Houghton Mifflin Harcourt Algebra 1, 2015
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Houghton Mifflin Harcourt Algebra 1, 2015 View details
2. Using Intercepts
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Exercise Lesson Performance Task Page 178

Practice makes perfect
a We have the linear equation 10x+3y =240. To find the intercepts, recall that the x-intercept is the x-coordinate of the point where the graphs intercepts the x-axis. When this happens the y-coordinate is always 0, therefore, we can substitute the value y=0 in our equation and solve for the corresponding x value.
10x+3y =240
10x+3( 0) =240
10x = 240
10x/10 = 240/10
x =24
We can see that the x-intercept is 24. Now we will find the y-intercept. Recall that the y-intercept is the y-coordinate of the point where the graphs intercepts the y-axis. When this happens the x-coordinate is always 0, therefore, this time we will substitute x=0 in our equation and solve for the corresponding y value.
10x+3y =240
10( 0)+3y =240
3y = 240
3y/3 = 240/3
y =80
We found that the y-intercept is 80. Knowing both intercepts, we can graph the shape of the sail. For this, we just plot the points (24,0) and (0,80). The line joining them will be our function. Since the function gives the values in feet, we can see that the sail is a triangle with a base of 24 feet and 80 feet tall. Recall that the area of a triangle is given by A= 12bh. Let's calculate the area of the sail.
12bh
12( 24)( 80)
960
Therefore, the area of the sail is 960 ft^2
b This time we are working with the generic equation Ax + By = C. In this equation A and B cannot both be 0, and A, B, and C are positive real numbers. Just as we did in Part A, since the x-intercept the y-coordinate is always 0 we can substitute the value y=0 in our equation and solve for the corresponding x value.
Ax + By = C
Ax + B( 0) = C
Ax = C
Ax/A = C/A
x = C/A

Therefore, the x-intercept is given by the quotient CA. Similarly, we can find the y-intercept by substituting this time x=0, since when the y-intercept happens the x-cooordinate is always 0.
Ax+By= C
A( 0) +By = C
By = C
By/B = C/B
y = C/B
Then, the y-intercept is given by the quotient CB. In this case the triangular sail would have a base of CA ft and a height of CB ft. We can now calculate its area.
12bh
12( CA)( CB)
C^2/2AB
As we can see, the area for this sail will depend on the values of the parameters of the line equation A, B, and C. It is C^22AB ft^2.