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

The correct option is D $\frac{x}{2}-\frac{y}{3}=1$ or $\frac{x}{-2}+\frac{y}{1}=1$

Line is $\frac{x}{a}+\frac{y}{b}=1$

The line passing through the point $(4,3)\equiv (x,y)$

$\Rightarrow \frac{4}{a}+\frac{3}{b}=1$

the sum of intercept is $-1$

$a+b=-1$

$\Rightarrow \frac{4}{a}+\frac{3}{b}=1$

$\Rightarrow \frac{4b+3a}{ab}=1$

$\Rightarrow 3a+3b+b=ab$

$\Rightarrow 3(a+b)+b=ab$

$\Rightarrow 3(-1)+b=ab$

$\Rightarrow -3=ab-b$

$\Rightarrow -3=b(-1-b)$

$\Rightarrow -3+b=b-b^2$

$\Rightarrow b^2+2b-3=0$

$\Rightarrow b^2+3b-b-3=0$

$\Rightarrow b(b+3)-1(b+3)=0$

$\Rightarrow (b-1)(b+3)=0$

$b-1=0$ $b+3=0$

$b=1$ $b=-3$

point is $(-2,1)(2,-3)$

Line is

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

$\therefore$ so the answer is D.