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The straight line L_(1) passes through the points with coordinates (4,6) and (12,2) The straight line L_(2) passes through the origin and has gradient -3 The lines L_(1) and L_(2) intersect at point P. Find the coordinates of P.

1 Answer

1 vote

Answer:


P=(-(16)/(5) ,(48)/(5) )

Step-by-step explanation:

equation of L_(1):


(y-y1)/(y2-y1) = (x-x1)/(x2-x1)


(y-6)/(2-6) =(x-4)/(12-4)


(y-6)/(-4) =(x-4)/(8)

8(y - 6) = -4(x - 4)

-2(y - 6) = x - 4

-2y + 12 = x - 4

x + 2y = 16 ... [1]

equation of L_(2):


y-y1=m(x-x1)

y - 0 = -3(x - 0)

y = -3x

3x + y = 0 ... [2]

[1] & [2]

x + 2y = 16 ⇔ x + 2y = 16

3x + y = 0 ⇔ 6x + 2y = 0

------------------ (-)

-5x = 16

x =
-(16)/(5)

[2]

3x + y = 0


3(-(16)/(5) ) + y =0


y=(48)/(5)

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