Question

8.sp.a.2 additional practice (cont’d)
for numbers 4 - 6, write the equation for the line of best - fit.
4)
(0,1)
(11,12)
y = 1x + 1 or y = x + 1
\frac{12 - 1}{11 - 0}=\frac{11}{11} or 1
5)
(0,26)
(24,0)
\frac{26 - 0}{0 - 24}=\frac{26}{-24}
6)
(5,16)
(17,12)
\frac{12 - 16}{17 - 5}=\frac{-4}{12}

Explanation:

Step1: Recall the slope - intercept form

The equation of a line is $y = mx + b$, where $m$ is the slope and $b$ is the y - intercept.

Step2: Calculate the slope $m$ for problem 4

Using two points $(x_1,y_1)=(0,1)$ and $(x_2,y_2)=(11,12)$, the slope $m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{12 - 1}{11-0}=\frac{11}{11}=1$.

Step3: Find the y - intercept $b$ for problem 4

Substitute one of the points, say $(0,1)$ into $y=mx + b$. When $x = 0$ and $y = 1$, and $m = 1$, we get $1=1\times0 + b$, so $b = 1$. The equation of the line is $y=x + 1$.

Step4: Calculate the slope $m$ for problem 5

Using two points $(x_1,y_1)=(0,26)$ and $(x_2,y_2)=(24,0)$, the slope $m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{0 - 26}{24-0}=-\frac{13}{12}$.

Step5: Find the y - intercept $b$ for problem 5

Since the line passes through $(0,26)$, the y - intercept $b = 26$. The equation of the line is $y=-\frac{13}{12}x+26$.

Step6: Calculate the slope $m$ for problem 6

Using two points $(x_1,y_1)=(5,16)$ and $(x_2,y_2)=(17,12)$, the slope $m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{12 - 16}{17 - 5}=-\frac{4}{12}=-\frac{1}{3}$.

Step7: Find the y - intercept $b$ for problem 6

We use the point - slope form $y - y_1=m(x - x_1)$. Using the point $(5,16)$ and $m =-\frac{1}{3}$, we have $y-16=-\frac{1}{3}(x - 5)$. Expand to get $y-16=-\frac{1}{3}x+\frac{5}{3}$, then $y=-\frac{1}{3}x+\frac{5}{3}+16=-\frac{1}{3}x+\frac{5 + 48}{3}=-\frac{1}{3}x+\frac{53}{3}$.

Answer:

1. $y=x + 1$
2. $y=-\frac{13}{12}x+26$
3. $y=-\frac{1}{3}x+\frac{53}{3}$