Question

LOTS OF POINTS GIVING BRAINLIEST I NEED HELP PLEASEE

Answer 1

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:

$\displaystyle\mathsf{\overline{BC}\:\: and\:\:\overline{EF}}$

$\displaystyle\mathsf{\overline{BA}\:\: and\:\:\overline{ED}}$

$\displaystyle\mathsf{\overline{AC}\:\: and\:\:\overline{DF}}$

We must determine the slope of segment BC from ΔABC, which corresponds to segment EF from ΔDEF.

Slope of Segment BC:

In order to solve for the slope of segment BC, we can use the following slope formula:

$\displaystyle\mathsf{Slope\:(m)\:=\:\frac{y_2 \:-\:y_1}{x_2 \:-\:x_1}} }$

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:

$\displaystyle\mathsf{Slope\:(m)\:=\:\frac{y_2 \:-\:y_1}{x_2 \:-\:x_1}}\:=\:\frac{1\:-\:4}{1\:-\:(-2)}\:=\:\frac{-3}{1\:+\:2}\:=\:\frac{-3}{3}\:=\:-1}$

Slope of Segment EF:

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:

$\displaystyle\mathsf{Slope\:(m)\:=\:\frac{y_2 \:-\:y_1}{x_2 \:-\:x_1}}\:=\:\frac{2\:-\:4}{6\:-\:4}\:=\:\frac{-2}{2}\:=\:-1}$

Our calculations show that segment BC and EF have the same slope of -1.  In geometry, we know that two nonvertical lines are parallel if and only if they have the same slope.  

Since segments BC and EF have the same slope, then it means that  $\displaystyle\mathsf{\overline{BC}\:\: | |\:\:\overline{EF}}$.

Slope-intercept form:

Segment BC:

The y-intercept 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 slope-intercept form (y = mx + b) to solve for the y-intercept, b.

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 y-intercept of segment BC is: b = 2.

Therefore, the linear equation in slope-intercept form of segment BC is:

⇒  y = -x + 2.

Segment EF:

Using the slope of segment EF, m = -1, and the coordinates of point E, (4, 4), substitute these values into the slope-intercept form to solve for the y-intercept, b.

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 y-intercept of segment BC is: b = 8.

Therefore, the linear equation in slope-intercept form of segment EF is:

⇒  y = -x + 8.