Find the equation of the straight line passing through the intersection of the lines `x-2y=1` and `x+3y=2` and parallel to `3x+4y=0.`
Find the equation of the straight line passing through the intersection of the lines `x-2y=1` and `x+3y=2` and parallel to `3x+4y=0.`
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Correct Answer - 3x+4y-5=0
Point of intersection of x-2y=1 and x+3y=2 is (7/5, 1/5)
Any line parallel to 3x+4y=0 is 3x+4y+K=0.
As this line passes through (7/5, 1/5), we get
`(21)/(5) + (4)/(5)+K =0`
or K=-5
Therefore, the required line is
3x+4y-5=0
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