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What is the equation of a line that passes through the point (6,-8) and is perpendicular to the line whose equation is Y=2/3 x+3

User Ronenz
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Question-

What is the equation of a line that passes through the point (6,-8) and is perpendicular to the line whose equation is Y=2/3 x+3

Answer-

y = x/2 + 3......x/2 is the same as 1/2x....so u have a slope of 1/2.

A perpendicular line will have a negative reciprocal slope. To find the negative reciprocal, flip the slope and change the sign. So we have 1/2....and we flip it making it 2/1....and we change the sign making it -2/1 or just -2. So ur perpendicular line will have a slope of -2.


y = mx + b

slope(m) = -2

(1,8)...x = 1 and y = 8

now we sub and find b, the y int

8 = -2(1) + b

8 = -2 + b

8 + 2 = b

10 = b

so ur perpendicular equation is : y = -2x + 10 <==


Hope i helped have a wonderful day...

User Cody
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