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What is an equation of the line that passes through the point (1,6) and is perpendicular to the line x+3y=27

User Tarc
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1 Answer

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So here, we have an equation in standard form.

x+3y=27

We need to solve this for y.

Subtracting x from both sides we get

3y=27-x

Dividing both sides by 3 gives us

y=9-(x/3)


Given that the slope is -1/3, we need the slope perpendicular to that. Therefore, the slope perpendicular to -1/3 will be 3. Thus, our slope will be 3. Now we want it to go through (1,6) thus, we can use the point slope formula where we are given (x1,y1) and we have our perpendicular slope, m=3. The point slope formula is defined as

y-y1=m(x-x1) where (1,6) can be (x1,y1) and m=3

Thus we have

y-6=3(x-1)

Using distributive property on the right side of the equation, we get

y-6=3x-3

Adding 6 to both sides we get

y=3x+3

So that is your equation. Hope this helps
User TheGrandWazoo
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