# problem with an equation

#### thetomps

##### Junior Member
Find an equation for a line containing the points (3, 4) and (-2, -7).

In this problem I am given some options such as:

a. 3x + y = 13
b. 3x + 5y = 29
c. 11x - 5y = 13
d. 11x - y = 29
e. none of these

By the looks of it none of them work, is that true or am I looking at this wrong?

#### happy

##### Full Member
Find an equation for a line containing the points (3, 4) and (-2, -7).

In this problem I am given some options such as:

a. 3x + y = 13
b. 3x + 5y = 29
c. 11x - 5y = 13---answer because 33-20= 13 and -22+35= 13
d. 11x - y = 29
e. none of these

By the looks of it none of them work, is that true or am I looking at this wrong?

#### TchrWill

##### Full Member
Posted: Wed Oct 05, 2005 10:40 pm Post subject: problem with an equation

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Find an equation for a line containing the points (3, 4) and (-2, -7).

In this problem I am given some options such as:

a. 3x + y = 13
b. 3x + 5y = 29
c. 11x - 5y = 13
d. 11x - y = 29
e. none of these

By the looks of it none of them work, is that true or am I looking at this wrong?

The slope of the line is m = (4 - (-7)) / 3 - (-2)) = 11/5

Substituting into y = mx + b, 4 = (11/5)3 + b making b = -(13/5)

In the standard format, y = (11/5)x - (13/5)

Converting to y our format,, 11x - 5y = 13

#### soroban

##### Elite Member
Hello, thetomps!

Find an equation for a line containing the points (3, 4) and (-2, -7)
We're expected to know the Slope Formula . . .

Given points P(x<sub>1</sub>,y<sub>1</sub>) and Q(x<sub>2</sub>,y<sub>2</sub>), the slope of the line through P and Q

. . . . . . . . . . . . . . . . . . .y<sub>2</sub> - y<sub>1</sub> . <-- (difference of the y's)
. . is given by: . m . = . ----------
. . . . . . . . . . . . . . . . . . .x<sub>2</sub> - x<sub>1</sub> . <-- (difference of the x's)

We should also know the Point-Slope Formula . . .

Given a line containing point (x<sub>1</sub>,y<sub>1</sub>) and with slope m,

. . . the equation is: . y - y<sub>1</sub> .= .m(x - x<sub>1</sub>)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We are given two points: .P(3, 4) and Q(-2, -7)

. . . . . . . . . . . . . . . . . . . . . .- 7 - 4 . . . . -11 . . . . .11
The slope of PQ is: . m . = . -------- . = . ----- . = . ----
. . . . . . . . . . . . . . . . . . . . . . -2 - 3 . . . . .-5 . . . . . .5

We have point P(3, 4) and m = 11/5.

The equation of the line is: . y - 4 .= .(11/5)(x - 3)

Multiply through by 5: . 5(y - 4) .= .11(x - 3)

. . and we get: . 5y - 20 .= .11x - 33

. . which simplifies to: . 11x - 5y .= .13 . . . Answer (c)