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Find the equation of a line perpendicular to y=x-3 that contains the point (2,1)

1 Answer

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Answer:

y = -x + 3

Explanation:

y = x - 3 This equation has a slope of 1

A line that would be perpendicular to the above line would then need a slope of -1

If the line has a slope of -1 and passes through (2, 1) then we can plug into

y = mx + b to get the 'b' value

1 = (-1)(2) + b

1 = -2 + b

3 = b

Equation would be y = -x + 3

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