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Question
Find the equation of the line passing through the point of intersection of 2x + y = 5 and x + 3y + 8 = 0 and parallel to the line 3x + 4y = 7.
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Solution
Given that: 2x + y = 5 .....(i)
x + 3y + 8 = 0 ......(ii)
3x + 4y = 7 ......(iii)
Equation of any line passing through the point of intersection of equation (i) and equation (ii) is
(2x + y – 5) + λ(x + 3y + 8) = 0 ......(iv) (λ = constant)
⇒ 2x + y – 5 + λx + 3λy + 8λ = 0
⇒ (2 + λ)x + (1 + 3λ)y – 5 + 8λ = 0
Slope of line m1 (say) = `(-(2 + lambda))/(1 + 3lambda)` .....`[because "m" = (-"a")/"b"]`
Now slope of line 3x + 4y = 7 is m2 (say) = `- 3/4`
If equation (iii) is parallel to equation (iv) then
m1 = m2
⇒ `(-(2 + lambda))/(1 + 3lambda) = - 3/4`
⇒ `(2 + lambda)/(1 + 3lambda) = 3/4`
⇒ 8 + 4λ = 3 + 9λ
⇒ 9λ – 4λ = 5
⇒ 5λ = 5
⇒ λ = 1
On putting the value of λ in equation (iv) we get
(2x + y – 5) + 1(x + 3y + 8) = 0
⇒ 2x + y – 5 + x + 3y + 8 = 0
⇒ 3x + 4y + 3 = 0
Hence, the required equation is 3x + 4y + 3 = 0.
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