Find the slope-intercept form of the equation of the line passing through the points. Sketch the line.
- (-8,1),(-8,7)
- (-1,4),(6,4)
The slope that I get for number 1 is infinity. I cannot proceed because I don't know how the infinity can be calculated.
Hope you can help me!
On
If (x1,y1) and (x2,y2)be two points in the plane.Then the equation of the line joining these two points is $(y-y1) (x1-x2)=(x-x1)(y1-y2)$. From this equation you will get slope intercept form.
On
The formula for slope is $$\begin{array}{llllr}m=\frac{\Delta y}{\Delta x}&&&&(1)\end{array}$$ To find the equation of a line passing through 2 points the formula is $$\begin{array}{llllr}m=\frac{\Delta y}{\Delta x}&&&&(2)\end{array}$$ For example, if we wanted to find the equation of a line passing through points $(-2, 11)$ and $(3,-4)$ we would first use equation $(1)$ to find the slope.
$$\begin{array}{lllr} m&=&\frac{\Delta y}{\Delta x}\\ m&=&\frac{\color{blue}{11} - (-4)}{\color{blue}{-2}-(3)}\\ m&=&\frac{15}{-5}\\ m&=&-3 \end{array}$$
By equation $(2)$ we have
$$\begin{array}{lllr} -3&=&m\\ -3&=&\frac{\Delta y}{\Delta x}\\ -3&=&\frac{\color{blue}{y}-(-4)}{\color{blue}{x}-(3)}\text{ or alternatively, } \frac{y-11}{y-(-2)}\\ -3&=&\frac{y+4}{x-3} \end{array}$$
Next, we have to express the equation in the form $y=mx+b$
$$\begin{array}{lllr} -3&=&\frac{y+4}{x-3}\\ \frac{y+4}{x-3}&=&-3\\ y+4&=&-3(x-3)\\ y&=&-3(x-3)-4\\ y&=&-3x+9-4\\ y&=&-3x+5 \end{array}$$
We call this slope-intercept form because the $-3$ represents the slope, and the $5$ represents the y-intercept. Notice that you can also express the equation in this form. $$\begin{array}{lllr} y&=&-3x+5\\ y-5&=&-3x\\ \frac{y-5}{-3}&=&x\\ x&=&\frac{y-5}{-3}\\ x&=&-\frac{1}{3}y+\frac{5}{3}\\ \end{array}$$ Where $-1/3$ is the slope's reciprocal (i.e. $1/m$), and $5/3$ is the $x-intercept$.
By the way, did anyone notice that $(1$) and $(2)$ are the same formula?
HINT:
Let the slope intercept from be $$y=mx+c$$
Set $(-8,1);(-8,7)$ to find two linear simultaneous equation in $m,c$