Question

From the choices given below, choose the equation whose graphs are given in Fig. (i) and Fig. (ii).

For fig. (i)	For fig. (ii)
(i) $y=x$	(i) $y=x+2$
(ii) $x+y=0$	(ii) $y=x-2$
(iii) $y=2x$	(iii) $y=-x+2$
(iv) $2+3y=7x$	(iv) $x+2y=6$

Answer 1

In the given fig. (i), the solutions of the equation are $(-1,1),(0,0)$ and $(1,-1)$.

$\therefore$ the equation which satisfies these solutions is the correct equation.

Equation (ii) $x+y=0$, satisfies these solutions.

Proof:

If we put the value of $x=-1$ and $y=1$ in the equation $x+y=0$

$=x+y=-1+1=0$

$\therefore L.H.S=R.H.S$

if we put the the value of x = 0 and y = 0

$=x+y=0+0=0$

$\therefore L.H.S=R.H.S$

If we put the value of $x=1$ and $y=-1$

$=x+y=1+(-1)=1–1=0$

$L.H.S=R.H.S$

Hence, option (ii) $x+y=0$ is correct.

In the given Fig.(ii) the solutions of the equation are $(-1,3),(0,2)$ and $(2,0)$.

$\therefore$The equation which satisfies these solutions is the correct equation.

Equation (iii) $y=-x+2$ , satisfies these solutions.

Proof:

If we put the value of $x=-1$ and $y=3$ in the equation $y=-x+2$

$y=-x+2$

$3=-(-1)+2$

$3=3$

$\therefore L.H.S=R.H.S$

if we put the the value of $x=0$ and $y=2$

$y=-x+2$

$2=-0+2$

$2=2$

$\therefore L.H.S=R.H.S$

If we put the value of $x=2$ and $y=0$

$y=-x+2$

$0=-2+2$

$0=0$

$L.H.S=R.H.S$

Hence, option (iii) $y=-x+2$ is correct.