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)