Question

Which equations represent the line that is perpendicular to the line 5x − 2y = −6 and passes through the point (5, −4)? Check all that apply. y = –2/5x – 2 2x + 5y = −10 2x − 5y = −10 y + 4 = –2/5(x – 5) y – 4 =5/2 (x + 5)

Answer 1

Let's rewrite each equation in the Slope-Intercept Form of the Equation of a Line. First, let's start with the main equation:

$\bullet \ 5x-2y=-6 \therefore y=\frac{5}{2}x+3$

Then, our options are the following:

$A) \ y = -\frac{2}{5}x-2 \\ \\ B) \ 2x+5y=-10 \therefore y=-\frac{2}{5}x-2 \\ \\ C) \ 2x-5y=-10 \therefore y=\frac{2}{5}x+2 \\ \\ D) \ y+4=-\frac{2}{5}(x-5) \therefore y=-\frac{2}{5}x-2 \\ \\ E) \ y-4=\frac{5}{2}(x + 5) \therefore y=\frac{5}{2}x+\frac{33}{2}$

For two perpendicular lines it is true that the product of its slopes is:

$m_{1}m_{2}=-1$

$m_{1}m_{2}=-1 \\ \\ m_{1} \ is \ the \ slope \ of \ y=\frac{5}{2}x+3, \ that \ is, \ m_{1}=\frac{5}{2} \\ \\ Then, \ the \ slope \ of \ a \ perpendicular \ line \ is: \\ \\ m_{2}=-\frac{2}{5}$

According to this, only A) B) and D) might be the perpendicular lines we are looking for. Notice that these lines are the same. The other condition is that the line must pass through the point (5, -4). By substituting this point in the equation, we have:

$y = -\frac{2}{5}x-2 \\ \\ -4=-\frac{2}{5}(5)-2 \\ \\ -4=-2-2 \\ \\ \boxed{-4=-4} \ True!$

Finally, the right answer are:

$A) \ y = -\frac{2}{5}x-2 \\ \\ B) \ 2x+5y=-10 \\ \\ D) \ y+4=-\frac{2}{5}(x-5)$

Answer 2

The perpendicular equation is y = -2/5x - 2

You can find this by first realizing that perpendicular lines have opposite and reciprocal slopes. So since it starts at 5/2 we flip it and make it a negative and the new slope is -2/5. Now we can use that and the point to get the y intercept using slope intercept form.

y = mx + b

-4 = (-2/5)(5) + b

-4 = -2 + b

-2 = b

And now we can use our new slope and new intercept to model the equation.

y = -2/5x - 2

This equation is also equal to 2x + 5y = -10 and y + 4 = -2/5(x - 5)