Question

Write down the equation of a line whose slope is $\frac{3}{2}$ and which passes through point $P$, where $P$ divides the line segment $AB$ joining $A(-2,6)$ and $B(3,-4)$ in the ratio $2:3$

Answer 1

Here, $A\equiv (-2,6)=(x_1,y_1)$
$B\equiv (3,-4)=(x_2,y_2)$
$P$ divides $segAB$ internally in the ratio $2:3$
$\therefore m:n=2:3$
Let $P\equiv (x,y)$
We have,
$x=\frac{m{x}_2+n{x}_1}{m+n}=\frac{2\left(3\right)+3(-2)}{2+3}=\frac{0}{5}=0$
and $y=\frac{m{y}_2+n{y}_1}{m+n}=\frac{2(-4)+3\left(6\right)}{2+3}=\frac{10}{5}=2$
$\therefore P\equiv (0,2)$
Now, the equation of the line having slope $\frac{3}{2}$ and passing through $P(0,2)$ is
$y-2=\frac{3}{2}(x-0)$
i.e. $2y-4=3x$
$3x-2y+4=0$
Thus, the equation of the required line is
$3x-2y+4=0$.