InterviewSolution
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InterviewSolution
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The point `A (2, 1)` is shifted by `3sqrt2` unit distance parallel to the line `x + y = 1` in the direction of increasing ordinate to reach a point B. Find the image of B by the line `x+y= 1`.A. (5,-2)B. (-3,2)C. (5,4)D. (-1,4) |
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Answer» The equation of a line passing through (2,1) and parallel to x + y = 1 is `(x - 2)/(cos 3pi //4) = (y-1)/(sin pi//4) or , (x -1)/(-1 //sqrt2) = (y-1)/(1//sqrt2)` The coordinates of points on this line which are at a distance `3sqrt2` from (2,1) are given by `(x-2)/(-1//sqrt2) = (y-1)/(1//sqrt2) = pm 3 //sqrt2 implies x = 2 pm 3 , y = 1 pm 3` But the ordinate of Q is more than that of P (2,1) . So , the coordinates of Q are (-1,4). The image of Q(-1,4) in the line x + y = 1 is given by `(x +1)/(1) = (y-4)/(1) = -2((-1 + 4 -1)/(1^(2) + 1^(2)))` `implies (x +1)/(1) = (y-4)/(1) = -2 implies x = -3 , y =2` Thus , the coordinates of the required point are (-3,2) |
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