Question

A line passes through the point $(3,4)$ and the point of intersection of the lines $\frac{x}{3}+\frac{y}{4}=1$ and $\frac{x}{4}+\frac{y}{3}=1$ and makes intercepts of lengths $a$ and $b$ on the coordinate axes then

• A. $a+b=\frac{25}{12}$
• B. $|a-b|=\frac{7}{12}$
• C. $ab=-1$
• D. $ab=9$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $a+b=\frac{25}{12}$
B $|a-b|=\frac{7}{12}$
C $ab=-1$

Given line

$\frac{x}{3}+\frac{y}{4}=1$

$4x+3y=12$----(1)

$\frac{x}{4}+\frac{y}{3}=1$

$3x+4y=12$----(2)

On solving equation (1) and (2) we get

$x=\frac{12}{7},y=\frac{12}{7}$

Point of intersection of given lines $(\frac{12}{7},\frac{12}{7})$

Line from Points $(3,4)$ and $(\frac{12}{7},\frac{12}{7})$

$y-4=\frac{\frac{12}{7}-4}{\frac{12}{7}-3}(x-3)$

$y-4=\frac{16}{9}(x-3)$

$9y-36=16x-48$

$16x-9y=12$

$\frac{x}{\frac{3}{4}}+\frac{y}{-\frac{4}{3}}=1$

On comparing above eq with intercept form of a line

$a=\frac{3}{4},b=-\frac{4}{3}$

$|a+b|=\frac{3}{4}-\frac{4}{3}=\frac{7}{12}$

$|a-b|=|\frac{3}{4}+\frac{4}{3}|=\frac{25}{12}$

$ab=\frac{3}{4}\times -\frac{4}{3}=-1$