Find the equations of the lines through the point of intersection of the lines x – y + 1 = 0 and 2x – 3y + 5 = 0 and whose distance from the point (3, 2) is .

Given two lines are:

x – y + 1 = 0 …(i)

and 2x – 3y + 5 = 0 …(ii)

Now, point of intersection of these lines can be find out as:

Multiplying eq. (i) by 2, we get

2x – 2y + 2 = 0 …(iii)

On subtracting eq. (iii) from (ii), we get

2x – 2y + 2 – 2x + 3y – 5 = 0

⇒ y – 3 = 0

⇒ y = 3

On putting value of y in eq. (ii), we get

2x – 3(3) + 5 = 0

⇒ 2x – 9 + 5 = 0

⇒ 2x – 4 = 0

⇒ 2x = 4

⇒ x = 2

So, the point of intersection of given two lines is:

(x, y) = (2, 3)

Let m be the slope of the required line

∴ Equation of the line is

y – 3 = m(x – 2)

⇒ y – 3 = mx – 2m

⇒ mx – y – 2m + 3 = 0 …(i)

Since, the perpendicular distance from the point (3, 2) to the line is then

Squaring both the sides, we get

⇒ 49(m² + 1) = 25(m + 1)²

⇒ 49m² + 49 = 25(m² + 1 + 2m)

⇒ 49m² + 49 = 25m² + 25 + 50m

⇒ 25m² + 25 + 50m – 49m² – 49 = 0

⇒ – 24m² + 50m – 24 = 0

⇒ – 12m² + 25m – 12 = 0

⇒ 12m² – 25m + 12 = 0

⇒ 12m² – 16m – 9m + 12 = 0

⇒ 4m (3m – 4) – 3(3m – 4) = 0

⇒ (3m – 4)(4m – 3) = 0

⇒ 3m – 4 = 0 or 4m – 3 = 0

⇒ 3m = 4 or 4m = 3

or

Putting the value of in eq. (i), we get

⇒ 4x – 3y + 1 = 0

Putting the value of in eq. (i), we get

⇒ 3x – 4y + 6 = 0

hence, the required equation are 4x – 3y + 1 = 0 and 3x – 4y + 6 = 0