Respuesta :
The general equation of a line passing through two points (xb₁,y₁)Pxb₂,y₂) is expressed as
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ \text{where} \\ m\Rightarrow slope\text{ of the line, expr}essed\text{ as }m\text{ = }\frac{y_2-y_1}{x_2-x_1} \\ (x_1,y_1)\Rightarrow coordinate_{}\text{ of point P} \\ (x_2,y_2)\Rightarrow coordinate_{}\text{ of point Q} \end{gathered}[/tex]Given that the coordinates of the two points are (-2, -3) and (3, -5), we have
[tex]\begin{gathered} (x_1,y_1)\Rightarrow(-2,\text{ -3)} \\ (x_2,y_2)\Rightarrow(3,\text{ -5)} \end{gathered}[/tex]Step 1:
Evaluate the slope o the line.
The slope is thus evaluated as
[tex]\begin{gathered} m\text{ = = }\frac{y_2-y_1}{x_2-x_1} \\ \text{ = }\frac{\text{-5-(-3)}}{3-(-2)} \\ =\frac{-5+3}{3+2} \\ \Rightarrow m\text{ = -}\frac{2}{5} \end{gathered}[/tex]Step 2:
Substitute the values of x₁,
Thus, we have
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ x_1=-2 \\ y_1=-3 \\ m\text{ =- }\frac{2}{5} \\ \text{thus,} \\ y-(-3)\text{ = -}\frac{2}{5}(x-(-2)) \\ y+3\text{ =- }\frac{2}{5}(x+2) \end{gathered}[/tex]Step 3:
Make .
[tex]\begin{gathered} y+3\text{ =- }\frac{2}{5}(x+2) \\ \text{Multiply both sides of the equation by 5 } \\ 5(y+3)\text{ = -2(x+2)} \\ \text{open brackets} \\ 5y\text{ + 15 =- 2x - 4} \\ \Rightarrow5y\text{ =- 2x - 4 -15} \\ 5y\text{ = -2x-1}9 \\ \text{divide both sides of the equation by the coefficient of y, which is 5.} \\ \text{thus,} \\ \frac{5y}{5}=\frac{-\text{2x-1}9}{5} \\ \Rightarrow y\text{ =- }\frac{2}{5}x\text{ - }\frac{19}{5} \end{gathered}[/tex]Hence, the equation of the line is
[tex]y\text{ = -}\frac{2}{5}x\text{ - }\frac{19}{5}[/tex]y₁ and m into the general equation of the line.
