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Question

Find the value of k so that the line 2kx+5y6=0 may be parallel to x+3ky1=0.

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Solution

Given:2kx+5y6=0 .......(1)
Slope of line (1)=coefficientofxcoefficientofy=2k5
Given:x+3ky1=0 .......(2)
Slope of line (2)=coefficientofxcoefficientofy=13k
As the lines (1) and (2) are parallel,2k5=13k
6k2=5
k2=56
k=±56

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