Question

Find the equation of the straight line through the intersection of the lines 5x – 6y – 1=0 and 3x + 2y +5 =0 and perpendicular to the line 3x – 5y +11 =0 1

Answer 1

The two given lines are

5x - 6y = 1 ...(i)

and

3x + 2y + 5 = 0

⇒ 3x + 2y = - 5 ...(ii)

Now, (i) × 3 and (ii) × 5 ⇒

15x - 18y = 3

15x + 10y = - 25

On subtraction, we get

- 18y - 10y = 3 + 25

⇒ - 28y = 28

⇒ y = - 1

Now, putting y = - 1 in (i), we get

5x - 6 (- 1) = 1

⇒ 5x + 6 = 1

⇒ 5x = 1 - 6

⇒ 5x = - 5

⇒ x = - 1

So, the intersection of the lines (i) and (ii)

is (- 1, - 1)

Any perpendicular line to the line 3x - 5y + 11 = 0 is

5x + 3y = k ...(iii)

The point (- 1, - 1) lies on the line (iii). So,

5 (- 1) + 3 (- 1) = k

⇒ - 5 - 3 = k

⇒ k = - 8

Hence, from (iii), we get the required line as

5x + 3y = - 8

i.e., 5x + 3y + 8 = 0