Question

If the line 2x-5y+1=0 is perpendicular to (p+1)x+ (2p+3)y+3=0,find p
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Answer 1

SOLUTION

TO DETERMINE

The value of p when the line 2x-5y+1=0 is perpendicular to (p+1)x+ (2p+3)y+3=0

EVALUATION

Here the given equations of the line are

2x-5y+1=0 - - - - (1)

(p+1)x+ (2p+3)y+3=0 - - - - - (2)

Slope of first line

$\displaystyle \sf{ = \frac{2}{5} }$

Slope of the second line

$\displaystyle \sf{ = - \frac{p + 1}{2p + 3} }$

So by the given condition

$\displaystyle \sf{ - \frac{p + 1}{2p + 3} \times \frac{2}{5} = - 1 }$

$\displaystyle \sf{ \implies - \frac{p + 1}{2p + 3} = - \frac{5}{2} }$

$\displaystyle \sf{ \implies \frac{p + 1}{2p + 3} = \frac{5}{2} }$

$\displaystyle \sf{ \implies 10p + 15 = 2p + 2 }$

$\displaystyle \sf{ \implies 8p = - 13 }$

$\displaystyle \sf{ \implies p = - \frac{13}{8} }$

Hence the required value

$\displaystyle \sf{ p = - \frac{13}{8} }$

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