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,

x^{2} + 25 = 169 ⇒ x^{2} = 169 – 25 = 144

⇒ x^{2} = (± 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 + x^{2} + 6x + 4 + y^{2} – 4y = 4 + x^{2}– 4x + 9 + y^{2} + 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)