TOPICS

Exercise - 7.2

Quadratic Equations

**Question-1 :-** Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3.

Let P(x, y) be the required point. Given that : x₁ = -1, y₁ = 7 ; x₂ = 4, y₂ = -3 ; m₁ = 2, m₂ = 3 Using the section formula, we get x = [(2x4 + 3x(-1))/(2 + 3)] = (8 - 3)/5 = 5/5 = 1 y = [(2x(-3) + 3x7)/(2 + 3)] = (-6 + 21)/5 = 15/5 = 3 Therefore, (1, 3) is the required point.

**Question-2 :-** Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).

Let P and Q be the points of trisection of AB i.e., AP = PQ = QB Therefore, P divides AB internally in the ratio 1 : 2. Therefore, the coordinates of P, by applying the section formula, are [(1x(-2) + 2x4)/(1 + 2), (1x(-3) + 2x(-1))/(1 + 2)] i.e., [(-2+8)/3, (-3-2)/3] i.e., (3, -5/3) Now, Q also divides AB internally in the ratio 2 : 1. So, the coordinates of Q are [(2x(-2) + 1x4)/(2 + 1), (2x(-3) + 1x(-1))/(2 + 1)] i.e., [(-4+4)/3, (-6-1)/3] i.e., (0, -7/3) Therefore, the coordinates of the points of trisection of the line segment joining A and B are (–1, 0) and (– 4, 2).

**Question-3 :-** To conduct Sports Day activities, in your rectangular shaped school ground ABCD,
lines have been drawn with chalk powder at a distance of 1m each.
100 flower pots have been placed at a distance of 1m from each other along AD, as shown in Fig. 7.12.
Niharika runs 1/4th the distance AD on the 2nd line and posts a green flag.
Preet runs 1/5th the distance AD on the eighth line and posts a red flag.
What is the distance between both the flags?
If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?

It can be observed that Niharika posted the green flag at 1/4 of the distance AD i.e., (1/4 x 100)m = 25m from the starting point of 2nd line. Therefore, the coordinates of this point G is (2, 25). Similarly, Preet posted red flag at 1/5 of the distance AD i.e., (1/5 x 100)m = 20m from the starting point of 8th line. Therefore, the coordinates of this point R are (8, 20). Distance between these flags by using distance formula = GR = √(8-2)² + (25-20)² = √(6)² + (5)² = √61 m The point at which Rashmi should post her blue flag is the mid-point of the line joining these points. Let this point be A (x, y). x = (2+8)/2, y = (25+20)/2 x = 5, y = 22.5 Hence, A(x, y) = (5, 22.5) Therefore, Rashmi should post her blue flag at 22.5m on 5th line

**Question-4 :-** Find the ratio in which the line segment joining the points (–3, 10) and (6, – 8) is divided by (– 1, 6).

Let the ratio in which the line segment joining (−3, 10) and (6, −8) is divided by point (−1, 6) be k : 1. Therefore, (-1, 6) = [(6k - 3)/(k + 1), (-8k + 10)/(k + 1)] -1 = (6k - 3)/(k + 1) -k - 1 = 6k - 3 -k - 6k = -3 + 1 -7k = -2 k = 2/7 Therefore, Required ratio is 2:7.

**Question-5 :-** Find the ratio in which the line segment joining A(1, – 5) and B(–4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

Let the ratio be k : 1. Then by the section formula, the coordinates of the point which divides AB in the ratio k : 1 are [(-4k + 1)/(k + 1), (5k - 5)/(k + 1)] This point lies on the x-axis, and we know that on the x-axis the ordinate is 0. Therefore, (5k - 5)/(k + 1) = 0 5k - 5 = 0 5k = 5 k = 1 i.e., the ratio is 1 : 1. Putting the value of k = 1, we get the point of intersection as = [(-4 + 1)/2, (5 - 5)/2] = (-3/2, 0)

**Question-6 :-** If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Let (1, 2), (4, y), (x, 6), and (3, 5) are the coordinates of A, B, C, D vertices of a parallelogram ABCD. Intersection point O of diagonal AC and BD also divides these diagonals. Therefore, O is the mid-point of AC and BD. If O is the mid-point of AC, then the coordinates of O are [(1+x)/2, (2+6)/2] = [(x+1)/2, 4] If O is the mid-point of BD, then the coordinates of O are [(4+3)/2, (5+y)/2] = [7/2, (y+5)/2] Since both the coordinates are of the same point O, (x+1)/2 = 7/2 ; 4 = (y+5)/2 x + 1 = 7 ; 8 = y + 5 x = 7 - 1 ; y = 8 - 5 x = 6; y = 3

**Question-7 :-** Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, –3) and B is (1, 4).

Let the coordinates of point A be (x, y). Mid-point of AB is (2, −3), which is the center of the circle. Therefore, (2, -3) = [(x+1)/2, (y+4)/2] 2 = (x+1)/2 ; -3 = (y+4)/2 4 = x + 1 ; -6 = y + 4 x = 4 - 1 ; y = -6 - 4 x = 3; y = -10 Hence, the coordinates of A are (3, -10).

**Question-8 :-** If A and B are (–2, – 2) and (2, – 4), respectively, find the coordinates of P such that AP = 3AB/7 and P lies on the line segment AB.

The coordinates of point A and B are (−2, −2) and (2, −4) respectively. AP = 3AB/7 AP/AB = 3/7 PB = AB - AP = 7 - 3 = 4 So, AP/PB = 3/4 Therefore, AP: PB = 3:4 Point P divides the line segment AB in the ratio 3:4. Coordinates of P = [(3x2 + 4x(-2))/(3 + 4), (3x(-4) + 4x(-2))/(3 + 4)] = [(6 - 8)/7, (-12 - 8)/7] = (-2/7, -20/7)

**Question-9 :-** Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts.

From the figure, it can be observed that points P, Q, R are dividing the line segment in a ratio 1:3, 1:1, 3:1 respectively. Now, Coordinates of P = [(1x2 + 3x(-2))/(1 + 3), (1x8 + 3x2)/(1 + 3)] = (-1, 7/2) Also, Coordinates of Q = [(1x2 + 1x(-2))/(1 + 1), (1x8 + 1x2)/(1 + 1)] = (0, 5) Also, Coordinates of R = [(3x2 + 1x(-2))/(3 + 1), (3x8 + 1x2)/(3 + 1)] = (1, 13/2)

**Question-10 :-** Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order.

Let (3, 0), (4, 5), (−1, 4) and (−2, −1) are the vertices A, B, C, D of a rhombus ABCD. Now, Length of diagonal of AC = √(3+1)² + (0-4)² = √(4)² + (4)² = √32 = 4√2 Also, Length of diagonal of BD = √(4+2)² + (5+1)² = √(6)² + (6)² = √72 = 6√2 Therefore, Area of Rhombus ABCD = 1/2 x 4√2 x 6√2 = 24 square units.

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