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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.

Respuesta :

nobillionaireNobley
[tex] \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)[/tex]
In standard form
[tex]y+2= \frac{4}{3} (x-3) \\ 3(y+2)=4(x-3) \\ 3y+6=4x-12 \\ -4x+3y=-18[/tex]
calculista

Answer:

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

Part B)  [tex]-4x+3y=-18[/tex] ---> negative [tex]4x[/tex] plus [tex]3y[/tex] 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

[tex]y-y1=m(x-x1)[/tex]

Find the slope

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


[tex]m=\frac{y2-y1}{x2-x1}[/tex]


we have


[tex]A(3,-2)\ B(6,2)[/tex]


Substitute the values


[tex]m=\frac{2+2}{6-3}[/tex]  

[tex]m=\frac{4}{3}[/tex]  

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

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

Part B) we know that

The equation of the line into standard form is equal to

[tex]Ax+By=C[/tex]

we have

[tex]y-2=\frac{4}{3}(x-6)[/tex] ------> convert to standard form

[tex]y=\frac{4}{3}x-8+2[/tex]

[tex]y=\frac{4}{3}x-6[/tex]

Multiply by [tex]3[/tex] both sides

[tex]3y=4x-18[/tex]    

[tex]-4x+3y=-18[/tex] ----> equation of the line in standard form

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