Question

Graph the line that contains the point P and has slope m. Find the point-slope form of the equation of the line.

Answer 1

Given point p and slope m, below

`\begin{gathered} P(1,3) \\ m=(-3)/(4) \end{gathered}`

The formula for finding point slope point of a straight-line graph is given below

`\begin{gathered} (y-y_1)/(x-x_1)=m \\ \text{where} \\ (x_1,y_1)is\text{ the point given} \end{gathered}`
`x_1=1,y_1=3,m=(-3)/(4)`

Substitute the given into the formula

`(y-3)/(x-1)=(-3)/(4)`
`\begin{gathered} \text{cross}-\text{mulyiply} \\ 4(y-3)=-3(x-1)_{} \\ 4y-12=-3x+3 \\ 4y+3x=+3+12 \\ 4y+3x=+15 \\ 4y+3x=15 \end{gathered}`

The graph is as shown below:

Hence, the slope-point form of the given point and slope is 4y+3x=15