Respuesta :
[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]
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]
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