The point (a, b) is the foot of the perpendicular drawn from the point (3,1) to the line x+3 y+4=0 . If (p, q) is the image of (a, b) with respect to the line 3 x-4 y+11= 0 , then p a + q b =

Step-by-step solution

Since, (a, b) is the foot of perpendicular drawn from (3,1) to the line x+3 y+4=0 . a+3 b+4=0...(i) Now, equation of a line passing through (3,1) and perpendicular to the given line is 3 x-y-8=0 so 3 a-b-8=0 ...(ii) After solving (i) and (ii), we get (a, b)=(2,-2) Let P= ( 2+p 2 , -2+q 2 ) Since, P lies on 3 x-4 y+11=0 3(2+p) 2 - 4(-2+q) 2 +11=0 3 p-4 q+36=0...(iii) Since slope of line 3 x-4 y+11=0 is 3 4 Now, 3 4 ( -2-q 2-p )=-1 2+q 2-p = 4 3 4 p+3 q-2=0....(iv) After solving (iii) and (iv) we get, p=-4, q=6 Now,