A trapezium ABCD, AB parallel to DC, BC is perpendicular to both AB and DC. A (4, 2), B (16, 8), C (13,14), D(5, 10). Lines AD and BC produced to meet at P. Find the coordinates of P.
OK, let's find an equation for both lines AD and BC.....then, we can find their intersection.
Equation of line AB
Slope = (10-2) /(5-4) = 8
So, writing in point/-slope form, we have
y - 2 = 8(x -4)
y - 2 = 8x - 32
y = 8x - 30
Equation of line BC
Slope = (14-8)/(13-16) = 6 / -3 = -2
So, writing in point-slope form, we have
y - 8 = -2(x - 16)
y - 8 - -2x + 32
y = -2x + 40
Setting these equal and solving for x, we have
8x - 30 = -2x + 40
10x = 70
x = 7
To find the y coordinate of the intersection point, "plug" the x value into one of the two equations. Using the first equation, we have
y = 8(7) -30
y = 56 -30
y = 26
Then point "P" = (7, 26)
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