Answer:
[tex]\text{point-slope form}\\\huge\boxed{y+3=\dfrac{7}{5}(x+7)}[/tex]
[tex]\text{slope-intercept form}\\\huge\boxed{y=\dfrac{7}{5}x+\dfrac{34}{5}}[/tex]
[tex]\text{standard form}\\\huge\boxed{7x-5y=-34}[/tex]
[tex]\text{general form}\\\huge\boxed{7x-5y+34=0}[/tex]
Step-by-step explanation:
The slope-point form of an equation of a line:
[tex]y-y_1=m(x-x_1)[/tex]
where
[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]
From the graph we have two points (-7, -3) and (-2, 4).
Substitute:
[tex]m=\dfrac{4-(-3)}{-2-(-7)}=\dfrac{4+3}{-2+7}=\dfrac{7}{5}\\\\y-(-3)=\dfrac{7}{5}(x-(-7))\\\\y+3=\dfrac{7}{5}(x+7)[/tex]
Convert to the slope-intercept form (y = mx + b):
[tex]y+3=\dfrac{7}{5}(x+7)\qquad|\text{use the distributive property}\\\\y+3=\dfrac{7}{5}x+\dfrac{49}{5}\qquad|\text{subtract}\ 3=\dfrac{15}{5}\ \text{from both sides}\\\\y=\dfrac{7}{5}x+\dfrac{34}{5}[/tex]
Convert to the standard form (Ax + By = C):
[tex]y=\dfrac{7}{5}x+\dfrac{34}{5}\qquad|\text{multiply both sides by 5}\\\\5y=7x+34\qquad\text{subtract}\ 7x\ \text{from both sides}\\\\-7x+5y=34\qquad|\text{change the signs}\\\\7x-5y=-34[/tex]
Convert to the general form (Ax + By + C = 0):
[tex]7x-5y=-34\qquad|\text{add 34 to both sides}\\\\7x-5y+34=0[/tex]
