CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
p=2,2
loader
C
p=1,3
loader
D
None of these
loader

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 perpendicular

So,
$$m_1\ m_2=-1$$

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
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image