If a line L passes through the point of inter | vedclass

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(D) First,find the point of intersection of the lines $2x + 3y + 1 = 0$ and $x + y - 3 = 0$. Solving the system: $x + y = 3 \Rightarrow y = 3 - x$. Su

Vedclass · Accepted Answer

$\frac{41}{6}$

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(D) First,find the point of intersection of the lines $2x + 3y + 1 = 0$ and $x + y - 3 = 0$. Solving the system: $x + y = 3 \Rightarrow y = 3 - x$. Substitute into the first equation: $2x + 3(3 - x) + 1 = 0$ $\Rightarrow 2x + 9 - 3x + 1 = 0$ $\Rightarrow -x + 10 = 0$ $\Rightarrow x = 10$. Then $y = 3 - 10 = -7$. The point of intersection is $(10, -7)$. The slope of the line $L$ is $m = 	an(	an^{-1} \frac{2}{3}) = \frac{2}{3}$. The equation of the line passing through $(10, -7)$ with slope $\frac{2}{3}$ is: $y - (-7) = \frac{2}{3}(x - 10)$ $\Rightarrow 3(y + 7) = 2(x - 10)$ $\Rightarrow 3y + 21 = 2x - 20$ $\Rightarrow 2x - 3y = 41$. To find the intercepts,write the equation in intercept form $\frac{x}{a} + \frac{y}{b} = 1$: $\frac{2x}{41} - \frac{3y}{41} = 1 \Rightarrow \frac{x}{41/2} + \frac{y}{-41/3} = 1$. Thus,$a = \frac{41}{2}$ and $b = -\frac{41}{3}$. The sum of the intercepts is $a + b = \frac{41}{2} - \frac{41}{3} = \frac{123 - 82}{6} = \frac{41}{6}$.

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