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5 votes
Find the equation of a line that passes through the points (-1,3) and (1,-5).

Leave your answer in the form as y=mx+c

User Igordc
by
8.2k points

2 Answers

5 votes

Answer: y+2x−1=0

Step-by-step explanation: Let's say A is the point (−1,3) and B is the point (3,−5)

The equation of a line that passes through two points is

y−y0=m(x−x0)

Replace x,x0,y and y0

by the coordinates of the two points to find your slope

⇒m

.

It doesn't matter which point you choose to replace

x,x0,y and y0 with as long as you pair xwith yand x0 with

y

0

.

m

=

y

y

0

x

x

0

=

5

3

3

(

1

)

=

5

3

3

+

1

=

2

Now, all you have to do is to choose either the coordinates of

A

or

B

to replace in the equation of a line that passes through two points

y

y

0

=

m

(

x

x

0

)

. You are only going to replace

x

0

and

y

0

.

I'm using the point

A

(

1

,

3

)

y

y

0

=

m

(

x

x

0

)

y

3

=

2

(

x

+

1

)

y

3

=

2

x

2

y

+

2

x

1

=

0

is your line.Let's say

A

is the point

(

1

,

3

)

and

B

is the point

(

3

,

5

)

The equation of a line that passes through two points is

y

y

0

=

m

(

x

x

0

)

Replace

x

,

x

0

,

y

and

y

0

by the coordinates of the two points to find your slope

m

.

It doesn't matter which point you choose to replace

x

,

x

0

,

y

and

y

0

with as long as you pair

x

with

y

and

x

0

with

y

0

.

m

=

y

y

0

x

x

0

=

5

3

3

(

1

)

=

5

3

3

+

1

=

2

Now, all you have to do is to choose either the coordinates of

A

or

B

to replace in the equation of a line that passes through two points

y

y

0

=

m

(

x

x

0

)

. You are only going to replace

x

0

and

y

0

.

I'm using the point

A

(

1

,

3

)

y

y

0

=

m

(

x

x

0

)

y

3

=

2

(

x

+

1

)

y

3

=

2

x

2

y

+

2

x

1

=

0

is your line.

User PausePause
by
8.7k points
4 votes

Answer:


y=-4x+7

Explanation:

According to the formula of the equation of the line passing through two given points


y-y1=m(x-x1)

where m=slope of the line


m=(y2-y1)/(x2-x1) ---->eq (II)

where
y2=-5 , y1=3 , x2=1 , x1=-1

Substituting the

values in eq(II)


m=(-5-3)/(1-(-1))=-8/2=-4

Substituting the value of m in eq(I)


y-3=-4(x-(-1))\\= > y=-4x+7

So the equation of the line passing through the given points is


y=-4x+7

User Harri
by
9.0k points

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