Given the points a(-8,11) and b(-5,8). a) find the equation of line ab. b) fine the line parallel to line ab that passes through (-7,1). c) fine the line perpendicular to the line ab that passes through (-2,5).

a) The slope of AB = (8-11)/(-5--8)

= -3/3

= -1

The equation of AB is given by:

y - y1 = m(x-x1)

Where m is the slope.

Using point (-8,11), we have;

y - 11 = -1(x--8)

y -11 = -1(x+8)

y -11 = -x -8

y = -x -8 + 11

» y = -x+3

b) A line parallel to AB also has a slope of -1.

If this line passes through (-7,1).

Then its equation is;

y - 1 = -1(x --7)

y - 1 = -1(x+7)

y -1 = -x-7

» y = -x-7+1

» y = -x -6

c) A perpendicular to AB will have slope -1/-1 = 1

If the perpendicular passes through (-2,5), then its equation is given by

y - 5 = 1(x--2)

» y-5 = 1(x+2)

» y -5 = x+2

y = x+7 .

answer is d-29

step-by-step explanation:

since you didn't include the picture, i can give you the description so you know how to pick the correct answer. the third quartile of a data is always shown in a box plot. in the middle of your graph should be a box. the third quartile is the value that goes with the far right edge of the box.

answer:   x = 3 + √11     and     x = 3 - √11

step-by-step explanation:

x² - 6x + 9 = 11

(x - 3)² = 11

x - 3 = ±√11

    x = 3 ± √11

x = 3 + √11     and     x = 3 - √11