Q11Long answer
= 0. Solution First we find the point of intersection of lines 5x – 6y – 1 = 0 and 3x + 2y + 5 = 0 which is (– 1, – 1). Also the slope of the line 3x – 5y + 11 = 0 is . Therefore, −5 the slope of the line perpendicular to this line is (Why?). Hence, the equation of the required line is given by −5 y+1= (x + 1) or 5x + 3y + 8 = 0 Alternatively The equation of any line through the intersection of lines 5x – 6y – 1 = 0 and 3x + 2y + 5 = 0 is 5x – 6y – 1 + k(3x + 2y + 5) = 0 (1) − (5 + 3k ) or Slope of this line is – 6 + 2k Also, slope of the line 3x – 5y + 11 = 0 is Now, both are perpendicular − (5 + 3k ) 3 so – 6 + 2k × 5 = –1 or k = 45 Therefore, equation of required line in given by 5x – 6y – 1 + 45 (3x + 2y + 5) = 0 or 5x + 3y + 8 = 0