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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.

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is it a homework? –  Yimin Feb 15 '13 at 3:44
    
It's for an upcoming test. But I don't understand. –  user62366 Feb 15 '13 at 3:48
    
Graph it. It should be obvious. –  Tpofofn Feb 15 '13 at 4:12

3 Answers 3

up vote 5 down vote accepted

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

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Thank you! This helped. But would this way only work with right angle triangles? –  user62366 Feb 15 '13 at 4:07
    
Ok.But what if the one side of the triangle did not lie on the same line? –  user62366 Feb 15 '13 at 4:21
    
It doesn't work only if the triangle is a right triangle: We don't know yet what y might be, or where the perpendicular line from $(x,y)$ intersects the line x=3, where the base lies (the perpendicular distance from the point (x, y) to the which is height), only that it must intersect the line $x = 3$. It would be a right triangle iff y = 2 or y = 8, $y$ being the value of the unknown point. –  amWhy Feb 15 '13 at 14:10
    
We only know that the unknown point $x, y$ is limited by a perpendicular distance of $8$ to the line $x = 3$, and that the point is somewhere to the right of the line $x = 3$, because the choices for the unknown x value are all positive and $> 3$. –  amWhy Feb 15 '13 at 14:13

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

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

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