7, y_3=-5,$
So,

$$\begin{gathered} 3\left(p+5\right)+5\left(-5-1\right)+7\left(1-p\right)=0 \\ \Rightarrow 3p+15-30+7-7p=0 \\ \Rightarrow -4p-8=0 \\ \Rightarrow p=-2 \end{gathered}$$

Answer:

We have three collinear points.

In general if are collinear then,

So,

So,

Therefore,

So the answer is (c)

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, 5) divides the points A(2,3) and B(7,8). Substituting these values in the above formula we have,

Equating the individual components we have,

Hence the correct choice is option (d).

Answer:

Let P be the point of intersection of x-axis with the line segment joining A (3, 6) and B (12, −3) which divides the line segment AB in the ratio.

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 as,

Now equate the y component on both the sides,

On further simplification,

So x-axis divides AB in the ratio

So the answer is (a)

Answer:

Given that, the three points A(1, 2), B(–5, 6) and C(a, –2) are collinear.

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

Here,
$\begin{array}{rcl}AB& =& \sqrt{{\left(-5-1\right)}^2+{\left(6-2\right)}^2}\\ & =& \sqrt{36+16}\\ & =& \sqrt{52}\end{array}$

$\begin{array}{rcl}BC& =& \sqrt{{\left(a+5\right)}^2+{\left(-2-6\right)}^2}\\ & =& \sqrt{a^2+25+10a+64}\\ & =& \sqrt{a^2+10a+89}\end{array}$

$\begin{array}{rcl}CA& =& \sqrt{{\left(1-a\right)}^2+{\left(2+2\right)}^2}\\ & =& \sqrt{a^2+1-2a+16}\\ & =& \sqrt{a^2-2a+17}\end{array}$

If they are collinear, then AB + BC = CA.

Thus,
$\sqrt{52}+\sqrt{a^2+10a+89}=\sqrt{a^2-2a+17}$

Squaring both sides,
$$\begin{gathered} \Rightarrow 52+a^2+10a+89+2\sqrt{52a^2+520a+4628}=a^2-2a+17 \\ \Rightarrow 12a+124+2\sqrt{52a^2+520a+4628}=0 \end{gathered}$$

Squaring both sides,
$$\begin{gathered} 6a+62=-\sqrt{52a^2+520a+4628} \\ \Rightarrow 36a^2+3844+744a=52a^2+520a+4628 \\ \Rightarrow 16a^2-224a+784=0 \\ \Rightarrow a^2-14a+49=0 \\ \Rightarrow a^2-7a-7a+49=0 \\ \Rightarrow \left(a-7\right)\left(a-7\right)=0 \\ \Rightarrow a=7 \end{gathered}$$

Hence, the correct answer is option (b).

Answer:

The ordinate of a point gives its distance from the x-axis.

So, the distance of (4, 7) from x-axis is

So the answer is (b)

Answer:

The distance of a point from y-axis is given by abscissa of that point.

So, distance of (4, 7) from y-axis is.

So the answer is (a)

Answer:

Given that, the point P divides the line segment joining the points A(1, 3) and B(4, 6) in the ratio 2 : 1. Let the coordinates of P be (x, y).

Using the section formula, the coordinates of the point O(x, y) which divides the line segment joining the points X(x₁, y₁) and Y(x₂, y₂), internally, in the ratio m₁ : m₂ are given by $\left(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2}\right)$.

Here, (x₁, y₁) = (1, 3), (x₂, y₂) = (4, 6) and m₁ : m₂ = 2 : 1

Thus,
$\begin{array}{rcl}\mathrm{Coordinates} \mathrm{of} P& =& \left(\frac{2\left(4\right)+1\left(1\right)}{2+1},\frac{2\left(6\right)+1\left(3\right)}{2+1}\right)\\ & =& \left(\frac{9}{3},\frac{15}{3}\right)\\ & =& \left(3,5\right)\end{array}$

Hence, the correct answer is option (b).
 

Answer:

Let A(−1, 0) and B(5, 0) be the given points. Suppose the required point on the x-axis be P(x, 0).

It is given that P(x, 0) is equidistant from A(−1, 0) and B(5, 0).

∴ PA = PB

⇒ PA² = PB²

$\Rightarrow {\left[x-\left(-1\right)\right]}^2+{\left(0-0\right)}^2={\left(x-5\right)}^2+{\left(0-0\right)}^2$                           (Using distance formula)

$$\begin{gathered} \Rightarrow {\left(x+1\right)}^2={\left(x-5\right)}^2 \\ \Rightarrow x^2+2x+1=x^2-10x+25 \\ \Rightarrow 12x=24 \\ \Rightarrow x=2 \end{gathered}$$

Thus, the required point is (2, 0).

Hence, the correct answer is option B.

Answer:

We have an equilateral triangle whose co-ordinates are A (0, 0); and.

Since the triangle is equilateral. So,

So,

Cancel out the common terms from both the sides,

Therefore,

So, the answer is (d)

Answer:

We have three collinear points.

In general if are collinear then, area of the triangle is 0.

So,

So,

Take out the common terms,

Therefore,

So the answer is (b)

Answer:

Let the point be A be equidistant from the two given points P (−3, 4) and Q (2, 5).

So applying distance formula, we get,

Therefore,

Hence the co-ordinates of A are

So the answer is- (D) none of these.

Answer:

Disclaimer: option (b) and (c) are given to be same in the book. So, we are considering option (b) as −2, −4
instead of −2, 4.

We have a right angled triangle whose co-ordinates are A (5, 3); B (11,−5);

. So clearly the triangle is, right angled at A. So,

Now apply Pythagoras theorem to get,

So,

On further simplification we get the quadratic equation as,

Now solve this equation using factorization method to get,

Therefore,

So the answer is (c)

Answer:

We have three non-collinear points.

In general if are non-collinear points then are of the triangle formed is given by-

So,

So the answer is (d)

Answer:

We have the co-ordinates of the vertices of the triangle aswhich has an area of 5 sq.units.

In general if are non-collinear points then area of the triangle formed is given by-,

So,

Simplify the modulus function to get,

Therefore,

So the answer is (a)

Answer:

We have three collinear points.

In general if are collinear then,

So,

So,

Divide both the sides by,

So the answer is (a)

Answer:

We have to find the co-ordinates of the third vertex of the given triangle. Let the co-ordinates of the third vertex be.

The co-ordinates of other two vertices are (4,−3) and (−9, 7)

The co-ordinate of the centroid is (1, 4)

We know that the co-ordinates of the centroid of a triangle whose vertices are is

So,

Compare individual terms on both the sides-

So,

Similarly,

So,

So the co-ordinate of third vertex is (8, 8)

In general if are non-collinear points then are of the triangle formed is given by-,

So,

So the answer is (b)

Answer:

We have to find the unknown co-ordinates.

The co-ordinates of vertices are

The co-ordinate of the centroid is (6, 3)

We know that the co-ordinates of the centroid of a triangle whose vertices are is

So,

Compare individual terms on both the sides-

So,

Similarly,

So,

So the answer is (d)

Answer:

GIVEN: If P is a point on x axis such that its distance from the origin is 3 units.

TO FIND: The coordinates of a point Q on OY such that OP= OQ.

On x axis y coordinates is 0. Hence the coordinates of point P will be (3, 0) as it is given that the distance from origin is 3 units.

Now then the coordinates of Q on OY such that OP = OQ

On y axis x coordinates is 0. Hence the coordinates of point Q will be (0, 3)

Hence correct option is (a)

Answer:

It is given that is equidistant to the point

So,

So apply distance formula to get the co-ordinates of the unknown value as,

On further simplification we get,

So,

Thus,

So the answer is (c)

Answer:

We have to find the unknown co-ordinates.

The co-ordinates of vertices are

The co-ordinate of the centroid is

We know that the co-ordinates of the centroid of a triangle whose vertices are is-

So,

Compare individual terms on both the sides-

So the answer is (c)

Answer:

We have to find the unknown co-ordinates.

The co-ordinates of vertices are

The co-ordinate of the centroid is (−2, 1)

We know that the co-ordinates of the centroid of a triangle whose vertices are is-

So,

Compare individual terms on both the sides-

So,

Similarly,

So,

It can be observed that (x, y) = (−3, 5) does not satisfy any of the relations 3x + 8y = 0, 3x − 8y = 0, 8x + 3y = 0 or 8x = 3y.

Answer:

We have to find the co-ordinates of forth vertex of the rectangle ABCD.

We the co-ordinates of the vertices as (0, 0); (2, 0); (0, 3)

Rectangle has opposite pair of sides equal.

When we plot the given co-ordinates of the vertices on a Cartesian plane, we observe that the length and width of the rectangle is 2 and 3 units respectively.

So the co-ordinate of the forth vertex is

So the answer is (c).

Answer:

Let C be the point of intersection of x-axis with the line segment joining and which divides the line segment PQ in the ratio.

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 as,

Now equate the y component on both the sides,

On further simplification,

So the answer is (b)

Answer:

Let P be the point of intersection of y-axis with the line segment joiningandwhich divides the line segment AB in the ratio.

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 as,

Now equate the x component on both the sides,

On further simplification,

So the answer is (a)

Answer:

Let O(−2, 5) be the centre of the given circle and A(2, 3) and B(x, y) be the end points of a diameter of the circle.

Then, O is the mid-point of AB.

Using mid-point formula, we have

$\therefore \frac{2+x}{2}=-2$ and $\frac{3+y}{2}=5$                  

$\Rightarrow 2+x=-4$ and $3+y=10$

$\Rightarrow x=-6$ and $y=7$ 

Thus, the coordinates of the other end of the diameter are (−6, 7).

Hence, the correct answer is option A.

Answer:

The coordinates of A are (1, 3).

∴ Distance of A from the x-axis, AD = y-coordinate of A = 3 units

The number of units between B and C on the x-axis are 5.

∴ BC = 5 units

Now,

Area of ∆ABC = $\frac{1}{2}\times \mathrm{BC}\times \mathrm{AD}=\frac{1}{2}\times 5\times 3=\frac{15}{2}=7.5$ square units

Thus, the area of ∆ABC is 7.5 square units.

Hence, the correct answer is option C.

Answer:

It is given that A(4, 9), B(2, 3) and C(6, 5) are the vertices of ∆ABC.



Let CD be the median of ∆ABC through C. Then, D is the mid-point of AB.

Using mid-point formula, we get

Coordinates of D = $\left(\frac{4+2}{2},\frac{9+3}{2}\right)=\left(\frac{6}{2},\frac{12}{2}\right)=\left(3, 6\right)$

∴ Length of the median, AD

$$\begin{gathered} =\sqrt{{\left(6-3\right)}^2+{\left(5-6\right)}^2} \left(\mathrm{Using} \mathrm{distance} \mathrm{formula}\right) \\ =\sqrt{3^2+{\left(-1\right)}^2} \\ =\sqrt{10} \mathrm{units} \end{gathered}$$

Thus, the length of the required median is $\sqrt{10}$ units.

Hence, the correct answer is option B.

Answer:

It is given that P(2, 4), Q(0, 3), R(3, 6) and S(5, y) are the vertices of a parallelogram PQRS.



Join PR and QS, intersecting each other at O.

We know that the diagonals of the parallelogram bisect each other. So, O is the mid-point of PR and QS.

Coordinates of mid-point of PR = $\left(\frac{2+3}{2},\frac{4+6}{2}\right)=\left(\frac{5}{2},\frac{10}{2}\right)=\left(\frac{5}{2},5\right)$

Coordinates of mid-point of QS = $\left(\frac{0+5}{2},\frac{3+y}{2}\right)=\left(\frac{5}{2},\frac{3+y}{2}\right)$

Now, these points coincides at the point O.

$$\begin{gathered} \therefore \left(\frac{5}{2},\frac{3+y}{2}\right)=\left(\frac{5}{2},5\right) \\ \Rightarrow \frac{3+y}{2}=5 \\ \Rightarrow 3+y=10 \\ \Rightarrow y=7 \end{gathered}$$

Thus, the value of y is 7.

Hence, the correct answer is option A.

Answer:

Let A(0, 4), O(0, 0) and B(3, 0) be the vertices of ∆AOB.

Using distance formula, we get

OA = $\sqrt{{\left(0-0\right)}^2+{\left(4-0\right)}^2}=\sqrt{16}=4$ units

OB = $\sqrt{{\left(3-0\right)}^2+{\left(0-0\right)}^2}=\sqrt{9}=3$ units

AB = $\sqrt{{\left(3-0\right)}^2+{\left(0-4\right)}^2}=\sqrt{9+16}=\sqrt{25}=5$ units

∴ Perimeter of ∆AOB = OA + OB + AB = 4 + 3 + 5 = 12 units

Thus, the required perimeter of the triangle is 12 units.

Hence, the correct answer is option D.

Answer:

Use section formula for finding out the ratio in which P divided the line segment AB.
$$\begin{gathered} \left(2,1\right)=\left(\frac{m_1\left(8\right)+m_2\left(4\right)}{m_1+m_2},\frac{m_1\left(4\right)+m_2\left(2\right)}{m_1+m_2}\right) \\ \Rightarrow 2=\frac{8m_1+4m_2}{m_1+m_2} ; 1=\frac{4m_1+2m_2}{m_1+m_2} \\ \Rightarrow 3m_1+m_2=0 \\ \Rightarrow \frac{m_1}{m_2}=-\frac{1}{3} \\ \therefore the point P divided AB in ration 1: 3 externally. \\ AP:PB=1:3 \\ \Rightarrow AP:AB=1:2 \\ \Rightarrow AP=\frac{1}{2}AB \end{gathered}$$

Answer:

Using the Section Formula, we have
$$\begin{gathered} k=\frac{1\times \left(-7\right)+2\times 2}{1+2} \\ \Rightarrow k=\frac{-7+4}{3} \\ \Rightarrow k=\frac{-3}{3} \\ \Rightarrow k=-1 \end{gathered}$$


Hence, the correct answer is option (d).

Answer:

A line intersects the y axis, then the coordinates of P are (0, y) and x axis then the coordinates are Q(x, 0).
Therefore by section formula,
$$\begin{gathered} \left(\frac{x+0}{2},\frac{0+y}{2}\right)=\left(2,-5\right) \\ \Rightarrow \left(\frac{x}{2},\frac{y}{2}\right)=\left(2,-5\right) \\ \Rightarrow \frac{x}{2}=2, \frac{y}{2}=-5 \\ \Rightarrow x=4, y=-10 \end{gathered}$$

Hence the coordinates of P are (0, −10) and that of Q are (4, 0).

Answer:

It is given that the point A(x, 4) is at a distance of 5 units from origin O.

So, apply the distance formula to get,

Therefore,

So,

So the answer is (b)

Answer:

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

In a square all the sides are equal to each other.

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

The vertex ‘A’ should be equidistant from ‘B’ as well as ‘D’

Let us now find out the distances ‘AB’ and ‘AD’.

These two need to be equal.

Equating the above two equations we have,

Squaring on both sides we have,

Hence the correct choice is option (c).

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 O(0,0), A(a,0) and B(0,b).

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,

Hence the correct choice is option (b).

Answer:

TO FIND: The coordinates of a point on x axis which lies on perpendicular bisector of line segment joining points (7, 6) and (−3, 4).

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

PA=PB

On x-axis y is 0, so substituting y=0 we get x= 3

Hence the coordinates of point is i.e. option (b) is correct

Answer:

It is given that distance between P (2,−3) and is 10.

In general, the distance between A and B is given by,

So,

On further simplification,

We will neglect the negative value. So,

So the answer is (a)

Answer:

Given that, the line segment formed by joining the points A(3, –4), and B(1, 2) is trisected at points P(a, –2) and $Q\left(\frac{5}{3}, b\right).$

Using the section formula, the coordinates of the point O(x, y) which divides the line segment joining the points X(x₁, y₁) and Y(x₂, y₂), internally, in the ratio m₁ : m₂ are given by $\left(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2}\right)$.

Now, point P divides AB in the ratio 1 : 2.
Thus,
$\begin{array}{rcl}\left(a,-2\right)& =& \left(\frac{1\left(1\right)+2\left(3\right)}{1+2},\frac{1\left(2\right)+2\left(-4\right)}{1+2}\right)\\ & =& \left(\frac{7}{3},-2\right)\end{array}$
$\Rightarrow a=\frac{7}{3}$

While point Q divides AB in the ratio 2 : 1.
Thus,
$\begin{array}{rcl}\left(\frac{5}{3},b\right)& =& \left(\frac{2\left(1\right)+1\left(3\right)}{2+1},\frac{2\left(2\right)+1\left(-4\right)}{2+1}\right)\\ & =& \left(\frac{5}{3},0\right)\end{array}$
$\Rightarrow b=0$

Hence, the correct answer is option (b).
 

Answer:

(i) The mid-point of a line segment joining the points A$\left(x_1,y_1\right)$ and B$\left(x_2,y_2\right)$ is given by the coordinates $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$.
Here, a line segment is formed by joining the point J(6, 17) and I(9, 16).
Thus,
$\begin{array}{rcl}\mathrm{Coordinates} \mathrm{of} \mathrm{mid}-\mathrm{point} \mathrm{of} IJ& =& \left(\frac{6+9}{2},\frac{17+16}{2}\right)\\ & =& \left(\frac{15}{2},\frac{33}{2}\right)\end{array}$

Hence, the correct answer is option (c).

(ii) In the front view, we see that point P is 5 cm away from the y-axis and has coordinates (5, 8). Since the scale is given as 1 cm = 1 m, the distance of point P from the y-axis is 5 m.

Hence, the correct answer is option (a).

Disclaimer: Refer to the front view for point P and not the top view.

(iii) In the front view, the coordinates of point A are (1, 8) and that of point S are (17, 8).
Thus, the distance between the two points on the graph A and S is 17 − 1 = 16 cm. That means that the distance between the two points is actually 16 m.

Hence, the correct answer is option (c).

(iv) Given that, the line segment joining the points A and B in the ratio 1 : 3 internally. Now, the coordinates of point A are (1, 8) and that of point B are (5, 10). Let the point that divides AB in ratio 1 : 3 be M.
Thus,
$\begin{array}{rcl}\mathrm{Coordinates} \mathrm{of} M& =& \left(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2}\right)\\ & =& \left(\frac{1\times 5+3\times 1}{1+3},\frac{1\times 10+3\times 8}{1+3}\right)\\ & =& \left(\frac{8}{4},\frac{34}{4}\right)\\ & =& \left(2.0,8.5\right)\end{array}$

Hence, the correct answer is option (d).

(v) Given that, a point (x, y) is equidistant from the Q(9, 8) and S(17, 8). Let the point be N(x, y).
Thus, NQ = SN.
$$\begin{gathered} \Rightarrow {\left(x-9\right)}^2+{\left(y-8\right)}^2={\left(17-x\right)}^2+{\left(8-y\right)}^2 \\ \Rightarrow x^2+81-18x+y^2+64-16y=289+x^2-34x+64+y^2-16y \\ \Rightarrow 81-18x=289-34x \\ \Rightarrow 34x-18x=289-81 \\ \Rightarrow 16x=208 \\ \Rightarrow x=13 \\ \Rightarrow x-13=0 \end{gathered}$$

Hence, the correct answer is option (b).

Answer:

(i) According to the given figure, point A is taken as the origin. Therefore, the coordinates of point A are (0, 0).

Hence, the correct answer is option (c).

(ii) According to the given figure, the abscissa for point P is 4 and its ordinate is 6. Therefore, the coordinates of point P are (4, 6).

Hence, the correct answer is option (a).

(iii) According to the given figure, the abscissa for point R is 6 and its ordinate is 5. Therefore, the coordinates of point R are (6, 5).

Hence, the correct answer is option (a).

(iv) According to the given figure, the abscissa for point D is 16 and its ordinate is 0 as it lies on the x-axis. Therefore, the coordinates of point D are (16, 0).

Hence, the correct answer is option (a).

(v) Given that, D is taken as the origin, DA along negative x-axis and DC along y-axis. Now, point P lies 12 units away from D towards the negative x-axis and 6 units away on the positive y-axis. Thus, the coordinates of P is (−12, 6).

Hence, the correct answer is option (b).

Answer:

(i) According to the given figure, the abscissa is 3 and ordinate is 4. Thus, the coordinates of point A are (3, 4).

Hence, the correct answer is option (b).

(ii) From the figure, we get the points as A(3, 4), B(6, 7), C(9, 4) and D(7, 2).
The distance between two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given as $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$.

Here,
$\begin{array}{rcl}AB& =& \sqrt{{\left(6-3\right)}^2+{\left(7-4\right)}^2}\\ & =& \sqrt{9+9}\\ & =& \sqrt{18}\\ & =& 3\sqrt{2} \mathrm{units}\end{array}$
$\begin{array}{rcl}BC& =& \sqrt{{\left(9-6\right)}^2+{\left(4-7\right)}^2}\\ & =& \sqrt{9+9}\\ & =& \sqrt{18}\\ & =& 3\sqrt{2} \mathrm{units}\end{array}$
$\begin{array}{rcl}CD& =& \sqrt{{\left(7-9\right)}^2+{\left(2-4\right)}^2}\\ & =& \sqrt{4+4}\\ & =& \sqrt{8}\\ & =& 2\sqrt{2} \mathrm{units}\end{array}$
$\begin{array}{rcl}DA& =& \sqrt{{\left(7-3\right)}^2+{\left(2-4\right)}^2}\\ & =& \sqrt{16+4}\\ & =& \sqrt{20}\\ & =& 2\sqrt{5} \mathrm{units}\end{array}$
Since opposite pair of sides of the quadrilateral formed are not equal, the figure is not a parallelogram. Since it is not a parallelogram, it cannot be a square or a rhombus.

Hence, the correct options are option (a), (b) and (c).

Disclaimer: The figure formed is not a parallelogram as per the above solution. Since it is not a parallelogram, it cannot be a square or a rhombus. Thus, the figure formed is a trapezium.

(iii) The coordinates of the mid-point of a line segment with coordinates of the endpoints as $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are given as $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$.

Now,
$\begin{array}{rcl}\mathrm{Coordinates} \mathrm{of} \mathrm{mid}-\mathrm{point} \mathrm{of} AC& =& \left(\frac{3+9}{2},\frac{4+4}{2}\right)\\ & =& \left(\frac{12}{2},\frac{8}{2}\right)\\ & =& \left(6,4\right)\end{array}$

$\begin{array}{rcl}\mathrm{Coordinates} \mathrm{of} \mathrm{mid}-\mathrm{point} \mathrm{of} BD& =& \left(\frac{6+7}{2},\frac{7+2}{2}\right)\\ & =& \left(\frac{13}{2},\frac{9}{2}\right)\end{array}$

Using the distance formula to find the distance between the two points,
$\begin{array}{rcl}d& =& \sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ & =& \sqrt{{\left(6-\frac{13}{2}\right)}^2+{\left(4-\frac{9}{2}\right)}^2}\\ & =& \sqrt{{\left(-\frac{1}{2}\right)}^2+{\left(-\frac{1}{2}\right)}^2}\\ & =& \frac{1}{\sqrt{2}} \mathrm{units}\end{array}$

Answer:

(i) Shruti runs ${\frac{1}{3}}^{\mathrm{rd}}$of the distance in the second line along AD and post her flag. Saanvi runs ${\frac{1}{5}}^{\mathrm{th}}$ of the distance AD in the eighth line and posts her flag. Thus, Shruti runs 40 m while Saanvi runs 24 m. Therefore, the coordinates of Shruti's flag is (2, 40) and that of Saanvi's flag is (8, 24).

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

Thus, distance between the two flags is given as:
$\begin{array}{rcl}d& =& \sqrt{{\left(8-2\right)}^2+{\left(24-40\right)}^2}\\ & =& \sqrt{36+256}\\ & =& \sqrt{292} \mathrm{m}\\ & =& 2\sqrt{73} \mathrm{m}\end{array}$

Hence, the correct answer is option (a).

(ii) Reena's flag is exactly halfway between the line segment joining the two flags.
The coordinates of the mid-point of a line segment with coordinates of the endpoints as $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are given as $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$.

Thus,
$\begin{array}{rcl}\mathrm{Coordinates} \mathrm{of} \mathrm{Reena}\text{'}\mathrm{s} \mathrm{flag}& =& \left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ & =& \left(\frac{2+8}{2},\frac{40+24}{2}\right)\\ & =& \left(\frac{10}{2},\frac{64}{2}\right)\\ & =& \left(5,32\right)\end{array}$

Hence, the correct answer is option (c).

(iii) Shruti runs ${\frac{1}{3}}^{\mathrm{rd}}$of the distance, i.e., 40 m in the second line along AD and post her flag. Therefore, the coordinates of Shruti's flag is (2, 40).

Hence, the correct answer is option (a).

(iv)  Saanvi runs ${\frac{1}{5}}^{\mathrm{th}}$ of the distance AD, i.e., 24 min the eighth line and posts her flag. Therefore, the coordinates of Saanvi's flag is (8, 24).

Hence, the correct answer is option (d).

Answer:

Statement-1 (Assertion): If a + b + c = 0, then the centroid of the triangle whose vertices are P(a, b), Q(b, c) and R(c, a) is at the origin.


The coordinates of the centroid of the triangle whose vertices are A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right).$ Now, let the vertices of $△$PQR be P(a, b), Q(b, c) and R(c, a).
Thus, 
$\begin{array}{rcl}\mathrm{Coordinates} \mathrm{of} \mathrm{its} \mathrm{centroid}& =& \left(\frac{a+b+c}{3}, \frac{b+c+a}{3}\right)\\ & =& \left(\frac{0}{3}, \frac{0}{3}\right) \left(\because a+b+c=0\right)\\ & =& \left(0,0\right)\end{array}$
So, its centroid is at the origin. Thus, Statement-1 is true.

Statement-2 (Reason): The coordinates of the centroid of the triangle whose vertices are A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right).$

Let the vertices of $△$ABC be A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃). Let AD be the median of the $△$ABC drawn from point A and D is the mid-point of BC. Now, the coordinates of the mid-point of a line segment with coordinates of the endpoints as $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are given as $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$.

Thus, $\mathrm{coordinates} \mathrm{of} D=\left(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}\right)$.

Now, let G be the centroid of the $△$ABC that divides the median AD in the ratio 2 : 1.
$\begin{array}{rcl}\mathrm{Coordinates} \mathrm{of} G& =& \left(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2}\right)\\ & =& \left(\frac{2\times \left({\displaystyle \frac{x_2+x_3}{2}}\right)+1\times x_1}{2+1},\frac{2\times \left({\displaystyle \frac{y_2+y_3}{2}}\right)+1\times y_1}{2+1}\right)\\ & =& \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\end{array}$

Thus, Statement-2 is also true and is the correct explanation of Statement-1.

Hence, the correct answer is option (a).