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प्रश्न
Find the equation of a straight line through the intersection of lines 7x + 3y = 10, 5x – 4y = 1 and parallel to the line 13x + 5y + 12 = 0
उत्तर
Given lines are
7x + 3y = 10 ...(1)
5x − 4y = 1 ...(2)
(1) × 4 ⇒ 28x + 12y = 40 ...(3)
(2) × 3 ⇒ 15x − 12y = 3 ...(4)
By adding (3) and (4) ⇒ 43x = 43
x = `43/43` = 1
Substitute the value of x = 1 in (1)
7(1) + 3y = 10
⇒ 3y = 10 – 7
y = `3/3` = 1
The point of intersection is (1, 1)
Equation of the line parallel to 13x + 5y + 12 = 0 is 13x + 5y + k = 0
This line passes through (1, 1)
13(1) + 5(1) + k = 0
13 + 5 + k = 0
⇒ 18 + k = 0
k = – 18
∴ The equation of the line is 13x + 5y – 18 = 0
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