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Write the equation of the line perpendicular to y=3x-1 that passes through the point (3,-4)

explanation or step by step pls

thanku

1 Answer

5 votes

Answer:


y=-(1)/(3)x - 3

Explanation:

Hi there!

We want to write an equation of the line that is perpendicular to y=3x-1 and passes through (3, -4)

Perpendicular lines have slopes that multiply to get the value of negative one.

Let's first figure out the slope of y=3x-1

The line is written in slope-intercept form, which is y=mx+b, where m is the slope and b is the y intercept

3 is in the place of where m is, so 3 is the slope of y=3x-1

Now let's find the slope of the line perpendicular to it.

We can use a formula like this:


m_1 * m_2= -1

If 3 is
m_1, then:


3*m_2=-1

Divide both sides by 3


m_2=-(1)/(3)

So the slope of the perpendicular line is -1/3

We can write this equation in slope-intercept form. So far, we know that the equation of the line is:

y=
-(1)/(3)x + b

So let's find b

Remember that we are given that the line passes through the point (3, -4). That means that point is a solution to the equation that we're trying to find. Therefore, the values of that point should create a true statement when plugged into the equation.

So we can substitute 3 as x and -4 as y to help solve for b.

-4=
-(1)/(3)(3)+b

Multiply

-4=
-(3)/(3) + b

Simplify

-4 = -1 + b

Add 1 to both sides to isolate b

-3 = b

Substitute -3 as b in the equation


y=-(1)/(3)x - 3

Hope this helps!

User Damoiskii
by
7.7k points

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