SChand CLASS 9 Chapter 20 Coordinates and Graphs of Simultaneous Linear Equation TEST

 TEST

Question 1

Ans: $(1.21$ and $(3,8)$ are the end point of a diagonal of a square

So length of diagonal $=\sqrt{(3-1)^{2}+(8-2)^{2}}$

$=\sqrt{2^{2}+6^{2}}=\sqrt{4+36}=\sqrt{60}$

So Area of square $=\frac{(\text { diagonal })^{2}}{2}$

$\frac{\sqrt{40} 1^{2}}{2}=\frac{40}{2}=20$ sq units (b)

Question 2

Ans: Distance between $(0,-5)$ and $(x, 0)=13$

$\begin{aligned}&\Rightarrow \sqrt{(x-0)^{2}+(0+5)^{2}}=13 \\&\Rightarrow \sqrt{x^{2}+25}=13\end{aligned}$

squaring both sides.

$\begin{aligned}&x^{2}+25=169 \Rightarrow x^{2}=169-25-144 \\&\Rightarrow x^{2}=1 \pm 121^{2} \\&\text { So } x=\pm 12 \text { (d) }\end{aligned}$

Question 4

Ans: Distance between origin $(0.0)$ is

(9) $(2,-3)=\sqrt{2^{2}+3^{2}}=\sqrt{4+9} \sqrt{13}$

(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 10 $(0,0)$

option (C) is correct

Question 5

Ans: The coordinates of the each of a side ay square are $(4,-3)$ and $(-1,-5)$

so length of side $=\sqrt{(-1-4)^{2}+(-5+3)^{2}}$.

$\begin{aligned}&=\sqrt{(-5)^{2}+(-2)^{2}} \\&=\sqrt{25+4}=\sqrt{29} \\&\text { so Area of square }=\text { (side }^{2} \\&\left.=(\sqrt{29})^{2} \text { sq units }(d)\right)\end{aligned}$

Question 7

Ans: $2 x-y-1=0$ and $2 x+y=9$ 

$\Rightarrow x=\frac{y+1}{2}$

Giving some different values of $y$, we ret corresponding values of $x$ as given below.

$\begin{array}{|c|c|c|c|}\hline x & 1 & 2 & 0 \\\hline y & 1 & 3 & -1 \\\hline\end{array}$

Plot the points $(1,1)(2,3)$ and $(0,-1)$ on the graph and Join them to get a line.

Similarly $2 x+y=y$

$y=y-2 x$

$\begin{array}{|c|c|c|c|}\hline x & 4 & 5 & 6 \\\hline y & 1 & -1 & -3 \\\hline\end{array}$

Plot the points (4,1)$1(5,-1)$ and $(6,-3)$ on the graph and Join them to get another line.

These two lines intersect each other at point

$\begin{aligned}&\left(\frac{5}{2}, 4\right) \\&\text { so } x=\frac{5}{2}, y=4\end{aligned}$

(IMAGE TO BE ADDED)

Question 8

Ans: 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 $x=-y$

$\Rightarrow x+y=0$ (b)

Question 9

Ans: The distance between the point $(0,0)$ and point af intersection of $x=3$ and $y=4$

i.e. $\left(3,4)\right.$ will be $=\sqrt{(3-0)^{2}+(4-0)^{2}}$

$\begin{aligned}&=\sqrt{3^{2}+4^{2}}=\sqrt{9+16} \\&=\sqrt{25}=5 \text { units }\end{aligned}$

Question 10

Ans: The pointe $(-4,0),(4,0),(0,3)$ are given Now,

$\begin{aligned}A B &=\sqrt{(4+4)^{2}+(0+0)^{2}}=\sqrt{8^{2}+0^{2}} \\&=\sqrt{64}=8 \\B C &=\sqrt{(0-4)^{2}+(3-0)^{2}}=\sqrt{\left.(-4)^{2}+13\right)^{2}} \\&=\sqrt{16+9}=\sqrt{25}=5 \\& A C=\sqrt{10-4)^{2}+(3-0)^{2}}=\sqrt{(-4)^{2}+(3)^{2}} \\&=\sqrt{16+9}=\sqrt{25}=5 \\& \text { So } B C=A C\end{aligned}$

SO the given points are the vertices of an isosceles triangle (b)