Respuesta :

[tex]\bf (\stackrel{x_1}{-5}~,~\stackrel{y_1}{7})\qquad (\stackrel{x_2}{-4}~,~\stackrel{y_2}{0}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{0-7}{-4-(-5)}\implies \cfrac{0-7}{-4+5}\implies \cfrac{-7}{1}\implies -7 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-7=-7[x-(-5)]\implies y-7=-7(x+5)[/tex]

The equation of the straight line joining both two points is given by

y-7= - 7( x + 5).

What is the slope of a line which passes through points ( p,q) and (x,y)?

Its slope would be:

[tex]m = \dfrac{y-q}{x-p}[/tex]

So the slope of the given points are;

[tex]m = \dfrac{y-q}{x-p} m = \dfrac{0-7}{-4+ 5}\\\\m = -7[/tex]

We have given points are (-5,7) and (-4,0), then the equation of the straight line joining both two points is given by

[tex](y - y_1) = \dfrac{y_2 - y_1}{x_2 - x_1} (x -x_1)\\(y - 7) = -7 (x -(-5))\\\\(y - 7) = -7 (x + 5)[/tex]

Learn more about slope here:

https://brainly.com/question/2503591

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