Triangle QRS is similar to triangle TUV. Write the equation, in slope-intercept form, of the side of triangle QRS that is parallel to UV. You must show all work to receive credit. (10 points)
The slope-intercept form of a line is given by [tex]y = mx + c[/tex] [tex]\text{where m is the slope and c is the y-intercept}[/tex]
The slope is calculated using the following formula [tex]m = \dfrac{y_2 - y_1}{x_2-x_1}[/tex] [tex]\text{where $(x_1, y_1)$ and $(x_2, y_2)$ are two points on the straight line}[/tex]
The side in ΔQRS that is parallel to UV is side RS
Therefore we have to find the slope-intercept form of line RS
Take two points on RS, say S( -8, 1) and R(- 2, 7) [tex]\text{Slope m = $\dfrac{7 - 1}{-2 - (-8)}$ }\\\\= $\dfrac{6}{-2 + 8}$\\\\= \dfrac{6}{6}\\\\m = 1[/tex]
So the equation of the line RS is y = 1x + c
To find the y-intercept c, take any point on the line (x, y) and substitute these values into the slope equation
Take Point R(-2, 7)
Substituting y = 7, x = -1 gives 7 = 1 (-2) + c 7 = -2 + c 7 + 2 = 2 + 2 + c (add 2 both sides) 9 = 0 + c 9 = c or c = 9
Therefore the equation of RS is y = 1x + 9
which can be written as [tex]\boxed{y = x + 9}[/tex]
