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Find the equation of the straight line passing through the intersectionof the lines `x-2y=1`and `x+3y=2`and parallel to `3x+4y=0.` |
Answer» 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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