Question

why did gyro go into a bakery? for each exercise below, find the equation of the line that has the given slope and passes through the given point. circle the letter next to the correct equation. then write this letter in each box at the bottom of the page that contains the number of that exercise.
① m = 2; (3, 2) q: y = 2x + 1 r: y = 2x - 4
② m = -3; (1, 4) o: y = -3x + 7 p: y = -3x + 2
③ m = -5; (-1, 3) m: y = -5x - 2 d: y = -5x + 6
④ m = 3; (-4, -7) v: y = 3x + 1 e: y = 3x + 5
⑤ m = -1; (5, -2) u: y = -x + 3 c: y = -x - 1
⑥ m = 1/2; (6, 1) w: y = 1/2x - 5 h: y = 1/2x - 2
⑦ m = -2/3; (3, 4) a: y = -2/3x - 7 i: y = -2/3x + 6
⑧ m = 4/3; (-2, 0) k: y = 4/3x + 5/2 f: y = 4/3x + 8/3
⑨ m = -1/4; (2, 1) j: y = -1/4x + 3/2 d: y = -1/4x - 3/8
⑩ m = 4; (-1, 1/2) a: y = 4x - 2/3 t: y = 4x + 9/2
⑪ m = -2; (0, 0) l: y = -2x b: y = -2x - 2
⑫ m = 0; (-5, 3/4) s: y = 3/4 n: y = -5x
9 5 12 10 8 2 1 10 6 4 12 3 4 11 11 2 8 7 10
ective 5 - n: to find an equation of a line
algebra with pizzazz! 15

Explanation:

Step1: Recall point - slope formula

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $m$ is the slope and $(x_1,y_1)$ is a point on the line.

Step2: Solve for equation of line in each case

For (1): $m = 2$, $(x_1,y_1)=(3,2)$

$y - 2=2(x - 3)$
$y - 2=2x-6$
$y = 2x-4$ (Answer is R)

For (2): $m=-3$, $(x_1,y_1)=(1,4)$

$y - 4=-3(x - 1)$
$y - 4=-3x + 3$
$y=-3x + 7$ (Answer is O)

For (3): $m=-5$, $(x_1,y_1)=(-1,3)$

$y - 3=-5(x + 1)$
$y - 3=-5x-5$
$y=-5x-2$ (Answer is M)

For (4): $m = 3$, $(x_1,y_1)=(-4,-7)$

$y+7 = 3(x + 4)$
$y+7=3x + 12$
$y=3x + 5$ (Answer is E)

For (5): $m=-1$, $(x_1,y_1)=(5,-2)$

$y+2=-(x - 5)$
$y+2=-x + 5$
$y=-x + 3$ (Answer is U)

For (6): $m=\frac{1}{2}$, $(x_1,y_1)=(6,1)$

$y - 1=\frac{1}{2}(x - 6)$
$y - 1=\frac{1}{2}x-3$
$y=\frac{1}{2}x-2$ (Answer is H)

For (7): $m=-\frac{2}{3}$, $(x_1,y_1)=(3,4)$

$y - 4=-\frac{2}{3}(x - 3)$
$y - 4=-\frac{2}{3}x+2$
$y=-\frac{2}{3}x + 6$ (Answer is I)

For (8): $m=\frac{4}{3}$, $(x_1,y_1)=(-2,0)$

$y-0=\frac{4}{3}(x + 2)$
$y=\frac{4}{3}x+\frac{8}{3}$ (Answer is F)

For (9): $m=-\frac{1}{4}$, $(x_1,y_1)=(2,1)$

$y - 1=-\frac{1}{4}(x - 2)$
$y - 1=-\frac{1}{4}x+\frac{1}{2}$
$y=-\frac{1}{4}x+\frac{3}{2}$ (Answer is J)

For (10): $m = 4$, $(x_1,y_1)=(-1,\frac{1}{2})$

$y-\frac{1}{2}=4(x + 1)$
$y-\frac{1}{2}=4x + 4$
$y=4x+\frac{9}{2}$ (Answer is T)

For (11): $m=-2$, $(x_1,y_1)=(0,0)$

$y-0=-2(x - 0)$
$y=-2x$ (Answer is L)

For (12): $m = 0$, $(x_1,y_1)=(-5,\frac{3}{4})$

$y-\frac{3}{4}=0(x + 5)$
$y=\frac{3}{4}$ (Answer is S)

Answer:

1. R
2. O
3. M
4. E
5. U
6. H
7. I
8. F
9. J
10. T
11. L
12. S