Answer:
Segment EF: y = -x + 8
Segment BC: y = -x + 2
Step-by-step explanation:
Given the two similar right triangles, ΔABC and ΔDEF, for which we must determine the slope-intercept form of the side of ΔDEF that is parallel to segment BC.
Upon observing the given diagram, we can infer the following corresponding sides:
![\displaystyle\mathsf{\overline{BA}\:\: and\:\:\overline{ED}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cmathsf%7B%5Coverline%7BBA%7D%5C%3A%5C%3A%20and%5C%3A%5C%3A%5Coverline%7BED%7D%7D)
We must determine the slope of segment BC from ΔABC, which corresponds to segment EF from ΔDEF.
<h2>Slope of Segment BC:</h2>
In order to solve for the slope of segment BC, we can use the following slope formula:
![\displaystyle\mathsf{Slope\:(m)\:=\:\frac{y_2 \:-\:y_1}{x_2 \:-\:x_1}} }](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cmathsf%7BSlope%5C%3A%28m%29%5C%3A%3D%5C%3A%5Cfrac%7By_2%20%5C%3A-%5C%3Ay_1%7D%7Bx_2%20%5C%3A-%5C%3Ax_1%7D%7D%20%20%7D)
Use the following coordinates from the given diagram:
Point B: (x₁, y₁) = (-2, 4)
Point C: (x₂, y₂) = ( 1, 1 )
Substitute these values into the slope formula:
![\displaystyle\mathsf{Slope\:(m)\:=\:\frac{y_2 \:-\:y_1}{x_2 \:-\:x_1}}\:=\:\frac{1\:-\:4}{1\:-\:(-2)}\:=\:\frac{-3}{1\:+\:2}\:=\:\frac{-3}{3}\:=\:-1}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cmathsf%7BSlope%5C%3A%28m%29%5C%3A%3D%5C%3A%5Cfrac%7By_2%20%5C%3A-%5C%3Ay_1%7D%7Bx_2%20%5C%3A-%5C%3Ax_1%7D%7D%5C%3A%3D%5C%3A%5Cfrac%7B1%5C%3A-%5C%3A4%7D%7B1%5C%3A-%5C%3A%28-2%29%7D%5C%3A%3D%5C%3A%5Cfrac%7B-3%7D%7B1%5C%3A%2B%5C%3A2%7D%5C%3A%3D%5C%3A%5Cfrac%7B-3%7D%7B3%7D%5C%3A%3D%5C%3A-1%7D)
<h2>Slope of Segment EF:</h2>
Similar to how we determined the slope of segment BC, we will use the coordinates of points E and F from ΔDEF to find its slope:
Point E: (x₁, y₁) = (4, 4)
Point F: (x₂, y₂) = (6, 2)
Substitute these values into the slope formula:
![\displaystyle\mathsf{Slope\:(m)\:=\:\frac{y_2 \:-\:y_1}{x_2 \:-\:x_1}}\:=\:\frac{2\:-\:4}{6\:-\:4}\:=\:\frac{-2}{2}\:=\:-1}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cmathsf%7BSlope%5C%3A%28m%29%5C%3A%3D%5C%3A%5Cfrac%7By_2%20%5C%3A-%5C%3Ay_1%7D%7Bx_2%20%5C%3A-%5C%3Ax_1%7D%7D%5C%3A%3D%5C%3A%5Cfrac%7B2%5C%3A-%5C%3A4%7D%7B6%5C%3A-%5C%3A4%7D%5C%3A%3D%5C%3A%5Cfrac%7B-2%7D%7B2%7D%5C%3A%3D%5C%3A-1%7D)
Our calculations show that segment BC and EF have the same slope of -1. In geometry, we know that two nonvertical lines are <u>parallel</u> if and only if they have the same slope.
Since segments BC and EF have the same slope, then it means that
.
<h2>Slope-intercept form:</h2><h3><u>Segment BC:</u></h3>
The <u>y-intercept</u> is the point on the graph where it crosses the y-axis. Thus, it is the value of "y" when x = 0.
Using the slope of segment BC, m = -1, and the coordinates of point C, (1, 1), substitute these values into the <u>slope-intercept form</u> (y = mx + b) to solve for the y-intercept, <em>b. </em>
y = mx + b
1 = -1( 1 ) + b
1 = -1 + b
Add 1 to both sides to isolate b:
1 + 1 = -1 + 1 + b
2 = b
Hence, the <u><em>y-intercept</em></u> of segment BC is: <em>b</em> = 2.
Therefore, the linear equation in <u>slope-intercept form of segment BC</u> is:
⇒ y = -x + 2.
<h3><u /></h3><h3><u>Segment EF:</u></h3>
Using the slope of segment EF, <em>m</em> = -1, and the coordinates of point E, (4, 4), substitute these values into the <u>slope-intercept form</u> to solve for the y-intercept, <em>b. </em>
y = mx + b
4 = -1( 4 ) + b
4 = -4 + b
Add 4 to both sides to isolate b:
4 + 4 = -4 + 4 + b
8 = b
Hence, the <u><em>y-intercept</em></u> of segment BC is: <em>b</em> = 8.
Therefore, the linear equation in <u>slope-intercept form of segment EF</u> is:
⇒ y = -x + 8.