Question

write the slope-intercept form of an equation for each line described. please show all of your work!13. the y-intercept is (0, 2) and it goes through the point (1, 5).if you want, you can use the graph and/or the table to assist you.equation: $\boldsymbol{y=3x + 2}$14. the y-intercept is (0, -5) and it goes through the point (-2, -2).equation: $\boldsymbol{y=-\frac{3}{2}x - 5}$15. write the equation of the line that goes through the points (2, 3) and (8, -6).equation: _____________16. write the equation of the line that goes through the points (-3, 4) and (2, 14).equation: _____________17. determine if all three points are on the same line or not. if they are, write the equation for that line. if they are not, explain how you know they are not. you must verify your claim by using all three pointsa) (-6, 1), (6, -17), and (10, -23)b) (-5, 26), (1, -4), and (11, -2)

Explanation:

Step1: Find slope for Q15

Slope formula: $m=\frac{y_2-y_1}{x_2-x_1}$
$m=\frac{-6-3}{8-2}=\frac{-9}{6}=-\frac{3}{2}$

Step2: Find y-intercept for Q15

Use $y=mx+b$, substitute $(2,3)$:
$3=-\frac{3}{2}(2)+b$
$3=-3+b$
$b=6$

Step3: Write Q15 equation

$y=-\frac{3}{2}x+6$

Step4: Find slope for Q16

Slope formula: $m=\frac{y_2-y_1}{x_2-x_1}$
$m=\frac{14-4}{2-(-3)}=\frac{10}{5}=2$

Step5: Find y-intercept for Q16

Use $y=mx+b$, substitute $(-3,4)$:
$4=2(-3)+b$
$4=-6+b$
$b=10$

Step6: Write Q16 equation

$y=2x+10$

Step7: Check collinearity for Q17A

First find slope of first two points:
$m=\frac{-17-1}{6-(-6)}=\frac{-18}{12}=-\frac{3}{2}$
Use $(-6,1)$ to find $b$:
$1=-\frac{3}{2}(-6)+b$
$1=9+b$
$b=-8$
Equation: $y=-\frac{3}{2}x-8$
Verify third point $(10,-23)$:
$y=-\frac{3}{2}(10)-8=-15-8=-23$, matches.

Step8: Check collinearity for Q17B

First find slope of first two points:
$m=\frac{-4-26}{1-(-5)}=\frac{-30}{6}=-5$
Use $(-5,26)$ to find $b$:
$26=-5(-5)+b$
$26=25+b$
$b=1$
Equation: $y=-5x+1$
Verify third point $(11,-2)$:
$y=-5(11)+1=-55+1=-54
eq-2$, so not collinear.

Answer:

1. $\boldsymbol{y=-\frac{3}{2}x+6}$
2. $\boldsymbol{y=2x+10}$

17A. The three points are on the same line. Equation: $\boldsymbol{y=-\frac{3}{2}x-8}$
17B. The three points are not on the same line. When substituting $(11,-2)$ into the line equation derived from the first two points, the calculated $y$-value does not match $-2$.