A line with slope equal to 1 and a line with slope equal to 3 intersect at the point P(1,6) What is the area of triangle PQR?
+128473 QP must lie on the line with a slope of 1
Call Q = (x1 , 0)
So
(6 - 0 ) / ( 1 - x1) = 1
6 = 1 - x1
x1 = 1 - 6
x1 = -5
So Q =( -5,0)
Likewiae RP must lie on the line with a slope of 3
Call R = (x2 , 0)
So
(6-0) / (1 - x2) = 3
6 = 3 ( 1 - x2)
2 = 1 - x^2
x2 = 1 - 2
x2 = -1
So R =( -1, 0)
So QR = 4= base of PQR
And the height of the triangle = 6
Area = (6 * 4) / 2 = 12
