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?

  • $\begingroup$ Consider formatting with MathJax. Here is a tutorial: meta.math.stackexchange.com/questions/5020/… $\endgroup$ – M47145 May 22 '16 at 22:06
  • $\begingroup$ @M47145 I edited. Could you please now take a look and tell me if my method is correct? I would appreciate it a lot. $\endgroup$ – user2547460 May 22 '16 at 22:13
  • $\begingroup$ 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 $\endgroup$ – user2547460 May 22 '16 at 22:24
  • $\begingroup$ what do you mean is not enough? $\endgroup$ – user2547460 May 22 '16 at 22:26
  • 1
    $\begingroup$ @user2547460 I would roll back the last edit, angle of view of a line segment is a perfectly understandable thing. $\endgroup$ – 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.

  • $\begingroup$ but if we assume that ABZ is a right triangle then would my method be correct? $\endgroup$ – user2547460 May 22 '16 at 22:32
  • $\begingroup$ 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 $\endgroup$ – user2547460 May 22 '16 at 22:35
  • $\begingroup$ 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. $\endgroup$ – chx May 22 '16 at 22:37
  • $\begingroup$ 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... $\endgroup$ – user2547460 May 22 '16 at 22:40
  • 1
    $\begingroup$ Pity, it's much more interesting with 45 degrees :) $\endgroup$ – 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

  • $\begingroup$ note x^2+y^2=z^2 iff x and y are legs to a right triangle with hypotenuse length z. $\endgroup$ – shai horowitz May 22 '16 at 22:21
  • $\begingroup$ 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. $\endgroup$ – user2547460 May 22 '16 at 22:22
  • $\begingroup$ 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 $\endgroup$ – 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 $A$.

  • $\begingroup$ 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? $\endgroup$ – user2547460 May 22 '16 at 22:58
  • $\begingroup$ @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. $\endgroup$ – A---B May 22 '16 at 23:03

So all the fuzz here resulted from some unsharp definitons (where is the angle, what angle).

the sitation

You can try out different $x$ values here.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.