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A line passes through (3, –2) and (8, 2).
a. Write an equation for the line in point-slope form.
b. Rewrite the equation in standard form using integers.

a. y – 2 = 4/5(x – 3); –4x + 5y = 22
b. y + 2 = 4/5(x + 3); –4x + 5y = –22
c. y – 3 = 4/5(x + 2); –4x + 5y = 23
d. y + 2 = 4/5(x - 3); -4x + 5y = -22



Sagot :

Naǫidn
The point-slope form:
[tex]y-y_1=m(x-x_1)[/tex]
m - slope
(x₁,y₁) - point

[tex](3,-2) \\ x_1=3 \\ y_1=-2 \\ \\ (8,2) \\ x_2=8 \\ y_2=2 \\ \\ \hbox{the slope:} \\ m=\frac{y_2-y_1}{x_2-x_1}=\frac{2-(-2)}{8-3}=\frac{2+2}{5}=\frac{4}{5} \\ \\ \hbox{the point-slope form:} \\ y-y_1=m(x-x_1) \\ y-(-2)=\frac{4}{5}(x-3) \\ \boxed{y+2=\frac{4}{5}(x-3)}[/tex]

[tex]\hbox{the standard form:} \\ y+2=\frac{4}{5}(x-3) \\ y+2=\frac{4}{5}x-\frac{12}{5} \\ -\frac{4}{5}x+y=-\frac{12}{5}-2 \ \ \ |\times 5 \\ -4x+5y=-12-10 \\ \boxed{-4x+5y=-22}[/tex]

The answer is D.
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