Precalculus (6th Edition) Blitzer

Point-slope form is $y+4=-\frac{3}{5}\left( x-10 \right)$. Slope-intercept form is $y=-\frac{2}{3}x+2$.
For a line with slope m and passing through the point $\left( {{x}_{1}},{{y}_{1}} \right)$ , the point slope form is given as: $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ Consider $\left( {{x}_{1}},{{y}_{1}} \right)=\left( 10,-4 \right)$ Change the values of m and $\left( {{x}_{1}},{{y}_{1}} \right)$ as: \begin{align} & y-\left( -4 \right)=-\frac{3}{5}\left( x-10 \right) \\ & y+4=-\frac{3}{5}\left( x-10 \right) \\ & y+4=-\frac{3}{5}x+6 \end{align} Subtract $4$ from both sides of the above equation: \begin{align} & y=-\frac{3}{5}x+6-4 \\ & =-\frac{3}{5}x+2 \end{align} The point-slope form of a line is $y=-\frac{3}{5}x+2$. The slope-intercept form of the equation of a line is given by $y=mx+b$ ; here, m is the slope and b is the y-intercept, and the y-intercept is the y-coordinate of a point where the line intersects the y-axis. Change the values of m and $\left( x,y \right)=\left( 10,-4 \right)$ in $y=mx+b$ to find the value of b: \begin{align} & -4=-\frac{3}{5}\left( 10 \right)+b \\ & b=2 \end{align} Now, change the values of b and m in $y=mx+b$. Thus $y=-\frac{3}{5}x+2$ The slope-intercept form of the line is $y=-\frac{3}{5}x+2$. Therefore, the point-slope form with $m=-\frac{3}{5}$ and passing through the point $\left( 10,-4 \right)$ is $y+4=-\frac{3}{5}\left( x-10 \right)$ , and the slope-intercept form is $y=-\frac{3}{5}x+2$.