CBSE Class 11-science Answered
Asked by | 18 Jan, 2014, 11:18: AM
Expert Answer
The equation of the given line is
4x + 7y + 13 = 0 ... (1)
Let Q(a, b) be the image of the point P(-8, 12) wrt line (1).
Then PQ is perpendicular to line (1) and PC = CQ
Equation of the line PC is:
(y - 12) = (7/4)(x + 8)
i.e., 7x - 4y + 104 = 0
Solving (1) and (2), we get,
X = -12 and y = 5
Thus, the coordinates of C are (-12, 5).
Now, C is the mid-point of PQ. Therefore,
-12 = (a - 8)/2 and 5 = (b + 12)2
Thus gives a = -16 and b = -2
Thus, the coordinates of point Q are (-16, -2).
Answered by | 18 Jan, 2014, 03:38: PM
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