Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
2. Section 5.2
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Exercise 115 Page 302

Practice makes perfect
a To find the equation of AB, we have to determine its slope m and y-intercept b.
y=mx+ b Examining the diagram, we can identify where AB intercepts the y-axis.

The line intercepts the y-axis at (0,3), which means b= 3. To find the slope we have to measure the vertical and horizontal distance between two points on the graph. The only points that lie on coordinates we can identify are the x- and y-intercept.

Traveling from the y-intercept to the x-intercept requires that we move 5 steps horizontally in the positive direction and 3 steps vertically in the negative direction. With this information we can find the slope. Rise/Run=-3/5 ⇔ m= - 3/5 Now that we have the slope and the y-intercept, we can form our final equation. y= mx+b y= - 3/5x+3

b Let's look closely at the given diagram.

The x- and y-intercepts are the points at which the graph crosses the x- and the y-axis, respectively. Looking at the graph, we can notice that (5,0) is the x-intercept and (3,0) is the y-intercept.

c Let's highlight the triangle in the diagram. Note that the vertical and horizontal axes are perpendicular, which means △ ABC must be a right triangle. The area of a triangle is its base multiplied by its height divided by 2.
A=1/2bh Since this is a right triangle, its legs will make up the triangle's height and base. Let's identify their lengths in the diagram. We have all the information we need to calculate the area of the triangle.
A=1/2bh
A=1/2( 5)( 3)
Simplify right-hand side
A=1/2* 15
A=15/2
A=7.5
The area of the triangle is 7.5. To find the perimeter of the triangle, we have to add all of its sides. Since we do not know the hypotenuse we have to calculate it. We can do this by using the Distance Formula.
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
d=sqrt(( 5- 0)^2+( 0- 3)^2)
Simplify right-hand side
d=sqrt(5^2+(-3)^2)
d=sqrt(25+9)
d=sqrt(34)
When we know the hypotenuse, we can finally calculate the perimeter by adding all of the sides. Perimeter: 5+3+sqrt(34)≈ 13.83
d To write the equation of a line perpendicular to AB, we first need to determine its slope. Two lines are perpendicular when their slopes are opposite reciprocals. This means that the product of a given slope and the slope of a line perpendicular to it will be -1.
m_1*m_2=-1From Part A, we know the slope of AB. By substituting the slope of AB into the equation, we can solve for the slope of the perpendicular line, m_2.
m_1 * m_2 = - 1
-3/5* m_2 = - 1
Solve for m_2
-3m_2/5=-1
3m_2=5
m_2 = 5/3
Any line perpendicular to the given equation will have a slope of 53. Since we know the line intercepts the y-axis at point A, we know that b=3. With this information we can write the equation of the line through A that is perpendicular to AB as y= 53x+3.