MATH SOLVE

2 months ago

Q:
# Which is an equation of the given line in standard form?A. -6x+5y=13B. -6x+7y=-11C. -6x+7y=11D. -7x+6y=-11

Accepted Solution

A:

The slope:

[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]

The coordinates of the points:

[tex](-3;\ -1)\to x_1=-3;\ y_1=-1;\ \left(\dfrac{1}{2};\ 2\right)\to x_2=\dfrac{1}{2};\ y_2=2[/tex]

Substitute:

[tex]m=\dfrac{2-(-1)}{\frac{1}{2}-(-3)}=\dfrac{3}{3\frac{1}{2}}=\dfrac{3}{\frac{7}{2}}=3\cdot\dfrac{2}{7}=\dfrac{6}{7}[/tex]

We have:

[tex]y=\dfrac{6}{7}x+b[/tex]

Substitute the coordinates of the point (-3; -1) to the equation of the line:

[tex]-1=\dfrac{6}{7}\cdot(-3)+b\\\\-1=-\dfrac{18}{7}+b\ \ \ |+\dfrac{18}{7}\\\\b=\dfrac{11}{7}[/tex]

Therefore we have:

[tex]y=\dfrac{6}{7}x+\dfrac{11}{7}\ \ \ |\cdot7\\\\7y=6x+11\ \ \ |-6x\\\\\boxed{-6x+7y=11}[/tex]

Answer: C. -6x + 7y = 11

[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]

The coordinates of the points:

[tex](-3;\ -1)\to x_1=-3;\ y_1=-1;\ \left(\dfrac{1}{2};\ 2\right)\to x_2=\dfrac{1}{2};\ y_2=2[/tex]

Substitute:

[tex]m=\dfrac{2-(-1)}{\frac{1}{2}-(-3)}=\dfrac{3}{3\frac{1}{2}}=\dfrac{3}{\frac{7}{2}}=3\cdot\dfrac{2}{7}=\dfrac{6}{7}[/tex]

We have:

[tex]y=\dfrac{6}{7}x+b[/tex]

Substitute the coordinates of the point (-3; -1) to the equation of the line:

[tex]-1=\dfrac{6}{7}\cdot(-3)+b\\\\-1=-\dfrac{18}{7}+b\ \ \ |+\dfrac{18}{7}\\\\b=\dfrac{11}{7}[/tex]

Therefore we have:

[tex]y=\dfrac{6}{7}x+\dfrac{11}{7}\ \ \ |\cdot7\\\\7y=6x+11\ \ \ |-6x\\\\\boxed{-6x+7y=11}[/tex]

Answer: C. -6x + 7y = 11