# Is my teacher's final answer wrong?

## Hi, for iii) my teacher got x = -2.. however I got x = 2.. if my teacher is correct, how did he get x = -2.. since the math shown kinda skips through that part..

Mar 15, 2018

#### Explanation:

Let's find the slope from points $A$ to $O$.

Just note here that the slop is also the slope from $A$ to $C$.

Since $O$ is at the origin, we have:

$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

$\implies m = \frac{- 3 - 0}{6 - 0}$

$\implies m = - \frac{3}{6}$

$\implies m = - \frac{1}{2}$

Since line $O B$ is perpendicular to $A C$, we find the negative reciprocal of our slope.

$\implies M = - 1 \cdot \left(1 \div i \mathrm{de} - \frac{1}{2}\right)$

$\implies M = - 1 \cdot \left(1 \cdot - \frac{2}{1}\right)$

$\implies M = - 1 \cdot \left(- 2\right)$

$\implies M = 2$

Also, since the triangle $A O B$ has an area of 15, and it is a right triangle, let's figure out the length of the side $A O$.

$d = \sqrt{{\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({x}_{2} - {x}_{1}\right)}^{2}}$

$\implies d = \sqrt{{\left(- 3 - 0\right)}^{2} + {\left(6 - 0\right)}^{2}}$

$\implies d = \sqrt{{\left(- 3\right)}^{2} + {\left(6\right)}^{2}}$

$\implies d = \sqrt{9 + 36}$

$\implies d = \sqrt{45}$

$\implies d = 3 \sqrt{5}$

If we set this as our base, then our hight is $O B$.

$A = \frac{b h}{2}$

$\implies 15 = \frac{3 \sqrt{5} \cdot h}{2}$

$\implies 30 = 3 h \sqrt{5}$

$\implies 10 = h \sqrt{5}$

$\implies \frac{10}{\sqrt{5}} = h$

$\implies 2 \sqrt{5} = h$

This is the length of $O B$

We can get two equations from this:

Let the coordinates of point $B$ be $\left(X , Y\right)$

We can come up with the following:

$m = \frac{Y - 0}{X - 0}$ We found $m$ long ago!

$\implies 2 = \frac{Y}{X}$

$\implies 2 X = Y$

We also use the distance formula using the fact that the length of segment $B O$ equals $2 \sqrt{5}$

$d = \sqrt{{\left(Y - 0\right)}^{2} + {\left(X - 0\right)}^{2}}$ Substitute $Y$ with $2 X$.

We also know $d$!

$\implies 2 \sqrt{5} = \sqrt{{\left(2 X\right)}^{2} + {\left(X\right)}^{2}}$

$\implies 2 \sqrt{5} = \sqrt{4 {X}^{2} + {X}^{2}}$

$\implies 2 \sqrt{5} = \sqrt{5 {X}^{2}}$

$\implies 2 \sqrt{5} = X \sqrt{5}$

$\implies 2 = X$ You can use this to find that the coordinates of $B$ are $\left(2 , 4\right)$

Now, the important part:

We know the distance from $A$ to $O$ is $3 \sqrt{5}$.

Using our information from the problem, we know that the distance from $O$ to $C$ is $\frac{3 \sqrt{5}}{3} \implies \sqrt{5}$

We also know that the slope from $A$ to $C$ is $- \frac{1}{2}$.

Since the line intersects at the origin, the slope is the $y$ coordinate divided by the $x$ coordinate.

Therefore,

$- \frac{1}{2} = \frac{Y}{X}$

$\implies - \frac{X}{2} = Y$

$\implies X = - 2 Y$

Using our information, we have:

$\sqrt{5} = \sqrt{{\left(Y - 0\right)}^{2} + {\left(X\right)}^{2}}$ Substitute $X$ with $- 2 Y$

$\sqrt{5} = \sqrt{{\left(Y - 0\right)}^{2} + {\left(- 2 Y\right)}^{2}}$

$\sqrt{5} = \sqrt{{Y}^{2} + 4 {Y}^{2}}$

$\sqrt{5} = \sqrt{5 {Y}^{2}}$

$\sqrt{5} = Y \sqrt{5}$

$1 = Y$ We can use this to find that $X = - 2$.

Therefore, $X = - 2$