Question

PLEASE HELP!!

Find the equation of the line that passes through (1,5) and (2,8).
Find the equation of the line that passes through (-2,-7) and (4,-10).

Answer 1

Any straight line can be expressed as

y = mx + b.

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

m= 8-5/2-1=3/1

m=3

y = mx + b.




y =3*x + b

X=2 Y=8

8=3(2)+b

8=6+b

8-6=b

b=2

Equation for the first one:

y = mx + b.

Y=3x+2

Answer 2

1. y=3x+2

2. y=-1/2x-8

Explanation:

1. Find the equation of the line that passes through (1,5) and (2,8).

slope-intercept form of the line is y=mx+b

the given points are (1,5) and (2,8)

the equation for slope (m) is (y2-y1)/(x2-x1)

label the points:

x1=1

y1=5

x2=2

y2=8

substitute into the equation

m=(8-5)/(2-1)

m=3/1

m=3

the slope is 3

use slope-point form, which is y-y1=m(x-x1)

once again, substitute into the equation:

y-5=3(x-1)

do distributive property

y-5=3x-3

add 5 to both sides

y=3x+2

the equation of the first line is y=3x+2

2. Find the equation of the line that passes through (-2,-7) and (4,-10).

we are given the points (-2,-7) and (4,-10)

label the points:

x1=-2

y1=-7

x2=4

y2=-10

substitute into the equation for slope:

m=(-10--7)/(4--2)

m=(-3)/6

m=-1/2

the slope is -1/2

use slope-point form again:

y--7=-1/2(x--2)

y+7=-1/2(x+2)

do distributive property

y+7=-1/2x-1

subtract 7

y=-1/2x-8

the second equation is y=-1/2x-8

hope this helps :D