Math Problem?

Question

Write the equation of a line that passes threw (-2,9) & (2,1). I have no clue how to do this problem on my homework. Please help, and tell me the answer!!! : )

Answer 1

slope formula

m = y₂- y₁/ x₂- x₁

Ordered Pairs

(-2, 9)(2, 1)

m = 1 - 9 / 2 - (- 2)

m = - 8 / 4

m = - 2

- - - - - - - - -

Slope intercept form

y = mx + b

Ordered Pair

(- 2, 9)

9 = - 2(- 2) + b

9 = 4 + b

9 - 4 = 4 + b - 4

5 = b

- - - - - - - -

y = mx + b

Ordered pair

(2, 1)

1 = - 2(2) + b

1 = - 4 + b

1 + 4 = - 4 + b + 4

5 = b

- - - - - - - --

The equation

y = - 2x + 5

- - - - - - - - - - -s-

Answer 2

y=-2x+5

if this is a line, the general equation would be y=mx+n. If these two points are on this line, it means that these points verify the equation. That is, let's write x=-2 and y=9 on y=mx+n and than x=2 and y=1 on the same equation and solve it simultaneously.

For (-2,9) if x=-2 and y=9 the equation is 9=-2m+n
For (2,1) if x=2 and y=1 the equation is 1=2m+n

If we solve 9=-2m+n and 1=2m+n simultaneously;

9=-2m+n
1=2m+n
---------------
2n=10
n=5 and m=-2 than the genaral equation of y=mx+n would be

y=-2x+5

note:
you do not need to memorize the formula for finding the equation of a line giving two points (y-y1/x-x1=y2-y1/x2-x1). You cannot memorize all formulas in analytical geometry!

Answer 3

x1 y1 x2, y2
(-2,9) (2,1)

the formular for finding the equation of a line giving two points is:
y-y1/x-x1=y2-y1/x2-x1
substitute the values given into the formular.
y-9/x--2=1-9/2--2
y-9/x+2=-8/2+2
y-9/x+2=-8/4
cross multiply
(y-9)4=(x+2)-8
4y-36=-8x-16
4y=-8x-16+36
4y=-8x+20
divide by 4
y=-2x+5

Answer 4

Answer; y= -2x + 5 , working : First find the gradient slope of the line:(9 -1)/(-2-2) ,and the use the equation y=mx + c.The m is the gradient of the slope and you can find c by substitution.Hope this helps.

Answer 5

you need to find the slope of the line which is:
m= (y2-y1)/(x2-X1)
m= (-2-2)/(9-1)
m= - (1/2)

the equation of the line is:
y-y1= m (x-x1)
that means: you can substitute any of the points you have:
y-2= - 0.5 ( x - 1)
or
y -(-2) = -0.5 ( x - 9)

at the 2 cases the equation will be:
2y + x = 5

Answer 6

Equation of the line can be found with two point formula

y-y1 = (y2-y1)/(x2-x1) (x-x1)

we have x1=-2, y1=9, x2=2, y2=1

put them in formula

y-9 = (1-9) / (2-(-2)) [x-(-2)]

y-9 = -8/4 (x+2)

y-9 = -2 (x+2)
y-9 = -2x-4
y+2x-5=0

Answer 7

y = mx + b

m = (9 - 1)/(-2 - 2) = -8 / 4 = -2

y = -2x + b

(1) = -2(2) + b
b = 5

Final equation:
y = -2x + 5

Answer 8

y = -2x +5:

first find the slope (= -4)
then use
y = mx+b

9 = -2*-2 +b (substitute)
9= 4 +b
5 = b
therefore,
y = -2x +5

Answer 9

Call point one (x1,y1).
Call point two (x2,y2).

Slope = (y2 - y1)/(x2 - x1) = (1-9)/(2--2) = -8/4 = -2.

Equation of line is (y-y1)=slope(x-x1):

y-9 = -2(x--2)
y-9 = -2x-4
y = -2x+5

Answer 10

Y = mX + C

m = gradient of the line
C = y-intercept

m = (y2 - y1) / (x2 - x1)