Question

Find the value of p for which the lines $2x+3y-7=0$ and $4y-px-12=0$ are perpendicular to each other. (3 marks)

Answer 1

Given, $2x+3y-7=0$
$\Rightarrow -3y=2x-7$
$\Rightarrow y=-\frac{2}{3}x+\frac{7}{3}$
Comparing with the slope intercept form of line $y=m_{1}x+c$, we get,
$\therefore$ Slope, $m_1=-\frac{2}{3}$ (1 mark)

Now, $4y-px-12=0$ [Given]
$\Rightarrow 4y=px+12$
$\Rightarrow y=\frac{px}{4}+\frac{12}{4}$
Comparing with the slope intercept form of line $y=m_{2}x+c$, we get,
$\therefore$ Slope, $m_2=\frac{p}{4}$ (1 mark)

Since both the lines are perpendicular to each other,
$\therefore m_1{m}_2=-1$
$\Rightarrow -\frac{2}{3}\times \frac{p}{4}=-1\Rightarrow -2p=-12$
$\Rightarrow p=6$ (1 mark)