Question

Please help me... thank you

Answer 1

Slope-intercept form:

y = mx + b

"m" is the slope, "b" is the y-intercept (the y value when x = 0) or (0,y)

For lines to be parallel, they have to have the SAME slope.

For lines to be perpendicular, their slopes have to be the opposite/negative reciprocals (flipped sign and number)

For example:

slope is 2

perpendicular line's slope is -1/2

slope is -2/3

perpendicular line's slope is 3/2

9.) First find the slope of line PQ. Use the slope formula and plug in the two points.

P = (4, 1)    (x₁ , y₁)

Q = (8, 4)       (x₂ , y₂)

$m= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$m=\frac{4-1}{8-4}$

$m=\frac{3}{4}$

Line RS is parallel to line PQ, so they have the same slope of 3/4

$y=\frac{3}{4}x+b$

To find "b", plug in the point R = (3, -2) into the equation

$y=\frac{3}{4}x+b$

$-2=\frac{3}{4}(3)+b$

$-2=\frac{9}{4}+b$      Subtract 9/4 on both sides

$-2-\frac{9}{4}=b$     Make the denominators the same

$-\frac{8}{4} -\frac{9}{4}=b$

$-\frac{17}{4}=b$

$y = \frac{3}{4}x- \frac{17}{4}$

10.) Find the slope of line PQ

$m= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$m=\frac{4-1}{8-4} =\frac{3}{4}$

Line RS is perpendicular to line PQ, so the slope of line RS is -4/3

y = -4/3x + b

Plug in the point R = (3, -2) into the equation to find "b"

y = -4/3x + b

-2 = -4/3(3) + b

-2 = -4 + b    Add 4 on both sides

2 = b

$y = -\frac{4}{3}x+2$