Can you work out a point from the length of two lines and the position they start at?

Question

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?

Answer 1

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 $y=\pm15$.

Answer 2

You could use circles, as both points lie on the radical axis of $\omega_1,\omega_2$ and are symm. in the x-axis

Answer 3

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

Answer 4

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 $(0,15)$.