A kite has vertices at (2, 4), (5, 4), (5, 1), and (0,

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.

<h2>Slope of Segment BC:</h2>

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

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

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:

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

<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:

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

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 $\displaystyle\mathsf{\overline{BC}\:\: | |\:\:\overline{EF}}$ .

<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.

8 0

2 years ago

-a^2+46a-480=0<br> a - ?

sattari [20]

a = 30
a = 16

-a² + 46a -480 = 0

(-a + 30) (a -16) = 0
(-a)(a) + (-a)(-16) + 30(a) + 30(-16) = 0
-a² + 16a + 30a -480 = 0
-a² + 46a - 480 = 0

(-a + 30) = 0 ; (a - 16) = 0
-a = -30 ; a = 16
a = 30

To check:
a = 30
-(30)² + 46(30) - 480 = 0
-900 + 1380 - 480 = 0
480 - 480 = 0
0 = 0

a = 16
-(16)² + 46(16) - 480 = 0
-256 + 736 - 480 = 0
480 - 480 = 0
0 = 0

5 0

2 years ago

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A kite has vertices at (2, 4), (5, 4), (5, 1), and (0, –1).