The area of the triangle with vertices $(3, 2), (3, 8)$, and $(x, y)$ is $24$. What is $x$?

Question

The area of the triangle with vertices $(3, 2), (3, 8)$, and $(x, y)$ is $24$. A possible value for $x$ is:

• a) $7$
• b) $9$
• c) $11$
• d) $13$
• e) $15$

Please show your work and explain.

Answer 1

One side of the triangle lies on the line $x = 3$, and is length $6$. Why?. Take that to be your base, $b$.

The area of a triangle is given by $$\text{Area}\;=\;\dfrac 12 bh$$ where $h$ is the height of the triangle measured from the base (connecting the third point perpendicular to the base, so $$\dfrac12(6)h = 24 \iff h = 8$$

Now, height, h, is the perpendicular distance from the base, which is on $x = 3$, and the only possible choices for $x$ that are given are all positive.

Hence $h = 8 \implies x = 3 + 8 = 11.$

Answer 2

Hinz Take the first twopoint as base line. It has length 6. Therefore the height must be 8.

Answer 3

We know the area of a triangle (Article#25) having vertices $(x_i,y_i)$ for $i=1,2,3$ is

$$\frac12\det\begin{pmatrix} x_1 & y_1 & 1\\x_2&y_2&1\\ x_3 & y_3 &1\end{pmatrix}$$

Now, $$\det\begin{pmatrix}x & y &1\\ 3 & 2 & 1 \\ 3 & 8 & 1 \end{pmatrix}$$

$$=\det\begin{pmatrix}x-3 & y-2 &0\\ 3 & 2 & 1 \\ 0 & 6 & 0 \end{pmatrix}\text { (Applying } R_3'=R_3-R_2,R_1'=R_1-R_2)$$

$$=6(x-3)$$

As we take the area in absolute value,the are here will be $\frac12\cdot6|x-3|=3|x-3|$

If $x\ge 3, 3(x-3)=24\implies x=11$

If $x<3, 3(3-x)=24\implies -5$