मराठी

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. - Mathematics

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प्रश्न

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.

बेरीज

उत्तर

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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Straight Lines - Equation of Family of Lines Passing Through the Point of Intersection of Two Lines
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पाठ 10: Straight Lines - Exercise [पृष्ठ १७८]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
पाठ 10 Straight Lines
Exercise | Q 8 | पृष्ठ १७८

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