A line contains the points (-6, -4) and (0, 4). What is the equation of this line in slope-intercept form? A. B. C. y = x – 4 D. y = x + 4

y = 4/3 x + 4

Step-by-step explanation:

The general form of the equation of a line is given as

y = mx + c where m is the slope and c is the intercept on the y axis

where the slope m = (y2 - y1)/(x2 - x1)

Hence given the points  (-6, -4) and (0, 4)

m = (4 --4) / (0 --6)

= (4+4)/(0+6)

= 8/6

= 4/3

For a line with given points (x1, y1), the equation is given as

y-y1 = m(x-x1)

Hence the equation of this line in slope-intercept form

y - -4 = 4/3(x --6)

y + 4 = 4/3(x + 6)

3y + 12 = 4x + 24

3y = 4x + 24 - 12

3y = 4x + 12

y = 4/3 x + 4

For a line with given points (x2, y2), the equation is given as

y-y2 = m(x-x2)

Hence the equation of this line in slope-intercept form

y-4 = 4/3(x - 0)

multiply through by 3

3y - 12 = 4x

3y = 4x + 12

y = 4/3 x + 4

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