Question

A straight line passes through the point P(2,1) and cuts the axes in points A, B. If BP: PA = 3:1. Find (i) coordinates of A and B. (ii) the equation of the line AB.​

Answer 1

$\underline{\textbf{Given:}}$

$\textsf{A straight line passes through the point P(2,1) and}$

$\textsf{cuts the axes in points A ,B and BP : PA=3:1}$

$\underline{\textbf{To find:}}$

$\textsf{(i) Co-ordinates of A and B}$

$\textsf{(ii) Equation of the line AB}$

$\underline{\textbf{Solution:}}$

$\textsf{Since the line cuts the co-ordinate axes at A and B,}$

$\textsf{take A and B as (a,0) and (0,b) respectively}$

$\mathsf{BP:PA=3:1\;\implies\;AP:BP=1:3}$

$\implies\textsf{P divides AB internally in the ratio 1:3}$

$\textsf{By Section formula, the co-ordinates of P is}$

$\mathsf{\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)=(2,1)}$

$\mathsf{\left(\dfrac{1(0)+3(a)}{1+3},\dfrac{1(b)+3(0)}{1+3}\right)=(2,1)}$

$\mathsf{\left(\dfrac{3a}{4},\dfrac{b}{4}\right)=(2,1)}$

$\implies\mathsf{\dfrac{3a}{4}=2\;\;\;and\;\;\;\dfrac{b}{4}=1}$

$\implies\mathsf{a=\dfrac{8}{3}\;\;\;and\;\;\;b=4}$

$\therefore\mathsf{A\left(\dfrac{8}{3},0\right)\;\;\;and\;\;\;B(0,4)}$

$\textsf{Using Intercept form, Equation of line AB is}$

$\mathsf{\dfrac{x}{a}+\dfrac{y}{b}=1}$

$\mathsf{\dfrac{x}{\dfrac{8}{3}}+\dfrac{y}{4}=1}$

$\mathsf{\dfrac{3x}{8}+\dfrac{y}{4}=1}$

$\mathsf{\dfrac{3x+2y}{8}=1}$

$\implies\boxed{\mathsf{3x+2y-8=0}}$