OP Malhotra Class 9 Maths Solutions Chapter 20 Coordinates and Graphs of Simultaneous Linear Equations Chapter Test

Regular engagement with ICSE S Chand Maths Class 9 Solutions Chapter 20 Coordinates and Graphs of Simultaneous Linear Equations Chapter Test can boost students’ confidence in the subject.

S Chand Class 9 ICSE Maths Solutions Chapter 20 Coordinates and Graphs of Simultaneous Linear Equations Chapter Test

Question 1.
If (1, 2) and (3, 8) are the extremes of a diagonal of a square, then the area of the square is
(a) 10 sq. units
(b) 20 sq. units
(c) 15 sq. units
(d) 8 sq. units
Solution:
(1, 2) and (3, 8) are the end points of a diagonal of a square
∴ Length of diagonal = \(\sqrt{(3-1)^2+(8-2)^2}\)
= \(\sqrt{2^2+6^2}\) = \(\sqrt{4+36}\) = \(\sqrt{40}\)
∴ Area of square = \(\frac{(\text { Diagonal })^2}{2}\)
\(\frac{(\sqrt{40})^2}{2}\) = \(\frac{40}{2}\) = 20 sq. units (b)

Question 2.
If the distance between two points (0, -5) and (x, 0) is 13 units, then x =
(a) 10
(b) ±10
(c) 12
(d) ± 12
Solution:
Distance between (0, -5) and (x, 0) = 13
⇒ \(\sqrt{(x-0)^2+(0+5)^2}\) = 13
⇒ \(\sqrt{x^2+25}\) = 13
Squaring both sides,
x² + 25 = 169 ⇒ x² = 169 – 25 = 144
⇒ x² = (± 12)²
∴ x = ± 12 (d)

Question 3.
If A (x, y) is equidistant from P (-3, 2) and Q (2, -3) then
(a) 2x = y
(b) x = -y
(c) x = 2y
(d) x = y
Solution:
A (x, y) is equidistant from P (-3, 2) and Q (2, -3)
Then AP = AQ
⇒ \(\sqrt{(-3-x)^2+(2-y)^2}\) = \(\sqrt{(2-x)^2+(-3-y)^2}\)
Squaring,
(-3 -x)² + (2 – y)² = (2 – x)² + (-3 -y)²
⇒ 9 + x² + 6x + 4 + y² – 4y = 4 + x²– 4x + 9 + y² + 6y
⇒ 6x – 4y + 13 = 6y – 4x + 13
⇒ 6x + 4x =6y + 4y ⇒ 10x = 10y
∴ x = y

Question 4.
The nearest point from the origin is
(a) (2,-3)
(b) (6,0)
(c) (-2,-1)
(d) (3,5)
Solution:
Distance between origin (0, 0) is
(a) (2, -3) = \(\sqrt{2^2+3^2}\) = \(\sqrt{4+9}\) = \(\sqrt{13}\)
(b) (6, 0) = \(\sqrt{6^2+0^2}\) =\(\sqrt{36}\) = 6
(c) (-2, -1) = \(\sqrt{(-2)^2+(-1)^2}\) = \(\sqrt{4+1}\) = \(\sqrt{5}\)
(d) (3, 5) = \(\sqrt{3^2+5^2}\) = \(\sqrt{9+25}\) = \(\sqrt{34}\)
It is clear that point (-2, -1) is nearest to (0, 0) (c)

Question 5.
The coordinates of the vertices of a side of square are (4, -3) and (-1, -5). Its area is
(a) \(2 \sqrt{29}\) sq. units
(b) \(\frac{\sqrt{89}}{2}\) sq. units
(c) 89 sq. units
(d) 29 sq. units
Solution:
The coordinates of the each of a side of square are (4, -3) and (-1, -5)
∴ Length of side = \(\sqrt{(-1-4)^2+(-5+3)^2}\)
= \(\sqrt{(-5)^2+(-2)^2}\)
= \(\sqrt{25+4}\) = \(\sqrt{29}\)
∴ Area of square = (Side)²
= (√29)² sq. units
= 29 sq. units (d)
Solve by graphing.

Question 6.
2x + y = -8
y = \(\frac { 1 }{ 3 }\)x – 1
Solution:
2x + y = -8
y = \(\frac { 1 }{ 3 }\)x – 1
2x + y = -8 ⇒ y = -2x – 8
Giving some different values to x, we get corresponding values of y as given below:

X	-2	-4	-5
y	-4	0	2

Now plot the points (-2, -4),(-4, 0),(-5, 2) on the graph and join them to get a line Similarly, y = \(\frac { 1 }{ 3 }\)x – 1

X	3	6	-3
y	0	1	-2

Plot the points (3, 0),(6, 1) and (-3, -2) on the graph and join them to get another line. We see that these two lines intersect each other at (-3, -2)
∴ x = -3, y = -2

Question 7.
Draw the graph of 2x – y – 1 = 0 and 2x + y = 9, on the same axes. Use 2 cm = 1 unit on both axes and plot only 3 points per line. Read the coordinates of their point of intersection.
Solution:
2x – y – 1 = 0 and 2x + y = 9
⇒ x = \(\frac{y+1}{2}\)
Giving some different values of y, we get corresponding values of x as given below :

X	1	2	0
y	1	3	-1

Plot the points (1, 1),(2, 3) and (0, -1) on the graph and join them to get a line.
Similarly 2x + y = 9
y = 9 – 2x

X	4	5	6
y	1	-1	-3

Plot the points (4, 1),(5, -1) and (6, -3) on the graph and join them to get another line.
These two lines intersect each other at point \(\left(\frac{5}{2}, 4\right)\)
∴ x = \(\frac { 5 }{ 2 }\), y = 4

Question 8.
Choose the equation whose graph is given here
(a) y = x
(b) x + y = 0
(c) y = 2x
(d) 2 + 3y = 7x

Solution:
From the given graph, Two points are given on it whose x and y are equal but in opposite signs
It is through the origin
Equation will be x = -y
⇒ x + y = 0

Question 9.
The distance between the piont (0, 0) and the intersecting point of the graphs of x = 3 and y = 4 is
(a) 4 units
(b) 3 units
(c) 2 units
(d) 5 units
Solution:
The distance between the point (0, 0) and point of intersection of x = 3 and y = 4
i.e., (3, 4) will be = \(\sqrt{(3-0)^2+(4-0)^2}\)
= \(\sqrt{3^2+4^2}\) = \(\sqrt{9+16}\)
= \(\sqrt{25}\) = 5 units

Question 10.
The points (-4, 0),(4, 0),(0, 3) are the vertices of a
(a) right triangle
(b) isosceles triangle
(c) equilateral triangle
(d) scalene triangle
Solution:
The points (-4, 0),(4, 0),(0, 3) are given Now,
AB = \(\sqrt{(4+4)^2+(0+0)^2}\) = \(\sqrt{8^2+0^2}\) = \(\sqrt{64}\) = 8
BC = \(\sqrt{(0-4)^2+(3-0)^2}\) = \(\sqrt{(-4)^2+(3)^2}\) = \(\sqrt{16+9}\) = \(\sqrt{25}\) = 5
AC = \(\sqrt{(0-4)^2+(3-0)^2}\) = \(\sqrt{(-4)^2+(3)^2}\) = \(\sqrt{16+9}\) = \(\sqrt{25}\) = 5
∴ BC = AC
∴ The given points are the vertices of an isosceles triangle. (b)