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2 years ago

Perpendicular lines AB and CD intersect at point E. If m2AED = x+ 20, what is the value of

Mathematics

1 answer:

Leona [35]2 years ago

5 0

Answer:

x = 70

Step-by-step explanation:

Since the lines are perpendicular then ∠ AED = 90°, thus

x + 20 = 90 ( subtract 20 from both sides )

x = 70

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3 years ago

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Which equations represent the line that is perpendicular to the line 5x − 2y = −6 and passes through the point (5, −4)? Check al

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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:

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

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