What is the area of triangle whose vertices are p(-1,4), q(3,2), and r(3, -4)?

Area of triangle = 1/2 x base x height.
Let line PQ be the base, then the height is a line passing through R and perpendicular to line PQ.

PQ = sqrt((3 - (-1))^2 + (2 - 4)^2) = sqrt(4^2 + (-2)^2) = sqrt(16 + 4) = sqrt(20) = 2sqrt(5)
Equation of line PQ => y - 4 = (2 - 4)/(3 - (-1)) (x - (-1))
y - 4 = -1/2(x + 1)
y = -1/2x + 7/2
A line perpendicular to PQ will have a slope of 2 and since it passes through R(3, -4), the equation is y - (-4) = 2(x - 3)
y + 4 = 2x - 6
y = 2x - 10
The line will meet PQ at the point where -1/2x + 7/2 = 2x - 10
5/2x = 27/2
x = 27/5
y = 2(27/5) - 10 = 54/5 - 10 = 4/5
The line will meet PQ at the point (27/5, 4/5)
Length of the line joining point (27/5, 4/5) and (3, -4) is sqrt((3 - 27/5)^2 + (-4 - 4/5)^2) = sqrt(144/25 + 576/25) = sqrt(144/5) = 12/5 sqrt(5)

Therefore, base = 2sqrt(5) and height = 12/5 sqrt(5)
Area = 1/2 x 2sqrt(5) x 12/5 sqrt(5) = 12 square units.

Just divide the x axis (the top) by 2 or multiply the y axis (the bottom) by 2