Pearson Geometry Common Core, 2011
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Pearson Geometry Common Core, 2011 View details
1. Areas of Parallelograms and Triangles
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Exercise 35 Page 621

Practice makes perfect
a We want to graph the three lines on the same coordinate plane. Let's start with the vertical and horizontal lines.
Vertical Line:& x=4 Horizontal Line:& y=- 2 The vertical line intercepts the x-axis at 4, and the horizontal line intercepts the y-axis at - 2.

Let's now draw the graph of the third line. Note that this line is already written in slope-intercept form. Therefore, we can identify the slope and the y-intercept. y=3/4x-2 ⇕ y=3/4x+(- 2) Let's use the y-intercept - 2 and the slope 34 to graph this line.

b We will now find the area of the triangle enclosed by the lines. Note that the base has a length of 4 units and the height has a length of 3 units.
We can substitute these values into the formula for the area of a triangle and simplify. Let's do it!
A=1/2bh
A=1/2( 4)( 3)
â–Ľ
Evaluate right-hand side
A=1/2(12)
A=12/2
A=6
The area of the triangle enclosed by the given lines is 6 square units.