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).
A) y = -2/3x – 4
B) y=-2/33-7
C) y = 2/3x + 1
D p = 2/3x +9
E) y = -3/23-4

Answer 1

The equation of the line passing through (-6,-3) and perpendicular to JK is A) y = -2/3x - 4.

Step-by-step explanation:

To find the equation of a line passing through the point (-6,-3) and perpendicular to line JK, we need to find the slope of JK and then find the negative reciprocal of it.

The slope of JK can be found using the formula:

slope = (y2 - y1) / (x2 - x1)

Plugging in the values, we get:

slope = (5 - (-7)) / (6 - (-2)) = 12/8 = 3/2

The negative reciprocal of 3/2 is -2/3. So, the equation of the line passing through (-6,-3) and perpendicular to JK is y = -2/3x - 4.

Answer 2

sorry I can't answer because I can't find the answer