Given point p and slope m, below
[tex]\begin{gathered} P(1,3) \\ m=\frac{-3}{4} \end{gathered}[/tex]
The formula for finding point slope point of a straight-line graph is given below
[tex]\begin{gathered} \frac{y-y_1}{x-x_1}=m \\ \text{where} \\ (x_1,y_1)is\text{ the point given} \end{gathered}[/tex][tex]x_1=1,y_1=3,m=\frac{-3}{4}[/tex]
Substitute the given into the formula
[tex]\frac{y-3}{x-1}=\frac{-3}{4}[/tex][tex]\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}[/tex]
The graph is as shown below:
Hence, the slope-point form of the given point and slope is 4y+3x=15
