Asked by  | 18th 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  | 18th Jan, 2014, 03:38: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.