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Question

(A) Let the equation of the line passing through $(2, -3)$ with slope $m$ be $y + 3 = m(x - 2)$.This can be written as $mx - y = 2m + 3$.If $m 
eq 0$

Vedclass · Accepted Answer

$x + y + 1 = 0$ or $3x - 2y = 12$

Vedclass · Answer

(A) Let the equation of the line passing through $(2, -3)$ with slope $m$ be $y + 3 = m(x - 2)$.This can be written as $mx - y = 2m + 3$.If $m 
eq 0$,the intercept form is $\frac{x}{(2m+3)/m} + \frac{y}{-(2m+3)} = 1$.The intercepts are $a = \frac{2m+3}{m}$ and $b = -(2m+3)$.Given $a + b = -2$,we have $\frac{2m+3}{m} - (2m+3) = -2$.$2m + 3 - 2m^2 - 3m = -2m \implies 2m^2 - m - 3 = 0$.Factoring gives $(2m - 3)(m + 1) = 0$,so $m = \frac{3}{2}$ or $m = -1$.For $m = -1$,the equation is $y + 3 = -1(x - 2) \implies x + y + 1 = 0$.For $m = \frac{3}{2}$,the equation is $y + 3 = \frac{3}{2}(x - 2) \implies 2y + 6 = 3x - 6 \implies 3x - 2y = 12$.

Find the equation of the line passing through the point $(2, -3)$ such that the sum of its intercepts on the axes is $-2$.

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