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How do you write an equation for the line perpendicular to #y=2x-5# that contains (-9,6)?

Answer 1

Eva Adamson

The perpendicular would have its slope= #-1/2#. Since it contains (-9,6), its equation would be y-6= #-1/2 (x+9)#.

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Answer 2

Eva Adamson

First, determine the slope of the line y = 2x - 5. Since it's in the form y = mx + b, the slope is 2. The slope of a line perpendicular to this line will be the negative reciprocal of 2, which is -1/2.

Next, use the point-slope form of a line (y - y₁ = m(x - x₁)), where (x₁, y₁) is the given point (-9, 6) and m is the slope (-1/2).

Substitute the values into the equation:

y - 6 = (-1/2)(x - (-9))

Simplify:

y - 6 = (-1/2)(x + 9)

Multiply both sides by -2 to clear the fraction:

-2(y - 6) = x + 9

Distribute -2:

-2y + 12 = x + 9

Rearrange to isolate x:

x = -2y + 3

Therefore, the equation for the line perpendicular to y = 2x - 5 that contains the point (-9,6) is x = -2y + 3.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.