If a straight line perpendicular to 2x - 3y + | vedclass

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(B) The given line is $2x - 3y + 7 = 0$.Any line perpendicular to the given line is of the form $3x + 2y + k = 0$.To find the intercepts,set $y = 0$ t

Vedclass · Accepted Answer

$3x + 2y = \pm 6$

Vedclass · Answer

(B) The given line is $2x - 3y + 7 = 0$.Any line perpendicular to the given line is of the form $3x + 2y + k = 0$.To find the intercepts,set $y = 0$ to get $3x = -k$,so $x = -k/3$.Set $x = 0$ to get $2y = -k$,so $y = -k/2$.The area of the triangle formed with the coordinate axes is given by $\frac{1}{2} |x| |y| = 3$.$\frac{1}{2} |-\frac{k}{3}| |-\frac{k}{2}| = 3$.$\frac{k^2}{12} = 3$ $\Rightarrow k^2 = 36$ $\Rightarrow k = \pm 6$.Substituting $k$ back into the equation,we get $3x + 2y = \pm 6$.

If a straight line perpendicular to $2x - 3y + 7 = 0$ forms a triangle with the coordinate axes whose area is $3 \text{ sq. units}$,then the equation of the straight line is:

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