InterviewSolution
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InterviewSolution
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The equation of straight line equally inclined to the axes and equidistant from the point `(1, -2)` and `(3,4)` is:A. x + y + 1 = 0B. x + y + 2 = 0C. x - y - 2 = 0D. x -y - 1 = 0 |
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Answer» Correct Answer - D The slope of a line equally inclined with the co-ordinate axes is `pm` 1 . So , let its equation be `x pm y + lambda = 0 " " … (i)` CASEI When ` x + y + lambda = 0 ` is the line : It is equidistant from (1,-2) and (3,4) `therefore |(1-2 + lambda)/(sqrt2)| = |(3 + 4 + lambda)/(sqrt2)|` `implies | lambda -1| = |lambda + 7| implies lambda - 1 = - (lambda + 7) implies lambda = -3` So , the required line is x + y - 3 = 0. CASEII When ` x - y + lambda = 0` is the line : It is equidistant from (1, -2) and (3,4) `therefore |(1+2 + lambda)/(sqrt2)| = |(3 - 4 + lambda)/(sqrt2)|` `implies } lambda + 3| = |lambda - 1| implies lambda + 3 = - (lambda -1) implies lambda = -1` Thus , the required line is x - y - 1 = 0 |
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