Question: In the figure, point A divides the line segment CD in the ratio of 3:2 and BA _|_ CD. Find the equation of the line AB.

Solution:

Given,




CD is a straight line whose coordinates are C(-4,0) and D(1,-5). CD is divided by point A in the ratio m:n = 3:2.

Let C(-4,0) = ($x_1,x_2$) and D(1,-5) = ($x_2,y_2$)

Using section formula of internal division,
A (x,y) = $\dfrac{mx_2 + nx_1}{m+n} , \dfrac{my_2 + ny_1}{m+n}$

$= \left ( \dfrac{3×1 + 2×(-4)}{3+2}, \dfrac{3×(-5) + 2×0}{3+2} \right )$

$= \left ( \dfrac{3-8}{5}, \dfrac{-15+0}{5} \right )$

$= (-1, -3)$

So, coordinate of point A are (-1,-3)

Also,
Slope of line CD = $\dfrac{y_2-y_1}{x_2-x_1}$
$= \dfrac{-5-0}{1-(-4}$
$= \dfrac{-5}{1+4}$
$= \dfrac{-5}{5}$
$= -1$

We know,
Line AB and Line CD are perpendicular. So, the product of their slopes result negative 1.
$or, m_1 × m_2 = -1$
$or, -1 × m_2 = -1$
$\therefore m_2 = 1$

Now,
Equation of line AB when ($x_1,y_1$) = (-1,-3) and $m_2$ = 1 is given by:
$or, y - y_1 = m_2 (x -x_1)$

$or, y -(-3) = 1 \{x -(-1) \}$

$or, y +3 =  x +1$

$or 0 = x +1 -3 -y$

$or, x - y -2 = 0$ is the required equation of AB.

Therefore, the required equation of the line AB is x - y - 2 = 0.