Respuesta :
[tex]\bf (\stackrel{x_1}{8}~,~\stackrel{y_1}{-8})~\hspace{10em} slope = m\implies -\cfrac{5}{4} \\\\\\ \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-(-8)=-\cfrac{5}{4}(x-8)\implies y+8=-\cfrac{5}{4}(x-8)[/tex]
The equation in point-slope form for this line is 4y = -5x + 8.
What is point-slope form of equation of straight line ?
The equation of a straight line in the form y = mx + c where m is the slope of the line and c is its y-intercept is known as the point-slope form. Here both the slope (m) and y-intercept (c) have real values. It is known as point-slope form as it gives the definition of both the slope, y-intercept and the points mentioned in the line.
How to form the given equation of straight line ?
It is given that the line passes through the point (8.-8) and has a slope of -5/4.
The general representation for straight line is y = mx + c .
Here, m = -5/4 , x = 8 and y = -8 .
Thus we have ,
⇒ -8 = -5/4 * 8 + c
⇒ c = 10 - 8
∴ c = 2
The y-intercept is 2.
The equation of straight line becomes,
⇒ y = -(5/4)x + 2
∴ 4y = -5x + 8
Therefore, the equation in point-slope form for this line is 4y = -5x + 8.
To learn more about point-slope form, refer -
https://brainly.com/question/18617367
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