Asked by JaNelle

How do write the equation of a line passing through the point (-6, -3) and perpendicular to JK with J(-2,-7) and K(6,5)
Answered by oobleck
JK has slope (5+7)/(6+2) = 3/2
so we want slope -2/3
y+3 = -2/3 (x+6)
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