Answer:
[tex]\large\boxed{y=-\dfrac{2}{3}x+\dfrac{13}{3}}[/tex]
Step-by-step explanation:
[tex]\text{The slope-intercept form of an equation of a line:}\\\\y=mx+b\\\\m-slope\\b-y-intercept\\\\\text{The formula of a slope:}\\\\m=\dfrac{y_2-y_1}{x_2-x_1}\\\\===============================[/tex]
[tex]\text{We have the point:}\\\\(5,\ 1)\ \text{and}\ (-4,\ 7).\ \text{Substitute:}\\\\m=\dfrac{7-1}{-4-5}=\dfrac{6}{-9}=-\dfrac{6:3}{9:3}=-\dfrac{2}{3}\\\\\text{We have the equation in form:}\\\\y=-\dfrac{2}{3}x+b\\\\\text{Put the coordinates of the point (5, 1) to the equation:}\\\\1=-\dfrac{2}{3}(5)+b\\\\1=-\dfrac{10}{3}+b\qquad\text{add}\ \dfrac{10}{3}\ \text{to the both sides}\\\\\dfrac{3}{3}+\dfrac{10}{3}=b\to b=\dfrac{13}{3}\\\\\text{Finally:}\\\\y=-\dfrac{2}{3}x+\dfrac{13}{3}[/tex]
