+26393 How do u put this in slope form
(10,33) (5,15)
ex: y=mx+b
\(\text{Point 1: } ( x_1 = 10,\ y_1 = 33) \\ \text{Point 2: }(x_2 = 5,\ y_2 = 15 )\)
\(\begin{array}{rcll} m &=& \frac{y_2-y_1}{x_2-x_1} \\\\ m &=& \frac{15-33}{5-10} \\\\ m &=& \frac{-18}{-5} \\\\ m &=& \frac{18}{5} \\\\ \mathbf{m} &\mathbf{=}& \mathbf{3.6} \\\\ \hline \\ \frac{y-y_1}{x-x_1} &=& m\\\\ \frac{y-33}{x-10} &=& 3.6 \qquad & | \qquad \cdot (x-10) \\\\ y-33 &=& 3.6\cdot( x-10 ) \qquad & | \qquad +33 \\\\ y &=& 3.6\cdot( x-10 ) +33 \\\\ y &=& 3.6\cdot x -3.6\cdot 10 +33 \\\\ y &=& 3.6\cdot x -36 +33 \\\\ \mathbf{y} &\mathbf{=}& \mathbf{ 3.6\cdot x -3 } \end{array}\)
+26393 How do u put this in slope form
(10,33) (5,15)
ex: y=mx+b
\(\text{Point 1: } ( x_1 = 10,\ y_1 = 33) \\ \text{Point 2: }(x_2 = 5,\ y_2 = 15 )\)
\(\begin{array}{rcll} m &=& \frac{y_2-y_1}{x_2-x_1} \\\\ m &=& \frac{15-33}{5-10} \\\\ m &=& \frac{-18}{-5} \\\\ m &=& \frac{18}{5} \\\\ \mathbf{m} &\mathbf{=}& \mathbf{3.6} \\\\ \hline \\ \frac{y-y_1}{x-x_1} &=& m\\\\ \frac{y-33}{x-10} &=& 3.6 \qquad & | \qquad \cdot (x-10) \\\\ y-33 &=& 3.6\cdot( x-10 ) \qquad & | \qquad +33 \\\\ y &=& 3.6\cdot( x-10 ) +33 \\\\ y &=& 3.6\cdot x -3.6\cdot 10 +33 \\\\ y &=& 3.6\cdot x -36 +33 \\\\ \mathbf{y} &\mathbf{=}& \mathbf{ 3.6\cdot x -3 } \end{array}\)
