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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`
Solution
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
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