Find the distance between the two points rounding to the nearest tenth (if necessary). (-1,51 and (-7, -1)

Given a straight line defined by the points (1,51) and (3,91),

x1 = 1, y1 = 51, x2 = 3 and y2 = 91

Slope of the line m = ∆y/∆x

m = y2-y1/x2-x1

m = 91-51/3-1

m = 40/2

m = 20

Hence the slope of the line is 20.

Given the equation of a line y = mx+c

To get the y intercept of the line c, we will substitute one of the give points and the slope into the equation as shown;

Using the point (1,51) and m = 20,

Substitute into the equation y = mx+c

51 = 20(1)+c

51 = 20+c

c = 51-20

c = 31

Hence the slope is 20 and the y intercept of the line is 31

Step-by-step explanation:

I hope this helps:)

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

sorry can't

step-by-step explanation:

i googled though and something came up but it was in science langueage not common english sooo