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 the 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

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

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

Since this point is at a distance of $3$ units from points $(-1,2)$, accordingly 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.