Question

a line passes through (3, –2) and (6, 2). a. write an equation for the line in point-slope form. b. rewrite the equation in standard form using integers. a. y minus 2 equals four thirds times the quantity x minus 3 end of quantity; negative 4 x plus 3 y equals 18. b. y plus 2 equals four thirds times the quantity x plus 3 end of quantity; negative 4 x plus 3 y equals negative eighteen. c. y minus 3 equals four thirds times the quantity x plus 2 end of quantity; negative 4 x plus 3 y equals 17. d. y plus 2 equals four thirds times the quantity x minus 3 end of quantity; negative 4 x plus 3 y equals negative eighteen.

Answer 1

Answer:

Part A) $y+2=\frac{4}{3}(x-3)$ ----> y plus $2$ equals four thirds times the quantity x minus $3$ end of quantity

Part B)  $-4x+3y=-18$ ---> negative $4x$ plus $3y$ equals negative eighteen

The answer is the option D

Step-by-step explanation:  

Part A) we know that

The equation of a line into point slope for is equal to

$y-y1=m(x-x1)$

Find the slope

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

we have

$A(3,-2)\ B(6,2)$

Substitute the values

$m=\frac{2+2}{6-3}$  

$m=\frac{4}{3}$  

With the slope m and the point A find the equation of the line

$y+2=\frac{4}{3}(x-3)$ -----> equation of the line into point slope form

Part B) we know that

The equation of the line into standard form is equal to

$Ax+By=C$

we have

$y-2=\frac{4}{3}(x-6)$ ------> convert to standard form

$y=\frac{4}{3}x-8+2$

$y=\frac{4}{3}x-6$

Multiply by $3$ both sides

$3y=4x-18$    

$-4x+3y=-18$ ----> equation of the line in standard form

Answer 2

$\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1} \\ \frac{y-(-2)}{x-3} = \frac{2-(-2)}{6-3} \\ \frac{y+2}{x-3} = \frac{2+2}{6-3} \\ \frac{y+2}{x-3} = \frac{4}{3} \\ y+2= \frac{4}{3} (x-3)$
In standard form
$y+2= \frac{4}{3} (x-3) \\ 3(y+2)=4(x-3) \\ 3y+6=4x-12 \\ -4x+3y=-18$