What is the equation of a line in standard form, that combines the points (-7,9) and (3, -7)?

Question

A. 3x-7y=22
B. 5x+8y=-11
C. 8x-10y=22
D. 5y+8x=-11

Answer 1

The answer is D : 8x + 5y = -11

Line : y = mx + c

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

Taking the point (3,-7) into y = mx + c

-7 = (-8/5)3 + c

-7 = (-24/5) =+ c

c = (24/5) - 7 = 24/5 - 35/5 = -11/5

c = -11/5 so

y = (-8/5)x - 11/5

5y = -8x - 11

8x + 5y = -11

Answer 2

Yeah, you multiplied both the numerator and denominator by three which gave you -15/24 when it should have been -15/8. :)

y = -5/8x + b
-7 = (-5/8) * 3 + b
-7 = -15/8 + b
-7 + 1 7/8 = b
-5 1/8 = b

So the answer would be,

y = -5/8 * x - 41/8

Answer 3

the second guy almost got it except that the slope = the opposite or -8/5 (because you subtract the y's on top and the x's on the bottom)

Then plug it in and get -7 = -8/5(3) + b
-7 = -24/5 +b
-7 + 24/5 = b
-11/5 = b

So the equation is y = -8/5x + -11/5

Answer 4

to figure out the slope, it's

x2-x1 / y2-y1
=3-(-7) / -7-9
=10 / -16
= -5/8

Therefore you need to sub that into the standard form

y=mx+b

Using one of the points and your slope, solve for b.

-7=-5/8(3)+b
-7=-15/24 +b
-7 + 15/24 = b
-7/1 + 15/24 = b
-105/24 + 15/24 = b
-90/24 = b
-30/8 = b
-15/4 = b

Therefore y = -5/8x - 15/4

lol i probably made a huge mistake.

Answer 5

points (-7,9) and (3, -7) are on the line,
m=(9-(-7))/(-7-3)
m= -16/10
m= -8/5

point (3, -7) is on the line,
(y-(-7))= -8/5(x-3)
5y+8x= -11

Therefore, D is the answer.

Answer 6

y= x -7
the slope is zero (0/-6), so it would be -7 = 3(0) + b, which means be = -7