Question

How do you determine the linear function whose graph is a line that contains the points $\left(-1,-8\right)$ and $\left(2,10\right)$ ?

Answer 1

Hint: The equation of a line that passes through two points can be calculated by using the two-point form of the equation of a line. We will substitute the given two points into the two-point form of a line. Then after simplifying we can determine the required linear function.

Formula used:
The two-point form of a line passing through the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by ${\displaystyle \frac{y-y_1}{x-x_1}}={\displaystyle \frac{y_2-y_1}{x_2-x_1}}$ .

Complete step-by-step answer:
As we know we can determine the equation of a line that passes through two points by using the two-point form of a line.

Now, the two given points are
$A=\left(-1,-8\right)$ and $B=\left(2,10\right)$ .
Now, we know that the first coordinate is the x-coordinate and the second coordinate is the y-coordinate.
Here,
For point $A$, $x_1=-1$ and $y_1=-8$
For point $B$, $x_2=2$ and $y_2=10$
Now, as we know that, the two-point form of a line is
${\displaystyle \frac{y-y_1}{x-x_1}}={\displaystyle \frac{y_2-y_1}{x_2-x_1}}$ .
Now, we can determine the linear function by substituting the $x_1=-1$ , $x_2=2$ , $y_1=-8$ and $y_2=10$ in the two-point form of the equation of a line.
After substituting the values, we get
${\displaystyle \frac{y-(-8)}{x-(-1)}}={\displaystyle \frac{10-(-8)}{2-(-1)}}$
Now, after opening the brackets of numerators as well as denominators on both sides, we get
$\Rightarrow {\displaystyle \frac{y+8}{x+1}}={\displaystyle \frac{10+8}{2+1}}$
$\Rightarrow {\displaystyle \frac{y+8}{x+1}}={\displaystyle \frac{18}{3}}$
On Right-hand side, simplifying by dividing $18$ by $3$ , we have
$\Rightarrow {\displaystyle \frac{y+8}{x+1}}={\displaystyle \frac{6}{1}}$
Now, by cross multiplying, we get
$1(y+8)=6(x+1)$
$\Rightarrow y+8=6x+6$
Now, by subtracting $8$ on both sides, we get
$\Rightarrow$$y+8-8=6x+6-8$
$\Rightarrow y=6x-2$

∴ $y=6x-2$ is the required linear function whose graph is a line that passes through the points $(-1,-8)$ and $(2,10)$.

Note:
There is an alternate way to prove the two points form of the equation of a straight line. Consider the point-slope form of the equation of a line,
we have, $y-y_1=m(x-x_1)$ - - - - - - $(1.)$
Since the line is passing through the point $\left(x_1,y_1\right)$ in $\left(1.\right)$ and the slope of the line is $m={\displaystyle \frac{y_2-y_1}{x_2-x_1}}$ . So, $(1.)$ becomes $y-y_1=\left({\displaystyle \frac{y_2-y_1}{x_2-x_1}}\right)\left(x-x_1\right)$.