Question

Find the values of $p$ for which the straight lines $8px+(2-3p)y+1=0$ and $px+8y-7=0$ are perpendicular to each other.

• A. $p=1,2$
• B. $p=2,2$
• C. $p=1,3$
• D. None of these

Answer 1

Answer: (A)

$p=1,2$

Consider the given equations.

$8px+(2-3p)y+1=0$ $............\left(1\right)$

$Slope{m}_1=-\frac{8p}{2-3p}$

$px+8y-7=0$ $.............\left(2\right)$

$Slope{m}_2=-\frac{p}{8}$

Since, both lines are perpendicular

So,

$m_1{m}_2=-1$

Therefore,

$-\frac{8p}{2-3p}\times -\frac{p}{8}=-1$

$\frac{8p^2}{2-3p}=-8$

$8p^2=-16+24p$

$8p^2-24p+16=0$

$p^2-3p+2=0$

$p^2-2p-p+2=0$

$p(p-2)-1(p-2)=0$

$(p-1)(p-2)=0$

$p=1,2$

Hence, this is the answer.