# 10th Maths Chapter 5 Coordinate Geometry Exercise 5.5

10th Standard Maths Chapter 5 Coordinate Geometry Exercise 5.5 Guide. TN SSLC Samacheer Kalvi Guide Chapter 5 Exercise 5.5 Book Back Answers & Solutions. 10th All Subject Guide – Click Here. Class 1 to 12 All Subject Book Back Answers – Click Here.

### Multiple choice questions:

#### 5. The point of intersection of 3x – y = 4 and x + y = 8 is (1) (5, 3) (2) (2, 4) (3) (3, 5) (4) (4, 4)Solution: (3) (3, 5)] Hint:

∴ The point of intersection is (3, 5)

(1) 1
(2) 4
(3) -5
(4) 2
Solution:
(4) 2
Hint:

#### 10. The equation of a line passing through the origin and perpendicular to the line 7x – 3y + 4 = 0 is (1) 7x – 3y + 4 = 0

(2) 3x – 7y + 4 = 0
(3) 3x + 7y = 0
(4) 7x – 3y = 0
Solution:
(3) 3x + 7y = 0

#### 12. A straight line has equation 8y = 4x + 21. Which of the following is true (1) The slope is 0.5 and the y-intercept is 2.6 (2) The slope is 5 and the intercept is 1.6 (3) The slope is 0.5 and the y-intercept is 1.6 (4) The slope is 5 and the intercept is 2.6Solution: (1) The slope is 0.5 and the y-intercept is 2.6

13. When proving that a quadrilateral is a trapezium, it is necessary to show ………………
(1) Two sides are parallel.
(2) Two parallel and two non-parallel sides.
(3) Opposite sides are parallel.
(4) All sides are of equal length.
(2) Two parallel and two non-parallel sides.

14. When proving that a quadrilateral is a parallelogram by using slopes you must find
(1) The slopes of two sides
(2) The slopes of two pairs of opposite sides
(3) The lengths of all sides
(4) Both the lengths and slopes of two sides
Solution:
(2) The slopes of two pair of opposite sides

15. (2, 1) is the point of intersection of two lines.
(1) x – y – 3 = 0; 3x – y – 7 = 0
(2) x + y = 3; 3x + y = 7
(3) 3x + y = 3; x + y = 7
(4) x + 3y – 3 = 0; x – y – 7 = 0
(2) x + y = 3; 3x + y = 7
Hint:
Substitute the value of x = 2 and y = 1 in the given equation.
(1) ⇒ x – y – 3 = 0
2 – 1 – 3 = 0
2 – 4 = 0
– 2 ≠ 0
not true

3x – y – 7 = 0
3(2) – 1 – 7 = 0
6 – 8 = 0
-2 ≠ 0
not true

(2) ⇒ x + y = 3
2 + 1 = 3
3 = 3
True

3x + y = 7
3(2) + 1 = 7
6 + 1 = 7
7 = 7
True
∴ (2, 1) is the point of intersection

(3) ⇒ 3x + y = 3
3(2) + 1 = 3
6 + 1 = 3
7 = 3
not true

x + y = 7
2 + 1 = 7
3 = 7
not true

(4) ⇒ x + 3y – 3 = 0
2 + 3 – 3 = 0
5 – 3 = 0
2 ≠ 0
not true

x – y – 7 = 0
2 – 1 – 7 = 0
2 – 8 = 0
-6 ≠ 0
not true

UNIT 5 – EXERCISE 5

### 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 midpoints of PQ, QR, RS & SP respectively.

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

### 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 (x1, y1),

Solution:

### 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.
Solution:

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

Solution:

The slope of AB = Slope of CD
The slope of BC = Slope of DA
Hence ABCD forms a parallelogram.

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

Let the intercepts be x1, y1 respectively

### 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:

Solution:

### 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

Solution: