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

Solution

The correct option is C $$p=1,2$$Consider the given equations.$$8px+(2-3p)y+1=0$$               $$............(1)$$$$Slope\ m_1=-\dfrac{8p}{2-3p}$$$$px+8y-7=0$$              $$.............(2)$$$$Slope\ m_2=-\dfrac{p}{8}$$Since, both lines are perpendicularSo,$$m_1\ m_2=-1$$Therefore,$$-\dfrac{8p}{2-3p}\times -\dfrac{p}{8}=-1$$$$\dfrac{8p^2}{2-3p}=-8$$$$8p^2=-16+24p$$$$8p^2-24p+16=0$$$$p^2-3p+2=0$$$$p^2-2p-p+2=0$$$$p(p-2)-1(p-2)=0$$$$(p-1)(p-2)=0$$$$p=1, 2$$Hence, this is the answer.Mathematics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More