Question

write an equation of a line passing through the point (-6,-3) and perpendicular to JK with J (-2, -7) and K (6,5)

Answer 1

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:

`\text{Slope}=((y_2-y_1))/((x_2-x_1))`

Replacing the ordered pairs J=(-2,-7) and K=(6,5) give us the slope:

`\text{Slope}=((5-(-7)))/((6-(-2)))=(12)/(8)=(3)/(2)`

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