The point slope form of the equation of the line that passes through (-5-1) and (10.-7) isstandard form of the equation for this line?

Answer:The standard form of the equation for this line can be:l: 2x + 5y = -15.Step-by-step explanation:Start by finding the slope of this line. For a line that goes through the two points [tex](x_0, y_0)[/tex] and [tex](x_1, y_1)[/tex],[tex]\displaystyle \text{Slope} = \frac{y_{1} - y_{0}}{x_{1} - x_{0}}[/tex].For this line,[tex]\displaystyle \text{Slope} = \frac{(-1) - (-7)}{(-5) - 10} = -\frac{2}{5}[/tex].Find the slope-point form of this line's equation using[tex]\displaystyle \text{Slope} = -\frac{2}{5}[/tex], andThe point [tex](-5, -1)[/tex] (using the point [tex](10, -7)[/tex] should also work.)The slope-point form of the equation of a line with slope [tex]m[/tex] andpoint [tex](x_{0}, y_{0})[/tex] should be [tex]l:\; y - y_{0} = m(x - x_0)[/tex].For this line,[tex]\displaystyle m = -\frac{2}{5}[/tex], and[tex]x_0 = -5[/tex], and[tex]y_0 = -1[/tex].The equation in slope-point form will be [tex]\displaystyle l:\; y - (-1) = -\frac{2}{5}(x - (-5))[/tex].The standard form of the equation of a line in a cartesian plane is[tex]l: \; ax + by = c[/tex] where[tex]a[/tex], [tex]b[/tex], and [tex]c[/tex] are integers. [tex]a \ge 0[/tex].Multiply both sides of the slope-point form equation of this line by [tex]5[/tex]:[tex]l:\; 5 y + 5 = -2x -10[/tex].Add [tex](2x-5)[/tex] to both sides of the equation:[tex]l: \; 2x + 5y = -15[/tex].Therefore, the equation of this line in standard form is [tex]l: \; 2x + 5y = -15[/tex].