# 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

Toolbox:
• Distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
Let $y=mx+c$ be the line through the point A(-1, 2)
$\therefore 2 = m(-1)+c$
$\Rightarrow 2 = -m+c$
or $c=m+2$
Hence $y=mx+m+2$------(1)
Equation of the given line is
$x+y=4$------(2)
Let us solve equation (1) and (2) for $x$ and $y$.
$x + mx+m+2=4$
$\Rightarrow x(1+m)=2-m$
$\therefore x = \large\frac{2-m}{1+m}$
also $y = \large\frac{5m+2}{1+m}$
Hence $B \bigg( \large\frac{2-m}{1+m}$$, \large\frac{5m+2}{1+m} \bigg) is the point of intersection. It is given that the distance between the two points A and B is 3 units. \therefore \overline{AB} = \sqrt{ \bigg( \large\frac{2-m}{1+m}+1 \bigg)^2 + \bigg( \large\frac{5m+2}{1+m}-2 \bigg)^2} \Rightarrow 3 = \sqrt{ \bigg[ \large\frac{2-m+1+m}{1+m} \bigg]^2+ \bigg[ \large\frac{5m+2-2-2m}{1+m} \bigg]^2} Squaring on both sides we get, 9= \large\frac{9}{(1+m)^2}$$ + \large\frac{9m^2}{(1+m)^2}$