Using vertices to determine possible coordinates of a triangle

Question

Two of the vertices of triangle MNO are O(0,0), and M(3,-4). What are possible coordinates for N if triangle MNO is a) isosceles? b) equilateral? Please provide a thorough explanation. (Analytical geometry, grade 10)

Answer 1

Hint:

a) isoceles. There are three ways it can be isosceles.

a1) $NO = NM$. In this case you must find all points where $N=(x,y)$ and $d(N,O) = d(N,M)$. Or in this case $\sqrt{(x-0)^2 + (y-0)^2} = \sqrt{(x-3)^2 + (y+4)^2}$. (Hint: there are an infinite number of these. You must state a condition that will describe them all.)

a2) $MO = ON$. So you must find all points $N=(x,y)$ where $\sqrt{(x-0)^2 + (y-4)^2} = \sqrt {(3-0)^2 + (-4 - 0)^2}$.(Hint: ditto)

a3) $NM = MO$. So you must find all .... yadda yadda yadda.

b) equalateral . There is only one way that can occur is if

$NO = NM = MO$. So you must fiand all points $N=(x,y)$ where $\sqrt{(x-0)^2 + (y-0)^2} = \sqrt{(x-3)^2 + (y+4)^2}=\sqrt {(3-0)^2 + (-4 - 0)^2}$ (Hint: there are exactly two of them.)

Answer 2

Hint: Find the perpendicular bisector of $O$ and $M$. You know one point on the perpendicular bisector (the midpoint) and you know its slope (it has to be perpendicular to $\vec{OM}$), so you should be able to calculate it.

If the third vertex is on the perpendicular bisector, the triangle will always be an isosceles triangle. But there are two particular points on the perpendicular bisector that will give you equilateral triangles and we can find them.

Calculation: We know that $m=(\frac{0+3}{2},\frac{0-4}{2})=(\frac{3}{2},-2)$ is the midpoint of $O$ and $M$ which is a point on the perpendicular bisector by definition. Now we want to find the equation of a line passing through $m$ with its slope equal to $-(\frac{-4-0}{3-0})^{-1}=\frac{3}{4}$, because it has to be perpendicular to $\vec{OM}$. Hence, the equation of the desired line is:

$$\mathcal{l}=\{(x,y) \in \mathbb{R}^2: y+2 = \frac{3}{4}(x-\frac{3}{2})\}$$

Any third vertex with the coordinates $(x,y)$ chosen on this line will give you an isosceles triangle. Now parametrize the line $\mathcal{l}$ with parameter $t$ and you will find two $t$'s that give you an equilateral triangle. If we parametrize with $t$, we see that we must have:

$$\|(t,\frac{3t-14}{4})\|^2=\|\vec{OM}\|^2=5$$

Now you have a quadratic polynomial in $t$ which can be solved easily to find its two solutions.

Answer 3

The locus of the points equidistant to the two given points is $$x^2+y^2=(x-3)^2+(y+4)^2$$ or

$$-6x+8y+25=0$$ which is a straight line.

The locus of the points which are at distance $3^2+(-4)^2=25$ from the first point or the second point respectively are

$$x^2+y^2=25,\\(x-3)^2+(y+4)^2=25,$$ two circles.

Now equilateral triangles are formed by points that verify both conditions,

$$x^2+y^2=(x-3)^2+(y+4)^2=25.$$

By eliminating $y$,

$$x^2+\left(\frac{25+6x}8\right)^2=25$$ yields

$$x=\frac{\pm4\sqrt3-3}2,$$ then

$$y=\frac{\pm3\sqrt3+4}2.$$