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3 years ago

10

Triangle JKL has vertices J(-4, -1), K(0, 4), and L(-4, -2). Graph The triangle and it’s image after a dilation with a scale fac

tor of 0.5.

1 answer:

Dafna1 [17]3 years ago

5 0

Answer:

J(-2, -0.5), K(0, 2), L(-2, -1)

Step-by-step explanation:

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The original line has the equation of y=-2x+8

enot [183]

The new equation of line that is perpendicular to the original that goes through the point (6, -1) in slope intercept form is $y = \frac{1}{2}x - 4$

<h3><u>Solution:</u></h3>

Given that original line has the equation of y = -2x + 8

We have to write a new equation that is perpendicular to the original that goes through the point (6, -1)

Let us first find slope of original line

<em><u>The slope intercept form of line is given as:</u></em>

y = mx + c ------ eqn 1

Where "m" is the slope of line and "c" is the y - intercept

On comparing the slope intercept form and given original equation, we get "m = -2"

Thus slope of original line "m" = -2

We know that product of slope of a line and slope of line perpendicular to it are always -1

slope of original line x slope of line perpendicular to it = -1

$\begin{array}{l}{-2 \times \text { slope of line perpendicular to it }=-1} \\\\ {\text { slope of line perpendicular to it }=\frac{1}{2}}\end{array}$

Let us find equation of line with slope m = 1/2 and passes through point (6, - 1)

Substitute $m = \frac{1}{2}$ and (x, y) = (6, -1) in eqn 1

$-1 = \frac{1}{2} \times 6 + c\\\\-1 = 3 + c\\\\c = -4$

<em><u>Thus the required equation of line is:</u></em>

Substitute "c" = -4 and $m = \frac{1}{2}$ in eqn 1

$y = \frac{1}{2}x - 4$

Thus the equation of line perpendicular to original line is found

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