Question

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 `7/5`

Answer 1

Given equations are x – y + 1 = 0   ......(i)

And 2x – 3y + 5 = 0   ......(ii)

Solving equation (i) and equation (ii) we get

   2x – 2y + 2 = 0
   2x – 3y + 5 = 0
(–)    (+)    (–)        
            y – 3  = 0

∴ y = 3

From equation (i) we have

x – 3 + 1 = 0

⇒ x = 2

So, (2, 3) is the point of intersection of equation (i) and equation (ii).

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 + 3 – 2m = 0

Since, the perpendicular distance from (3, 2) to the line is `7/5`

Then `7/5 = |(m(3) - 2 + 3 - 2m)/sqrt(m^2 + 1)|`

Since, the perpendicular distance from (3, 2) to the line is `49/25 = ((3m - 2 + 3 - 2m)^2)/(m^2 + 1)`

⇒ `49/25 = (m + 1)^2/(m^2 + 1)`

⇒ 49m² + 49 = 25m² + 50m + 25

⇒ 49m² – 25m² – 50m + 49 – 25 = 0

⇒ 24m² – 50m + 24 = 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 and 4m – 3 = 0

∴ m = `4/3,3/4`

Equation of the line taking m = `4/3` is 

y – 3 = `4/3(x - 2)`

⇒ 3y – 9 = 4x – 8

⇒  4x – 3y + 1 = 0

Equation of the line taking m = `3/4` is 

y – 3 = `3/4(x - 2)`

⇒ 4y – 12 = 3x – 6

⇒ 3x – 4y + 6 = 0

Hence, the required equations are 4x – 3y + 1 = 0 and 3x – 4y + 6 = 0