find the equation of the line, in standard from, that combines the points (-7, 9) and (3,-7)?

Answer 1

We can start by finding the slope, which is rise over run, or change in y over change in x. That's (9 - (-7)) / (-7 - 3) = 16 / -10 = -8 / 5. We can easily write this in point-slope form, using the point (-7, 9), giving us y - 9 = (-8 / 5)(x + 7).

Standard form requires integer coefficients for x and y, with their sum set equal to a constant. We'll need to multiply through by 5 to have an integer coefficient for x, and that gives us 5y - 45 = -8(x + 7) ==> 5y - 45 = -8x - 56 ==> 8x + 5y = -11.

Answer 2

First find the slope of the line = 9-(-7)/ -7 -3 = 16/-10 = -8/5

Next find the y intercept using the points (-7,9) or (3,-7). Using
(-7,9) the y intercept will be
9 = -7(-8/5) +c
45 = 56 + 5c
c= -11/5

Therefore the equation of the line will be
y = -8x/5 - 11/5
After normalization the equation will be
5y = -8x -11
or 8x + 5y = -11

Answer 3

(-7, 9) and (3,-7)

gradient, m = 9--7 / -7-3 = 16/-10 = -8/5

eqn of lin ---> y - y1 = m (x -x1) where y1 is any of the y cordinate and x1 is the coresponding x of the y1 coordinate

y - 9 = <-8/5> ( x +7)
y = (-8x/5) - 56/5 + 9
y = (-8x/5) - 11/5
5y = -8x - 11

Answer 4

Slope formula

m = y₂- y₁/ x₂- x₁

m = 9 - (- 7) / -7 - 3

m = 9 + 7 / - 10

m = - 16 / 10

m = - 8 / 5

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