I know two points in the plane, $(0,0)$ and $(20,0)$. I also know that a point between and above them(the apex of a triangle formed from these three points) is $15$ from $(0,0)$ and $25$ from $(20,0)$. But I can’t work out what this third point’s position is, though I’m sure it is possible to work it out. Can anyone help, and I’m a secondary student, so can the explanations not be too advanced?
$\begingroup$
$\endgroup$
2
-
1$\begingroup$ So I think what you're looking at is a triangle of side lengths 20, 15 and 25. From these figures, you can work out the angles and then pinpoint the third point. Also notice that in theory, there are two solutions: One that has a positive $y$-coordinate and another that has a negative $y$-coordinate. $\endgroup$– Matti P.Commented Mar 25, 2020 at 11:19
-
$\begingroup$ Note that this is really a 3-4-5 triangle (scaled up by a factor of 5), so by Pythagoras, it is a right triangle. $\endgroup$– NickDCommented Mar 25, 2020 at 18:08
Add a comment
|
4 Answers
$\begingroup$
$\endgroup$
2
You are looking for a point at distance $15$ from $(0,0)$ and $25$ from $(20,0)$. This means the point is on the circle of radius $15$ centered at $(0,0)$ and also on the circle of radius $25$ centered at $(20,0)$. A picture will show that these circles meet in two points!
Now to find the coordinates of these two points, you can determine the equations for the two circles and set them equal to eachother to find the intersection points. The equations of the circles are $$x^2+y^2=15^2\qquad\text{ and }\qquad (x-20)^2+y^2=25^2.$$ Subtracting one from the other shows that $$15^2-25^2=(x^2+y^2)-((x-20)^2+y^2)=40x-400,$$ which shows that $x=0$ and hence $y=\pm15$.
-
$\begingroup$ Thank you. I hadn’t thought of using circles. The fact that they meet twice surprises me, but makes sense. $\endgroup$ Commented Mar 25, 2020 at 11:25
-
$\begingroup$ @AlexBell I have added a derivation of the coordinates of the two points. $\endgroup$– ServaesCommented Mar 25, 2020 at 11:32
$\begingroup$
$\endgroup$
You could use circles, as both points lie on the radical axis of $\omega_1,\omega_2$ and are symm. in the x-axis
$\begingroup$
$\endgroup$
Since both points lie on the radical axis, compute where $Pow(P,\omega_1)=Pow(P,\omega_2)=0$ to find $x$ with Pythagorean theorem, then plug back in
$\begingroup$
$\endgroup$
As others have pointed out, the third vertex lies on the intersection of a pair of circles with radii equal to the distances from the two given points. However, you can save yourself a lot of work by recognizing that this is a 3-4-5 right triangle. Since the distance between the two given point is $20=5\cdot 4$ and the third vertex is at a distance of $15=5\cdot 3$ from the origin, the right angle must be at the origin, so the third vertex is at $(0,15)$.
