Find the equation of a line parallel to 2x - | vedclass

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Question

(A) Let the equation of the required line be $2x - 3y = k$,where $k$ is a constant.The $x$-intercept is found by setting $y = 0$,so $2x = k \Rightarro

Vedclass · Accepted Answer

$2x - 3y = 12$

Vedclass · Answer

(A) Let the equation of the required line be $2x - 3y = k$,where $k$ is a constant.The $x$-intercept is found by setting $y = 0$,so $2x = k \Rightarrow x = k/2$.The $y$-intercept is found by setting $x = 0$,so $-3y = k \Rightarrow y = -k/3$.The area of the triangle formed with the axes is given by $\frac{1}{2} 	imes |	ext{base}| 	imes |	ext{height}| = 12$.$\frac{1}{2} 	imes |\frac{k}{2}| 	imes |-\frac{k}{3}| = 12$.$\frac{k^2}{12} = 12$ $\Rightarrow k^2 = 144$ $\Rightarrow k = \pm 12$.Thus,the equations are $2x - 3y = 12$ or $2x - 3y = -12$.

Find the equation of a line parallel to $2x - 3y = 4$ which forms a triangle of area $12$ square units with the coordinate axes.

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