# Is this a correct way to solve this high school coordinate geometry question?

Here's the question:

Given point $A$: $(-3;-1)$
Given point $B$: $(3;7)$

Given point $Z$: $(x;0)$

Find the $x$ coordinate of point $Z$ so that the angle of view of AB segment is $90$ degrees at point Z, so Point Z has a $90$ degrees interior angle.

The answer is $4$ and $-4$.

Here's the way I did it, is it correct?

• $AB$ segment has length of $10$.
• So to make $ABZ$ a right triangle, one method we can use is that $AZ$ segment and $BZ$ segment must have the length of $\sqrt{50}$, because then $\sqrt{50}^2 + \sqrt{50}^2 = 10^2$. (Which is the length of $AB$)
• So in that case triangle $ABZ$ would be an isosceles right angled triangle and indeed there would be a $90$ degrees interior angle at point $Z$.
• Then, I used distance formula to calculate the $x$ coordinate of point $Z$ so that $AZ$ and $BZ$ segments would have a length of $\sqrt{50}$.

So that's how I got $4$ and $-4$, and these are the correct answers.

I know there are many possible solutions to solve this problem, but is my method correct?

• Consider formatting with MathJax. Here is a tutorial: meta.math.stackexchange.com/questions/5020/… – M47145 May 22 '16 at 22:06
• @M47145 I edited. Could you please now take a look and tell me if my method is correct? I would appreciate it a lot. – user2547460 May 22 '16 at 22:13
• sorry i meant segments, i edited. is it clearer now? and the question asked for an 45 degrees interior angle at point Z in the ABZ triangle – user2547460 May 22 '16 at 22:24
• what do you mean is not enough? – user2547460 May 22 '16 at 22:26
• @user2547460 I would roll back the last edit, angle of view of a line segment is a perfectly understandable thing. – chx May 22 '16 at 22:31

First of all, the angle of view of a line segment are two arcs and Z are on these arc and also it is on a line, that is the $y=0$ line. Your task is to figure out the formula for the arcs and then find the intersections of $y=0$ on it. Edit: hint, you need to use the Central Angle Theorem to find the centre of the arc/circle.

I do not understand why $ABZ$ is a right triangle in the first place.

You claim

So to make $ABZ$ a right triangle, one method we can use is that $AZ$ segment and $BZ$ segment must have the length of $\sqrt{50}$, because then $\sqrt{50}^2 + \sqrt{50}^2 = 10^2$. (Which is the length of $AB$)

This is wrong. If AZ and BZ equal length and the angle at Z is 45 degrees then the other two angles of triangle is (180-45)/2.

• but if we assume that ABZ is a right triangle then would my method be correct? – user2547460 May 22 '16 at 22:32
• the question said that the angle of view of a AB segment at point Z is a right angle so ABZ must be a right angled triangle – user2547460 May 22 '16 at 22:35
• BLINK what? The question says the angle of views is 45 degrees! Do you know what a right angle in degrees is? I edited my answer with a hint on how to solve this. – chx May 22 '16 at 22:37
• omg, I'am really sorry, i'm an idiot, I didn't copy the question correctly. So to make everything clear, the question said that the angle of views is 90 degrees so that would make ABZ triangle a right angled triangle. again sorry, it's 1am here... – user2547460 May 22 '16 at 22:40
• Pity, it's much more interesting with 45 degrees :) – chx May 22 '16 at 22:40

your method is correct. Good job sorry it took so long, it's hard to review things not in math Jax I highly recommend you learn

• note x^2+y^2=z^2 iff x and y are legs to a right triangle with hypotenuse length z. – shai horowitz May 22 '16 at 22:21
• Thank you. I know it's technically correct because I do indeed end up with the correct answer but is it mathematically correct too? The official answer to this question used the thales theorem, another answer used scalar vector multiplication method. – user2547460 May 22 '16 at 22:22
• as I said your method is correct not just your answer. you may want to include a note about the x^2+y^2=z^2 iff right triangle thing to make it more rigorous – shai horowitz May 22 '16 at 22:24

It is incorrect

First of all,

$Z(x, 0)$ Now for every possible value of x, a line can be made that is intersecting line AB at $45$ degree.

i think you mean point $A$ and $B$ is seen at $45$ degrees.

i got the answer this way.

$m_1$ = $\frac{7+1}{3+3}$

$m_1$ = $\frac{4}{3}$

$tan 45$ = $\frac{-m_1 + m_2}{1 + m_1*m_2}$

$1 + m_1*m_2$ = ${ + m_2 -m_1}$

$1 + \frac{4}{3}*m_2$ = ${+m_2 -\frac{4}{3}}$

$3 + 4*m_2$ = ${-4 + 3*m_2}$

$m_2$ = ${-7}$

$m_2$ = $-7$

$\frac{7-y}{3-x}$ = $-7$

$7x +y -28$ = $0$

Now it is given that y = $0$

$7x - 28$ = $0$

$x$ = $+4$

Do the same for $A$.

• Thanks, the answer sheet says i get partial credits for my answer if I solved the question using the AB segment/AB diameter. Do you think I would get this partial credit? – user2547460 May 22 '16 at 22:58
• @user2547460 Maybe not because if you use that then the answer would come by a slightly different way. not very different but just slightly different. – A---B May 22 '16 at 23:03

So all the fuzz here resulted from some unsharp definitons (where is the angle, what angle). You can try out different $x$ values here.

• thank you, but i know my answer is correct, i was asking if my method was correct to calculate the correct answer :) – user2547460 May 22 '16 at 22:44