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**Short Answer Type Questions**

**Q.1. Find the coordinates of a point A, where AB is diameter of a circle whose centre is (2, -3) and B is the point (1, 4). [Delhi 2019]Ans**. AB is diameter of the circle.

Let C be centre of circle, coordinates of C are (2, -3).

So, C is mid-point of AB (diameter).

Let coordinates of A are (x, y).

∴ the coordinates of A are (3, - 10)

ATQ

∴ Coordinates of the point P are (2, 0).

∴ Co-ordinate of A are (-7,0)

Ans.

By using distance formula,

PA = PB

Squaring both sides, we get

(x - 1)

x

-2x + 17 - 8y = 2x - 4y + 5

- 4x - 4y = -12

x + y = 3

∴ D is mid-point of BC

Coordinates of D are:

Ans.

Let, p(4, m) divides A(2, 3) and B(6, - 3) in the ratio m

[By using section formula,

∴ m = 0

Also,for diagonal BD,

Coordinates of mid-point O

Let A(3, 2), B (1, 0) and O (2, - 5) are coordinates.

Let coordinates of C are (a, b), and coordinates of D are (x, y). As diagonals of parallelogram bisect each other at O. So, O is mid point of AC and BD

Join AC.AC divides quadrilateral ABCD into ΔABC and ΔACD.

By using formula,

If coordinates are (x, 3), (4, 4) and (3, 5), then area of triangle

- x + 5 = 8 or - x + 5 = - 8

- x = 3 or - x = - 13

x = - 3 or x = 13

Using distance formula,

⇒

⇒

On squaring, we get

⇒ 25 - 9 = k

∴

∴ PA = PB

Squaring both sides, we get

Area of ΔABC

Hence, co-ordinates are (1, -12) and (5, -10).

Let P(x, 0) be any point on x-axis.

Now, P(x, 0) is equidistant from point A(2, -5) and B(-2, 9)

∴ AP = BP

Squaring both sides, we have

(x - 2)

⇒ x

∴ x = 56/-8 = -7

∴ The point on the x-axis equidistant from given point is (- 7, 0).

∴ Co-ordinates of B are (5, 0)

Let co-ordinates of C be (x.y)

AC

Using distance formula

(+ ve sign to be taken)

Given: ar(ΔABC) = 5 sq. units

∴ Area of triangle with vertices (a, a

∵ Area of A ≠ 0

∴ Given points are not collinear.

(a + b - x)

⇒ (a + b)

= (a - b)

⇒ 4ay = 4bx or bx = ay

Hence proved.

∵ C (-1, 2) divides AB in the ratio 3 : 4.

Ans.

which is independent of t.

Hence proved.

Let the point on y-axis be p (0, y) and AP : PB = k : 1**Q.28. Let P and Q be the points of trisection of the line segment joining the points A(2, - 2) and 5(-7, 4) such that P is nearer to A. Find the coordinates of P and Q. [CBSE (AI) 2016]****Ans. ** ∵ P divides AB in the ratio 1 : 2.

∵ Q is the mid-point of PB.

∴ **Q.29. Find the ratio in which the point (-3, k) divides the line-segment joining the points (-5, -4) and (-2, 3). Also find the value of k. [CBSE (F) 2016]****Ans. **Let Q, divide AB in the ratio of p : 1

⇒ -3p - 3 = -2p - 5 ⇒ p = 2

∴ Ratio is 2:1**Q.30. Show that the points A (a, a), B (- a, - a) and form an equilateral triangle. [Foreign 2015]****Ans.**

Since AB = BC = AC. ∴ ΔABC is equilateral.

**Long Answer Type Questions**

**Q.1. Find the area of a quadrilateral ABCD whose vertices are A(1, 1), B(7, -3), C(7, 21) and D(7, 21) [AI 2017(C)]****Ans. **ar(quad. ABCD) = ar(ΔABC) + ar(ΔADC) ...(i)**Q.2. In Fig. 6.32, the vertices of ΔABC are A(4, 6), B (1, 5) and C(7, 2). A line-segment DE is drawn to intersect the sides AB and AC at D and E respectively such that Calculate the area of ΔADE and compare it with area of ΔABC. [CBSE (AI) 2016]**

Area of ΔABC

Area of ΔADE : area oF ΔABC

∴ Coordinates of point B are (0, 3).

So, BC = 6 units

Let the coordinates of point A be (x, 0).

Using distance formula,

Also, AB = BC (∵ ΔABC is an equilateral triangle)

⇒

∴ Coordinates of points = (x, 0) =

Since BACD is a rhom bus.

∴ AB = AC = CD = DB

∴ Coordinates of point D

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