Question

Find the equation of the straight lines each of which passes through the points (3,2) and intersects the x-axis and y-axis in A,B respectively such that OA-OB=2

Answer 1

Let the equation of line be

$(y-y_1)=m(x-x_1)$

Given that the line passes through the point $(3,2)$, i.e., the point must satisfy the equation of line.

$\therefore (y-2)=m(x-3).....\left(1\right)$

The line intersects at $x$-axis and $y$-axis at $A$ and $B$ respectively.

For intersection at $x$-axis, we have

Susbstituting $y=0$ in ${eq}^n\left(1\right)$, we have

$0-2=m(x-3)$

$\Rightarrow x=3-\frac{2}{m}=OA$

Similarly for intersection at $y$-axis, we have

Substituting $x=0$ in ${eq}^n\left(1\right)$, we have

$y-2=m(0-3)$

$\Rightarrow y=2-3m=OB$

Given that $OA-OB=0$

$\therefore (3-\frac{2}{m})-(2-3m)=0$

$\Rightarrow 3-\frac{2}{m}-2+3m=0$

$\Rightarrow 3m-2-2m+3m^2=0$

$\Rightarrow 3m^2+m-2=0$

$\Rightarrow (m+1)(3m-2)=0$

$\Rightarrow m=-1,\frac{2}{3}$

Case I:- For $m=-1$

Substituting the value of $m$ in ${eq}^n\left(1\right)$, we have

$y-2=-1(x-3)$

$\Rightarrow y-2=-x+3$

$\Rightarrow x+y-5=0$

Hence for $m=-1$, the equation of line will be $x+y-5=0$.

Case I:- For $m=\frac{2}{3}$

Substituting the value of $m$ in ${eq}^n\left(1\right)$, we have

$y-2=\frac{2}{3}(x-3)$

$\Rightarrow 3y-6=2x-6$

$\Rightarrow 2x-3y=0$

Hence for $m=\frac{2}{3}$, the equation of line will be $2x-3y=0$.