Question

Find the direction in which a straight line must be drawn through the point $(1,2)$, so that its point of intersection with the line $x+y$ $=$ $4$ may be at a distance of $3$ units from this point.

Answer 1

Let $y$ $=mx+c$ be the line through point $(-1,2)$

Accordingly, $2$ $=m(-1)+c$

$\Rightarrow 2=-m+c$

$\Rightarrow c=m+2$

$\therefore y=mx+m+2$....(1)

The given line is

$x+y=4$....(2)

On solving equation (1) and (2) we obtain

$x=\frac{2-m}{m+1}\text{and}y=\frac{5m+2}{m+1}$

$\therefore \frac{2-m}{m+1},y=\frac{5m+2}{m+1}$ is the point of intersection of lines (1) and (2)

Since this point is at a distance of $3$ units from point $(-1,2)$, according to distance formula,

$\sqrt{{\{\frac{2-m}{m+1}+1\}}^2+{\{\frac{5m+2}{m+1}-2\}}^2}=3$

$\Rightarrow {\left\{\frac{2-m+m+1}{m+1}\right\}}^2+{\left\{\frac{5m+2-2m-2}{m+1}\right\}}^2=3^2$

$\Rightarrow \frac{9}{(m+1{)}^2}+\frac{9m^2}{(m+1{)}^2}=9$

$\Rightarrow \frac{1+m^2}{(m+1{)}^2}=1$

$\Rightarrow 1+m^2=m^2+1+2m$

$\Rightarrow 2m=0\Rightarrow m=0$

Thus, the slope of the required line must be zero i.e., the line must be parallel to the $x$-axis.