Question

The sum of intercepts made by a line passing through point (4, 3) on coordinate axes is -1. Its equation is -

• A. $\frac{x}{2}+\frac{y}{3}=1$ or $\frac{x}{2}+\frac{y}{1}=1$
• B. $\frac{x}{2}-\frac{y}{3}=-1$ or $\frac{x}{-2}+\frac{y}{1}=-1$
• C. $\frac{x}{2}+\frac{y}{3}=-1$ or $\frac{x}{-2}+\frac{y}{1}=-1$
• D. $\frac{x}{2}-\frac{y}{3}=1$ or $\frac{x}{-2}+\frac{y}{1}=1$

Answer 1

Answer: (D)

$\frac{x}{2}-\frac{y}{3}=1$ or $\frac{x}{-2}+\frac{y}{1}=1$

$Formulaused$
$slope\mathrm{int}erceptform:y=mx+c$
$equation\text{of line passing through two points:}$
${\text{y-y}}_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$
$Given$
$linepas\sin gthrough(3,4)$
$letthey\mathrm{int}eceptbej:(0,j)$
$x\mathrm{int}erceptbek:(k,0)$
$sumof\mathrm{int}ercepts:$
$j+k=-1$
$\Rightarrow k=-j-1.....\left(i\right)$
$slope=\frac{-j}{k}$
$u\sin gslope\mathrm{int}erceptform$
$equationofline$
$y=\frac{-j}{k}x+j$
$\Rightarrow y-j=\frac{-j}{-j-1}xfrom\left(i\right)$
$\Rightarrow y-j=\frac{jx}{j+1}$
$linepassesthrough(3,4)$
$4-j=\frac{3j}{j+1}$
$\Rightarrow (4-j)(j+1)=3j$
$\Rightarrow -j^2+3j+4=3j$
$\Rightarrow j^2=4$
$j=\pm 2$
$pluggingin\left(i\right)$
$k=-3,1$
$Hencepo\mathrm{int}sare:(2,0),(0,-3)and(-2,0),(0,1)$
$equationoflinepas\sin gthrough(2,0),(0,-3)$
$y-0=\frac{3}{2}(x-2)$
$\Rightarrow 2y=3x-6$
$\Rightarrow 3x-2y=6$

$\Rightarrow \frac{x}{2}-\frac{y}{3}=1$
$equationoflinepas\sin gthrough(-2,0),(0,1)$
$y-0=\frac{1}{2}(x+2)$
$\Rightarrow 2y=x+2$
$\Rightarrow -x+2y=2$
$\Rightarrow \frac{x}{-2}+\frac{y}{1}=1$
So, $\text{D}$ is the correct option.