# Thread: Area of triangle in 2D space

1. ## Area of triangle in 2D space

Hi,

I have a problem I'm having trouble with.

The points $\displaystyle P(ap^2, 2ap)$ and $\displaystyle Q(aq^2, 2aq)$ lie on the parabola with equation $\displaystyle y^2 = 4ax$. Show that the tangents to the parabola at P and Q meet at $\displaystyle R[apq, a(p+q)]$. Show further that the area of the triangle PQR is $\displaystyle \frac{1}{2}a^2 |p-q|^3$

I can do the first part about showing R, but I have no idea about finding the area of the triangle. Any help would be much appreciated

Stonehambey

2. Hello, Stonehambey!

The points $\displaystyle P(ap^2, 2ap)$ and $\displaystyle Q(aq^2, 2aq)$ lie on the parabola with equation $\displaystyle y^2 = 4ax$.

(a) Show that the tangents to the parabola at $\displaystyle P$ and $\displaystyle Q$ meet at $\displaystyle R[apq, a(p+q)]$.

(b) Show that the area of $\displaystyle \Delta PQR$ is: .$\displaystyle \tfrac{1}{2}a^2 |p-q|^3$
Lucky for us, there is a formula for the area . . .

The area of the triangle with vertices: $\displaystyle (x_1,y_1),\:(x_2,y_2),\:(x_3,y_3)$

. . . . . is given by: . $\displaystyle A \;=\;\frac{1}{2}\left|\begin{array}{ccc}x_2 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3& y_3 & 1 \end{array}\right|$

3. ## A specific case that can be generalized: q<0<p

Forgive me, I don't know any fancy geometry formulas for area, but this one has a nice graphical interpretation...

Without loss of generality, let $\displaystyle p>0$ and $\displaystyle q<0$ (we know $\displaystyle a>0$). The graph is a parabola opening from the origin to the right. P sits at an arbitrary point at the top, Q on the bottom. R is somewhere off in the 2nd or 3rd quadrant near the x-axis...

Cut the triangle in half and figure the areas of each separately. Draw a horizontal line through R at $\displaystyle y=a(p+q)$. Where does it intersect PQ?

Use the point-slope form of a line to connect P and Q via $\displaystyle y=2/(p+q)*(x-ap^2)+2ap$. Substitute $\displaystyle a(p+q)$ in for y and solve for x... $\displaystyle x=a(p^2+q^2)/2$, call this point M (NOT necessarily the midpoint of P and Q).

So the "base" of both triangles is segment RM, length $\displaystyle a(p^2+q^2)/2 - apq$, simplifies to $\displaystyle |RM|=a(p-q)^2/2$.

The height of triangle RPM is $\displaystyle |2ap-2a(p+q)|=-2aq$ (remember q is negative by definition)
The height of triangle RQM is $\displaystyle |2aq-2a(p+q)|=2ap$

So the area of RPM=$\displaystyle -q*a^2(p-q)^2/2$
And the area of RQM=$\displaystyle p*a^2(p-q)^2/2$

Whose sum is $\displaystyle (1/2)a^2(p-q)^3$

NOTE: THIS STILL WORKS IF WE ACCEPT THE IDEA OF "NEGATIVE AREA." IF p and q ARE BOTH POSITIVE OR BOTH NEGATIVE, THESE NUMBERS STILL WORK. EX: IF p=q, ACCORDING TO THE ABOVE EQN, AREA=0, AS EXPECTED. WE WOULD SIMPLY FIND THE AREA OF THE BIGGER HALF-TRIANGLE, AND SUBTRACT OFF THE AREA OF THE SMALLER HALF-TRIANGLE, SINCE THEY WOULD BE OVERLAPPING INSTEAD OF APPENDING EACH OTHER.

4. Formula works nicely. For the geometric interpretation, I was trying (without luck) a similar approach. So I'll keep reading over it until it's completely clear to me

thanks guys