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

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written 5.6 years ago by

teamques10 ★ 69k

modified 3.0 years ago by

pedsangini276 • 4.8k

engineering mathematics

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written 5.6 years ago by

teamques10 ★ 69k

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 …

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