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Question

Find the equation of straight line which passes through the point (2,-3) and the point of intersection of the lines x+y +4 = 0 and 3x-y-8=0


A

x-2y-7 = 0

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B

2x-y-7 = 0

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C

x-2y+7 = 0

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D

2x-y+7 = 0

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Solution

The correct option is B

2x-y-7 = 0


Any line through the intersection of the lines x + y + 4 = 0 and 3x - y - 8 = 0

Has the equation (x + y + 4) + λ (3x - y - 8) = 0-------------(1)

This equation also passes through the point (2,-3)

So, this point should satisfy the equation.

(2+3+4) + λ (6+3-8) = 0

3 + λ = 0

λ = -3

Putting the value of λ in equation 1 required equation is

(x + y + 4) -3(3x-y-8) = 0

x + y + 4 - 9x + 3y + 24 = 0

-8x + 4y + 28 = 0

-2x + y + 7 = 0

2x - y - 7 = 0

Alternative method

We can find the intersection between x+y+4 = 0 and 3x - y - 8 = 0

Adding both the equation 4x - 4 = 0

x = 1

Substituting x in equation (2)

1 + y + 4 = 0

y = -5

so, equation line which passes through (1,-5) and (2,-3)

y + 3 = 5+312(x2)

2x - y - 7 = 0


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