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Find the equation of the line through the point of intersection of lines, 4x+3y=8; and x+y=1 and parallel to the line 5x-7y=3

written 5.2 years ago by teamques10 ★ 68k

modified 2.6 years ago by pedsangini276 • 4.8k

engineering mathematics

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written 5.2 years ago by teamques10 ★ 68k

Solution:

4x+3y=8

Therefore, x+y=1

4x+3y=8

-4x+4y=4

-y = 4

y = -4

x - 4 = 1

x = 5

therefore, point of intersection = (5,-4)

Slope of the line 5x-7y=3 is,

$m = -\frac{a}{b} = -\frac{5}{-7} = \frac{5}{7}$

Slope of the required line is ,

$m = \frac{5}{7}$

Equation of the required line is,

$y - y_{1} = m(x-x_{1})$

$y+4 = \frac{5}{7}(x-5)$

5x-7y-53=0

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