# Find area of the triangle ABC, given the coordinates of vertices in plane

$A, B$ and $C$ are the points $(7,3), (-4,1)$ and $(-3,-2)$ respectively. Find the area of the triangle $ABC$.

I've worked out the lengths of each side of the triangle which are $AB=5\sqrt5$, $BC=\sqrt10$ and $AC=5\sqrt5$.

I know that the formula for the area of a triangle is $\frac12hb$ but when I checked the solutions the answer to the area of this triangle is $17\frac12$.

I do not understand how this answer is achieved.

• Every heard of the Shoelace formula? – suomynonA Oct 2 '16 at 15:34
• Another way would also be to draw a rectangle around the triangle and subtract each section not included in the triangle, but you would need the graph for that. – suomynonA Oct 2 '16 at 15:35
• Fine comments above. Still another way is to exploit en.wikipedia.org/wiki/Pick%27s_theorem – Jack D'Aurizio Oct 2 '16 at 15:36
• OP; stop changing it. it's fine. – suomynonA Oct 2 '16 at 15:39
• In order to not to waste your efforts, you may just apply Heron's formula to the computed side lengths, too. – Jack D'Aurizio Oct 2 '16 at 15:44

Follow-through

As you have said before, the side lengths of $\triangle ABC$ is $AB=AC=5\sqrt{5}$, $BC=\sqrt{10}$, using Heron's formula, we can compute the answer.

Heron's formula states that given side lengths $a,b,c$ of $\triangle ABC$, the area is given $$\sqrt{s(s-a)(s-b)(s-c)}\tag{1}$$ Where $s$ Is the semi perimeter. ($s=\frac {a+b+c}{2}$).

So in your case, we have $$a=5\sqrt{5},b=5\sqrt{5},c=\sqrt{10}\tag{2}$$ The semi perimeter is $$\frac {10\sqrt{5}+\sqrt{10}}{2}\tag{3}$$ and plugging in the values, we have $$\sqrt{\frac {10\sqrt{5}+\sqrt{10}}{2}\left(\frac {10\sqrt{5}+\sqrt{10}}{2}-5\sqrt{5}\right)\left(\frac {10\sqrt{5}+\sqrt{10}}{2}-5\sqrt{5}\right)\left(\frac {10\sqrt{5}+\sqrt{10}}{2}-\sqrt{10}\right)}=\boxed{17.5}\tag{4}$$

• I understand this now but how would I type this into a calculator to get the correct result ? Thanks. – Dan Oct 2 '16 at 16:32
• @Dan Hm... If I were to have to solve the messy radicals, I would first write it out, and then break down the parts. So I would first calculate what $\frac {10\sqrt{5}+\sqrt{10}}{2}-5\sqrt{5}$ is, then $\frac {10\sqrt{5}+\sqrt{10}}{2}-\sqrt{10}$ and so on... – Frank Oct 2 '16 at 16:37
• @Dan Also, note that on a calculator, if you press Alpha, then Y= and 1, then you get the option to write your fractions in $\frac ab$ form. Much more neater that way. – Frank Oct 2 '16 at 16:39
• @Frank - "on a calculator..." ?? Not on mine. I think you have made a generalization here that does not apply to many many calculators. – Michael Karas Oct 2 '16 at 18:04
• @MichaelKaras A standard TI-84+ calculator is able to do it. And I believe a standard TI-83 calculator is capable of doing that too. Unless I have mistaken for which type of calculator he used.. – Frank Oct 2 '16 at 18:06

Since you have obtained the length of each side, using Heron's Formula is a natural way to find the area. Let's consider the approach suomynonA suggested in the comments. Consider the figure below. We can find the area of $\triangle ABC$ by subtracting the sum of the areas of the three right triangles $ABD$, $ACF$, and $BCE$ from the area of rectangle $ADEF$. I will leave the details of the calculations to you.

Method-1 $$\Delta= \frac12\begin{vmatrix} x_1 & y_1 & 1\\ x_2& y_2 & 1\\ x_3& y_3 & 1 \end{vmatrix}$$ Method-2 This can also be used to find the area of polygon. $$\Delta= \frac12\begin{vmatrix} x_1 & y_1 \\ x_2& y_2 \\ x_3& y_3 \\ x_1 &y_1 \end{vmatrix}=\frac12\left ((x_1y_2+x_2y_3+x_3y_1)-(x_2y_1+x_3y_2+x_1y_3)\right)$$

You can use $BC$ as the base and measure the height with respect to it as the distance from a point to a line. If the line has equation $ax+by+c=0$, then the distance from the point $(x_0,y_0)$ to the line is $$\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$$

The line $BC$ has equation $$\frac{y-1}{-2-1}=\frac{x+4}{-3+4}$$ or, in simplified form, $$3x+y+11=0$$ The distance from $A$ to the line $BC$ is $$\frac{|3\cdot7+3+11|}{\sqrt{3^2+1^2}}=\frac{35}{\sqrt{10}}$$ Thus the area is $$\frac{1}{2}\cdot\sqrt{10}\cdot\frac{35}{\sqrt{10}}=\frac{35}{2}=17+\frac{1}{2}$$

Actually, there is a much simpler formula when one of the points is the origin. Suppose you want to determine the area of the triangle $OBC$, where $B(x_1,y_1)$ and $C(x_2,y_2)$. The line passing through $B$ and $C$ has equation $$(y_1-y_2)(x-x_2)-(x_1-x_2)(y-y_2)=0$$ or, in simplified form, $$(y_1-y_2)x-(x_1-x_2)y+x_1y_2-x_2y_1=0$$ By the above formula, the distance from $O$ to the line is obtained by using $x_0=0$ and $y_0=0$, so it is $$\frac{|x_1y_2-x_2y_1|}{\sqrt{(y_1-y_2)^2+(x_1-x_2)^2}}$$ and at the denominator you recognize the length of $BC$, so the area is $$\frac{1}{2}|x_1y_2-x_2y_1|= \frac{1}{2} \left| \det\begin{bmatrix} x_1 & x_2 \\ y_1 & y_2 \end{bmatrix} \right|$$ Well, your triangle can be seen as having the origin as one of its vertices, by the translation mapping $A$ to the origin! The new coordinates of $B$ and $C$ in the translated frame of reference are $$(-4-7,1-3)=(-11,-2) \qquad\text{and}\qquad (-3-7,-2-3)=(-10,-5)$$ so the area is $$\frac{1}{2} \left| \det\begin{bmatrix} -11 & -10 \\ -2 & -5 \end{bmatrix} \right|=\frac{1}{2}\lvert-55+20\rvert=\frac{35}{2}$$

In general, by implicitly doing the translation, the area of the triangle with vertices in $(x_0,y_0)$, $(x_1,y_1)$ and $(x_2,y_2)$ can be computed as $$\frac{1}{2} \left| \det\begin{bmatrix} x_1-x_0 & x_2-x_0 \\ y_1-y_0 & y_2-y_0 \end{bmatrix} \right|$$

When you have all three side lengths of a triangle then you can use Heron's formula to find the area:

$A=\sqrt{s(s-a)(s-b)(s-c)}$, where $s=\frac{a+b+c}{2}$

You considered using the formula $\text{area} = \frac{1}{2}(\text{height})(\text{base})$ here. While there are more specialized methods to this case, the approach you thought up works perfectly fine.

You just have to select a side to be the base, and compute the base and height lengths.

e.g. if you pick side $AB$ to be the base, you've already determined that it has length $5 \sqrt{5}$.

So to finish the problem, you need to find the height — that is, the distance from the line $AB$ to the point $C$. There are a number of ways to go about this, and the value turns out to be $7/\sqrt{5}$.

Thus, the area is $\frac{1}{2} \cdot (5 \sqrt{5}) \cdot \frac{7}{\sqrt{5}} = \frac{35}{2}$.

• This is wonderful, could you tell me a simple way to calculate the distance from the line AB to the point C please? I was seeing the diagram from N. F. Taussig's answer and thought BE is the distance you were referring to. But BE is only 3 units in the diagram. – Janaaaa Oct 3 '16 at 7:13
• In this case, you've proved that the triangle is isosceles, so it's easier to use BC as the base. Then the height is just the distance from A to the midpoint of BC, since the altitude and the median coincide. – saulspatz Oct 3 '16 at 11:06

I need anwer .A(6,2),B(2,4) and C(8,-4) are three in a plane in the line BC and Q is a point in the plane, such that PAQB is a rectangle. Find the (I) equation of the sides AP and BP. (ii) the coordinate of the points P (iii) the area of the rectangle

• To ask a new question use the "ask question" button. If you do, you should also explain what you have tried, why can't you solve it. – Rolazaro Azeveires Aug 14 '18 at 19:38