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Find the general equation of the straight line.?

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Find the general equation of the straight line passing through the origin and the point intersections of the lines:

5x-2y+3=0 and 3x+7y-1=0
asked Nov 2, 2013 in ALGEBRA 1 by mathgirl Apprentice

1 Answer

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First find the intersection point of the lines.

Multiply the equation 1 by 3 and second equation by 5 and subtract from each other

15x - 6y + 9 = 0
15x + 35y - 5= 0
----------------------
-41y + 14 = 0
-41y = -14

divide each side by -41
y = 14/41

substitute y = 14/41 in 5x-2y+3=0

5x -2 * (14/41) = -3

5x - 28/41 = -3

5x = 28/41 - 3 = -95/41

x = -19/41

(x, y) = (-19/41, 14/41)

Line passing through the origin so the line equation is y = mx and m is slope

find the slope of the line using the formula m = (y2 - y1) / (x2 - x1)

m = (14/41 - 0) / (-19/41 - 0)

m = -14/19

the line equation is y = -14/19x

general equation is 14x + 19 y = 0.

answered Nov 2, 2013 by steve Scholar

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