Our specific example, given the lines 2x - 3y + 13 = 0 and x + y - 1 = 0, we know that the line that passes through their point of intersection must be given by:

a) (2x - 3y + 13) + k(x + y - 1) = 0

b) (2x - 3y + 13) + k(x + y - 1) = (2 + k)x - (3 - k)y + (13 - k) = 0 from which we can easily form an expression for the gradient as m = (2 + k)/(3 - k)

Since it must be parallel to: 4x+3y-1=0

then m = -4/3

So,

m = (2+k)/(3 - k)

-4/3 = (2 + k)/(3 - k)

k = 18

Sub k value into (2x - 3y + 13) + k(x + y - 1)=0

20x + 15y - 5 = 0

Simplify 20x + 15y - 5 = 0

20x + 15y - 5 = 0

4x + 3y - 1 = 0

Hence, l3

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