The graphs of 3x+2y=1, y=2, and 3x - 4y=-29 contain the sides of a triangle. Find the measure of the area of the triangle.

Answer 1

Eq1: 3X + 2Y = 1.
Eq2: 3x - 4y = -29.
Eq3: Y = 2. This is a hor. line.

IN Eq1, substitute 2 for y:
3X + 2*2 = 1,
X = -1.
Solution: (X,Y) = (-1,2). = An end point of line3.

In Eq2: substitute 2 for Y:
3X - 4*2 = -29,
X = -7.
Solution: (x,y) = (-7,2).

Solve Eq1 and Eq2 using Elimination method and get:
Solution: (X,Y) = (-3,5).

Therefore the 3 vertices are:
A(-7,2), B(-1,2), C(-3,5).

tanA = (5-2) / (-3-(-7)) = 0.75,
A = 36.9deg.

(AC)^2 = (-3-(-7))^2 + (5-2)^2 = 25,
AC = 5.

Base = AB = -1 -(-7) = 6.

h = AC*sinA = 5*sin36.9 = = 3.

Area = bh/2 = 6 * 3 / 2 = 9 sq. Units.