EXPLANATION
Given the point: (-6,-3) and the vector JK with J=(-2,-7) K=(6,5)
First we need to the slope of the vector applying the slope formula:
[tex]\text{Slope}=\frac{(y_2-y_1)}{(x_2-x_1)}[/tex]
Replacing the ordered pairs J=(-2,-7) and K=(6,5) give us the slope:
[tex]\text{Slope}=\frac{(5-(-7))}{(6-(-2))}=\frac{12}{8}=\frac{3}{2}[/tex]
Now, we have the slope and we can use this to find the line that contains the point (-6, -3) applying the generic form:
y= -2x/3 + b where -2/3 is the negative and reciprocal slope perpendicular to the vector JK.
Finally, replacing the point (-6,-3) give us the y-intercept, b,
-3 = -2(-6)/3 + b
Multiplying terms:
-3 = 12/3 + b ---> -3 = 4 + b
Subtracting 4 to both sides:
-3 - 4 = b
Switching sides:
b= -7
The linear equation is y = (-2/3)x - 7 OPTION B
