A line with slope equal to $1$ and a line with slope equal to $2$ intersect at the point $P(1,6),$ What is the area of $\triangle PQR?$
+23246 The line RP is steeper than the line QP, so the slope of RP must be 2.
Using the point-slope form for a line, the equation of RP = y - 6 = 2(x - 1) ---> y - 6 = 2x - 2
---> y = 2x + 4
---> when y = 0: 0 = 2x + 4 ---> -4 = 2x ---> x = -2
The slope of PR is 1 ---> the equation of PR = y - 6 = 1(x - 1) ---> y - 6 = x - 1
---> y = x + 5
---> when y = 0: 0 = x + 5 ---> x = -5
The distance from Q to R is 3.
The base of triangle(PQR) = 3; its height = 6 ---> area = ½·3·6 = 9
