Pre-Alg Help Anyone?

Question

find the equation of the line that passes through each pair, write in standered form. (-1,6) (1,2)

Answer 1

Y = mx + b
Where m is the slope and b is the y-intercept.

Now you plug in the numbers given.

(-1,6)
6 = m(-1) + b
6= -m +b

(1,2)
2 = m(1) + b
2= m +b

If a line passes through both of these points then it must have the same slope and y-intercept. . . .

We eliminate b from the equation by subtracting one equation from the other.


6= -m +b
(-) 2= m +b

6= -m +b
-2= -m -b

4 = -2m
4/-2 = m
-2 = m

Now we know m.


To solve for b. . .
plug it into one of the original equations.

2= m +b
Where m = -2
2=-2 +b
4 = b

b is 4, in other words when x = 0, y = 4.

Standard form for linear equations is Ax + By = C

So plug in the numbers and rearrange.



y= m(x) + b

m = -2, b = 4

y = -2x +4

2x +y =4 (final answer)

Answer 2

ALL OTHER SOLUTIONS ON THIS PAGE ARE INCORRECT! TRUST ME, THIS IS THE CORRECT SOLUTION!

y=mx+b is slope-intercept form. Standard form is Ax+By=C, where A, B, and C are definite numbers, and x and y are variables.

To find the slope, we can do (y_2-y_1)/(x_2-x_1).
So: (2-6)/(1-(-1)), or -4/2, or -2. Now, if the slope, m, is -2, lets take the slope-intercept form first.
y=-2x+b
So, we need "b". How about we substitute one of the points in? how about (1,2)?
2= -2(1) +b
2= -2 +b
4 = b
so, y= -2x + 4. We have all information needed! Now lets change it to Ax+By=C Form: Standard Form. Just move -2x to the other side!
2x + y = 4
A is 2, B is 1, and C is 4.

I hope this helps! If you need more help, do not hesitate to email me at [e-mail]!

Answer 3

if the equation should be in the form y=mx+b
then say (Xa, Ya) = (-1, 6)
and (Xb, Yb)=(1,2)
then the slope of the line, m = (Yb-Ya) / (Xb-Xa)
or m= (2-6)/(1- neg1) = (2-6)/ (1+1)= -4/2 = -2

so y=-2x+b
b is the value where he line crosses the x-axis and we can find it by substituting the values for one of the points, let's use (1,2)
2=(-2)(1)+b
2= -2 +b
or b=4
so your equation is y= -2x + 4 or
or y=4-2x

Hope that helps

Answer 4

i feel the easiest solution would be
equation of a line passing through two given points is
(y-y1)/(y2-y1)=(x-x1)/(x2-x1) substituting
(y-6)/(2-6)=(x+1)/(1+1) =>(y-6)/-4=(x+1)/2 cross multiplying
2(y-6)=-4(x+1) =>2y-12=-4x-4 and transposing after dividing throughout by 2 the equation is y=-2x+4

Answer 5

equation of line passing through two points is:-

y-y1= (y2-y1)*(x-x1)
......................
(x2-x1)
here (x1,y1)= (-1,6) and (x2,y2)= (1,2)

so eaquation of line passing through(-1,6) and (1,2) is
y-6=(2-6)*{x-(-1)}
........................
{1-(-1)}
y-6 = -4(x+1)/ 2
y-6= -2(x+1)
y-6=-2x-2
2x-y+4=0

Answer 6

figure it out yourself