MHB Unknown_12's questions at Yahoo Answers regarding analytic geometry

AI Thread Summary
The discussion revolves around solving analytic geometry problems involving slopes and angles in triangles. The first question involves finding the interior angles of a triangle with given vertices, where the calculated angles A, B, and C sum to approximately 180 degrees. The second question focuses on determining the value of y in a line equation, which is solved by using the property that the product of the slopes of perpendicular lines equals -1. There is clarification on the use of the tangent formula for calculating angles, emphasizing the need for the angle-difference identity. Ultimately, the user expresses satisfaction upon resolving their queries.
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Here are the questions:

Question on analytic geometry about slopes?


Find the interior angle of the triangle with vertices (2, -5), (6,2), (4,1)

It must 180 degrees when calculated but stuck on the solution.

The line through (-2, y) and (2, 10) is perpendicular to a line through (-3, -7) and (5, -5) find y.

The formula will be like y2-y1 / x2- x1

and formula for angle tan m2-m1/1+m1(m2)

solution of the answer is appreciated.

I have posted a link there to this thread so the OP can see my work.
 
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Hello unknown_12,

1.) Let's take a look at the triangle in the plane with the given vertices:

https://www.physicsforums.com/attachments/1709._xfImport

The slope of line segment $a$ is:

$$m_a=\frac{2-1}{6-4}=\frac{1}{2}$$

The slope of line segment $b$ is:

$$m_b=\frac{1-(-5)}{4-2}=3$$

The slope of line segment $c$ is:

$$m_c=\frac{2-(-5)}{6-2}=\frac{7}{4}$$

Hence, angle $A$ is:

$$A=\tan^{-1}\left(m_b \right)-\tan^{-1}\left(m_c \right)=\tan^{-1}\left(3 \right)-\tan^{-1}\left(\frac{7}{4} \right)\approx0.197395559849881$$

Angle $B$ is:

$$B=\tan^{-1}\left(m_c \right)-\tan^{-1}\left(m_a \right)=\tan^{-1}\left(\frac{7}{4} \right)-\tan^{-1}\left(\frac{1}{2} \right)\approx0.588002603547568$$

Angle $C$ is:

$$C=\pi-\left(\tan^{-1}\left(m_b \right)-\tan^{-1}\left(m_a \right) \right)=\pi+\tan^{-1}\left(\frac{1}{2} \right)-\tan^{-1}(3)=\frac{3\pi}{4}$$

As a check, we see that:

$$A+B+C=\pi$$

2.) The slope of the first line is:

$$m_1=\frac{10-y}{2-(-2)}=\frac{10-y}{4}$$

The slope of the second line is:

$$m_2=\frac{-5-(-7)}{5-(-3)}=\frac{1}{4}$$

When two lines are perpendicular, the product of their slopes is $-1$, as proven here:

http://mathhelpboards.com/math-notes-49/perpendicular-lines-product-their-slopes-2953.html

Hence, we must have:

$$\frac{10-y}{4}\cdot\frac{1}{4}=-1$$

$$10-y=-16$$

$$y=26$$
 

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Hello, I'm the one posted from yahoo answer; is that same from the tangent formula that I've posted?

Because when I used the formula, it couldn't provide a 180 degrees.
 
Last edited:
jamescv31 said:
Hello, I'm the one posted from yahoo answer; is that same from the tangent formula that I've posted?

Hello and welcome to MHB, jamescv31! :D

I take it you are supposed to use the angle-difference identity for the tangent function instead of the product of the slopes.

$$\theta_1-\theta_2=\frac{\pi}{2}$$

Taking the tangent of both sides, we get:

$$\tan\left(\theta_1-\theta_2 \right)=\tan\left(\frac{\pi}{2} \right)$$

$$\frac{m_1-m_2}{1+m_1m_2}=\tan\left(\frac{\pi}{2} \right)$$

Since $$\tan\left(\frac{\pi}{2} \right)$$ is undefined, we see that we require:

$$1+m_1m_2=0$$

$$m_1m_2=-1$$
 
jamescv31 said:
...Because when I used the formula, it couldn't provide a 180 degrees.

This was added while I was composing my reply. I assume this refers to the first problem. You want the sum of the 3 angles to be $180^{\circ}=\pi$.

Do you understand how I computed the values of the individual angles?
 
Yes I'm referring on the interior problem, since our teacher provided this formula only for the discussion

the Tangent = m2+m1/ 1 + m1m2

Which the goal is to have an equal 180 degrees or 179.9 as possible.
 
Could I manage to have an exact 180 degrees with the formula of tangent
m2-m1/1+m1m2 ?

- - - Updated - - -

Well nevermind, got already.

Thank you for the time. :)
 
jamescv31 said:
Yes I'm referring on the interior problem, since our teacher provided this formula only for the discussion

the Tangent = m2+m1/ 1 + m1m2

Which the goal is to have an equal 180 degrees or 179.9 as possible.

Okay, well using the formula you originally gave, we find:

$$A=\tan^{-1}\left(\frac{m_b-m_c}{1+m_am_b} \right)=\tan^{-1}\left(\frac{3-\frac{7}{4}}{1+3\cdot\frac{7}{4}} \right)=\tan^{-1}\left(\frac{1}{5} \right)\approx0.197395559849881$$

$$B=\tan^{-1}\left(\frac{m_c-m_a}{1+m_am_c} \right)=\tan^{-1}\left(\frac{\frac{1}{2}-\frac{7}{4}}{1+\frac{1}{2}\cdot\frac{7}{4}} \right)=\tan^{-1}\left(\frac{2}{3} \right)\approx0.588002603547568$$

$$C=\pi+\tan^{-1}\left(\frac{m_a-m_b}{1+m_am_b} \right)=\pi+\tan^{-1}\left(\frac{\frac{1}{2}-3}{1+\frac{1}{2}\cdot3} \right)=\pi+\tan^{-1}\left(-1 \right)=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$$

Note: For angle $C$ it was necessary to add $\pi$ to get the angle in the correct quadrant.

If we convert the angles to degrees, we find:

$$A\approx11.31^{\circ}$$

$$B\approx33.69^{\circ}$$

$$C=135^{\circ}$$

We see then that:

$$A+B+C\approx180^{\circ}$$
 
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