logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XI  >>  Math  >>  Straight Lines
0 votes

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

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

Can you answer this question?
 
 

1 Answer

0 votes
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}$$=1$. This line meets the coordinates axes at A(a,0) and B(0,b) respectively. The coordinates of the point which divides the line joining A(a,0) and B(0, b) in the ratio 5:3 are
$ \bigg[ \large\frac{5 \times (0)+3(a)}{5+3}$$, \large\frac{5 \times (b) +3 (0)}{5+3} \bigg]$
(i.e) The coordinates are
$ \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$
answered Jul 1, 2014 by thanvigandhi_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...