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## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Unit Exercise 5

Question 1.

PQRS is a rectangle formed by joining the points P(-1, -1), Q(-1, 4) ,R(5, 4) and S(5, -1). A, B, C and D are the mid-points of PQ, QR, RS and SP respectively. Is the quadrilateral ABCD a square, a rectangle or a rhombus? Justify your answer.

Solution:

A, B, C and D are mid points of PQ, QR, RS & SP respectively.

∴ AB and BC are not perpendicular

⇒ ABCD is rhombus as diagonals are perpendicular and sides are not perpendicular.

Question 2.

The area of a triangle is 5 sq.units. Two of its vertices are (2, 1) and (3, -2). The third Vertex is (x, y) where y = x + 3 . Find the coordinates of the third vertex.

Solution:

Area of triangle formed by points (x_{1}, y_{1}),

Question 3.

Find the area of a triangle formed by the lines 3x + y – 2 = 0, 5x + 2y – 3 = 0 and 2x – y – 3 = 0

Solution:

Question 4.

If vertices of a quadrilateral are at A(-5, 7), B(-4, k) , C(-1, -6) and D(4, 5) and its area is

72 sq.units. Find the value of k.

Area (quadrilateral ABCD)

Question 5.

Without using distance formula, show that the points (-2, -1) , (4, 0) , (3, 3) and (-3, 2) are vertices of a parallelogram.

Solution:

Slope of AB = Slope of CD

Slope of BC = Slope of DA

Hence ABCD forms a parallelogram.

Question 6.

Find the equations of the lines, whose sum and product of intercepts are 1 and -6 respectively.

Let the intercepts be x_{1}, y_{1} respectively

Question 7.

The owner of a milk store finds that, he can sell 980 litres of milk each week at ₹ 14/litre and 1220 litres of milk each week at ₹ 16 litre. Assuming a linear relationship

between selling price and demand, how many litres could he sell weekly at ₹ 17/litre?

Solution:

Question 8.

Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.

Solution:

Question 9.

Find the equation of a line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.

Solution:

4x + 7y – 3 = 0

2x – 3y + 1 = 0

4x + 7y – 3 – 2(2x – 3y + 1) = 0

Question 10.

A person standing at a junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 seek to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find the equation of the path that he should follow.

Solution: