# If the intercept of a line between the coordinate axes is divided by the point (–5 4) in the ratio 1 : 2, then find the equation of the line.

$\begin {array} {1 1} (A)\;8x-5y-60=0 & \quad (B)\;8x+5y-60=0 \\ (C)\;5x-8y-60=0 & \quad (D)\;5x+8y-60=0 \end {array}$

Toolbox:
• Section formula : The coordinates of the point which divides the line joining the points $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m : n$ is $\bigg( \large\frac{mx_2+nx_1}{m+n}$$, \large\frac{my_2+ny_1}{m+n} \bigg) Step 1 : Let the equation of the line be \large\frac{x}{a}$$+\large\frac{y}{b}$$=1. The line meets the coordinate axes at A(a, 0) and B(0, b) respectively. The coordinates of the point which divides the line joining A(a,0) and B(0, b) in the ratio 1:2 are \bigg( \large\frac{1 \times 0 + 2 \times a }{1+2}$$, \large\frac{1 \times b + 2 \times 0}{1+2} \bigg)$
$\Rightarrow \bigg( \large\frac{2a}{3}$$, \large\frac{b}{3} \bigg) It is given that the point (-5, 4) divides AB in the ratio 1 : 2 \therefore \large\frac{2a}{3}$$=-5 \: and \: \large\frac{b}{3}$$=4 (i.e) a = -\large\frac{15}{2} and b = 12 Step 2 : Hence the equation of the required line is \large\frac{x}{-\Large\frac{15}{2}}$$+ \large\frac{y}{12}$$=1 \Rightarrow \large\frac{2x}{-15}$$+\large\frac{y}{12}$$=1$
$\Rightarrow 24x+15y=-180$
or $8x+5y-60=0$
would the equation of the straight line be correct if i use A(0,b) & B(a,0) as points of intercept in y and x axis where the point divides the line in 1:2 ratio? if not correct then why is it not correct?