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Find the value(s) of y such that the triangle with the given vertices has an area of 8 square units.
(Enter your answers as a comma-separated list.)
(−4, 5), (−3, 3), (−3, y)
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Let A = (-4,5)
B = (-3,3)
C = (-3,y)
Notice that the side BC is vertical line x= -3 of the length of |y-3| (absolute value).
Also notice that the distance of the point A from this line x= -3 is -3 - (-4) = -3 + 4 = 1 unit.
You may consider the side BC of the length |y-3| units as a base of the triangle ABC,
and consider the distance of the point A from the line x= 3 as the length of the altitude
of this triangle drawn to the base BC.
So, your equation for the triangle area is
= 8,
or
|y - 3| = 16.
This equation has two solutions y= 3+16 = 19 and y= 3-16 = -13.
ANSWER. There are two solutions for y: y= -13 or y= 19.
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