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# Find the equation of the line which passes through the point (– 4, 3) and the portion of the line intercepted between the axes is divided internally in the ratio 5 : 3 by this point

$\begin {array} {1 1} (A)\;9x+20y+96=0 & \quad (B)\;9x-20y+96=0 \\ (C)\;9x-20y-96=0 & \quad (D)\;9x+20y-96=0 \end {array}$

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A)
• 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}$$=1. This line meets the coordinates axes at A(a,0) and B(0,b) respectively. The coordinates of the point which divides the line joining A(a,0) and B(0, b) in the ratio 5:3 are \bigg[ \large\frac{5 \times (0)+3(a)}{5+3}$$, \large\frac{5 \times (b) +3 (0)}{5+3} \bigg]$
$\Rightarrow \bigg( \large\frac{3a}{8}$$, \large\frac{5b}{8} \bigg) Step 2 : It is given that the point (-4, 3) divides AB in the ratio 5 : 3 \therefore \large\frac{3a}{8}$$ = -4 \Rightarrow a = -\large\frac{33}{3}$
and $\large\frac{5b}{8}$$=3 \Rightarrow = \large\frac{24}{5} Hence the equation of the line is \large\frac{x}{\Large\frac{-32}{3}}$$+\large\frac{y}{\Large\frac{24}{5}}$$=1$
(i.e) $9x-20y=-96$
$\quad 9x-20y+96=0$
Hence the equation of the required line is $9x-20y+96=0$