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ohaa [14]
4 years ago
11

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 stan

dard 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.
Mathematics
2 answers:
svetlana [45]4 years ago
8 0

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}  

<u>With the slope m and the point A find the equation of the line</u>

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

Andreyy894 years ago
5 0
\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
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