Question

Match the entries of Column I with column II. under the following conditions:

Answer 1

(a) Solving, $\frac{9-3x}{4}=mx+1$ $\therefore x=\frac{5}{3+4m}$
As $x$ is an integer $\therefore 3+4m=\pm$ or $\pm 5$ or $4m=-3\pm 1$ or $-3\pm 5$ or $-2,-4,2,-8$.
Since $m$ is an integer
$\therefore 4m=-4,-8$ or $m=-1,-2$
$\therefore \left(a\right)\to (p,q)$
(b) $A(3,0),B(0,-2)$ $\therefore (a,b)=(3,-2)$
The line $y=mx+m^2$ passes through $(3,-2)$
$\therefore -2=3m+m^2$ or $m^2+3m+2=0$
or $(m+2)(m+1)=0$ $\therefore m=-1,-2$
$\therefore \left(b\right)\to (p,q)$
(c) Squaring and adding, we have $x^2+y^2=m^2+n^2$
Point (3, 4) lies on it $therefore{m}^2+n^2=25$
Also $3m-4n=0$
Solving, $m^2+{\left(\frac{3m}{4}\right)}^2=25$ or $25m^2=25\times 16$
$\therefore m^2=16$ or $m=\pm 4$
$\therefore \left(c\right)\to (r,s)$
(d) When $y=2,4x=4$ $x=1$ $\therefore (1,2)$
$y=mx-\frac{8}{m}$ passes through $(1,2)$.
$\therefore 2=m=\frac{8}{m}$ or $m^2-2m-8=0$ or $(m-4)(m+2)=0$
$\therefore m=4,-2$ $\therefore \left(d\right)\to (q,r)$