Question

Write a linear equation in slope intercept from that is perpendicular to x−3y=3 and passes through the point (5, −9).

Answer 1

Answer:

$y=-3x+6$

Step-by-step explanation:

First, let’s determine the slope of the original equation.

We have $x-3y=3$

Subtract x from both sides and then divide both sides by -3. So:

$-3y=-x+3\\y=\frac{1}{3}x-1$

Therefore, the slope of our original line is 1/3.

Remember that perpendicular lines have slopes that are negative reciprocals of each other.

Therefore, the slope of our new perpendicular line is the negative reciprocal of 1/3.

Thus, the slope of our new line will be -3. We flip the fraction and add a negative.

So, the slope of our new line is -3. We also know that it passes through the point (5, -9).

Now, we can use the point-slope form:

$y-y_1=m(x-x_1)$

Where m is the slope and (x₁, y₁) is a point.

So, substitute -3 for m and (5, -9) for (x₁, y₁). This yields:

$y-(-9)=-3(x-5)$

We want our line in slope-intercept form. So, distribute the right:

$y+9=-3x+15$

Subtract 9 from both sides. Therefore, our equation is:

$y=-3x+6$

Answer 2

Answer:

5

Step-by-step explanation: