Question

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

Answer 1

Answer:

Hence, the value of p is:

$p=6$

Step-by-step explanation:

The equation of first line is given as:

$2x+3y-7=0$

on converting this equation of line into slope intercept form of a line i.e.

$y=mx+c$

where m represents the slope of the line and c represents it's y-intercept.

⇒ $2x+3y-7=0\\\\3y=-2x+7\\\\y=\dfrac{-2}{3}x+\dfrac{7}{3}$

Hence the slope of line 1 is:

$m_1=\dfrac{-2}{3}$

Also equation of line 2 is:

$4y-px-12=0$

In slope intercept form it could be represented as:

$4y=px+12\\\\y=\dfrac{p}{4}x+\dfrac{12}{4}\\\\\\y=\dfrac{p}{4}x+3$

hence, slope of line 2 is:

$m_2=\dfrac{p}{4}$

for two line to be perpendicular we have the condition that:

$m_1\times m_2=-1$

This means that:

$\dfrac{-2}{3}\times \dfrac{p}{4}=-1\\\\\dfrac{-p}{6}=-1\\\\p=6$

Hence, the value of p such that two line are perpendicular is:

$p=6$