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Question

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

A

4

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B

-4

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C

6

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D

-6

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Solution

## The correct option is C 6 Given, 2x+3y−7=0 ⇒−3y=2x−7 ⇒y=−23x+73 Comparing with the slope intercept form of line y=m1x+c, we get, ∴ Slope, m1=−23 Now, 4y−px−12=0 [Given] ⇒4y=px+12 ⇒y=px4+124 Comparing with the slope intercept form of line y=m2x+c, we get, ∴ Slope, m2=p4 Since both the lines are perpendicular to each other, ∴m1m2=−1 ⇒−23×p4=−1⇒−2p=−12 ⇒p=6

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