Question

The equation of the straight line passing through the point of intersection of lines 3x – 4y – 7 = 0 and 12x – 5y – 13 = 0 and perpendicular to the line 2x – 3y + 5 = 0 is

• A. 33x + 22y + 13 = 0
• B. 33x + 22y – 13 = 0
• C. 33x – 22y + 13 = 0
• D. None of these

Answer 1

The correct option is A

33x + 22y + 13 = 0

Given lines are

3x – 4y – 7 = 0 …… (1)

12x – 5y – 13 = 0 …… (2)

and 2x – 3y + 5 = 0 …… (3)

equation of any line through the point of intersection of lines (1) and (2) is

(3x – 4y – 7) + k(12x – 5y – 13) = 0

or (3 + 12k)x – (4 + 5k)y – (7 + 13k) = 0 …… (4)

Slope of line (4) $=\frac{3+12k}{4+5k}$

Slope of line (3) $=\frac{2}{3}$

Since line (4) is perpendicular to line (3),

$\therefore \frac{3+12k}{4+5k}\times \frac{2}{3}=-1or6+24k=-12-15k$

or 39k = –18. $\therefore k=-\frac{6}{13}$

putting the value of k in equation (4), the equation of the required line is

$(3-\frac{72}{13})x-(4-\frac{30}{13})y-(7-6)=0$

or –33x – 22y – 13 = 0 or 33x + 22y + 13 = 0