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

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- 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$

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...