what is the area of the triangle formed by the intersection of y=2x-2, y=-1/2x+8 and y =0
OA is 45
what we do when y=0
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- ganeshrkamath
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I'm assuming the second equation is y = (-1/2)x + 8sana.noor wrote:what is the area of the triangle formed by the intersection of y=2x-2, y=-1/2x+8 and y =0
OA is 45
y = 2x - 2
y = 0
2x = 2
x = 1
So point A = (1,0)
Similarly point B = (16, 0)
point C = (4, 6)
AC and BC are perpendicular to each other. (Lines y = 2x - 2 and y = (-1/2)x + 8)
AC = 3 sqrt(5)
BC = 6 sqrt(5)
Area = 1/2 * AC * BC
= 0.5 * 3 sqrt(5) * 6 sqrt(5)
= 9 * 5
= 45 square units
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for the starters who cannot understand that how to reach at point C, here is the explanation.
put the two equations equal to each other
2x-2= (-1/2) =8 now solve for x
x= 4
put the value of x in any equation and one will get y= 6.
Now draw the diagram and find the area
put the two equations equal to each other
2x-2= (-1/2) =8 now solve for x
x= 4
put the value of x in any equation and one will get y= 6.
Now draw the diagram and find the area
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