The equation of a straight line,perpendicular | vedclass

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(B) The given equation of the line is $3x - 4y = 6$.Any line perpendicular to this line is of the form $4x + 3y = k$.The intercepts of this line on th

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$4x + 3y = 12$

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(B) The given equation of the line is $3x - 4y = 6$.Any line perpendicular to this line is of the form $4x + 3y = k$.The intercepts of this line on the coordinate axes are $x = \frac{k}{4}$ and $y = \frac{k}{3}$.The area of the triangle formed by the line with the coordinate axes is given by $	ext{Area} = \frac{1}{2} 	imes |	ext{base}| 	imes |	ext{height}|$.$\frac{1}{2} 	imes |\frac{k}{4}| 	imes |\frac{k}{3}| = 6$.$|\frac{k^2}{24}| = 6$.$k^2 = 144$.$k = \pm 12$.Thus,the required equations of the lines are $4x + 3y = 12$ or $4x + 3y = -12$.

The equation of a straight line,perpendicular to $3x - 4y = 6$ and forming a triangle of area $6 \text{ sq. units}$ with the coordinate axes,is

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