Is my teacher's final answer wrong?

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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..

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Answer 1 · 2018-03-15T01:38:16

Your teacher is correct...

Just note that I worked through all the parts to help you catch your mistake.

Explanation:

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

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

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

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ $m=(y_2-y_1)/(x_2-x_1)$

$\Rightarrow m = \frac{- 3 - 0}{6 - 0}$ $=>m=(-3-0)/(6-0)$

$\Rightarrow m = - \frac{3}{6}$ $=>m=-3/6$

$\Rightarrow m = - \frac{1}{2}$ $=>m=-1/2$

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

$\Rightarrow M = - 1 \cdot (1 \div - \frac{1}{2})$ $=>M=-1*(1divide-1/2)$

$\Rightarrow M = - 1 \cdot (1 \cdot - \frac{2}{1})$ $=>M=-1*(1*-2/1)$

$\Rightarrow M = - 1 \cdot (- 2)$ $=>M=-1*(-2)$

$\Rightarrow M = 2$ $=>M=2$

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

$d = \sqrt{{(y_{2} - y_{1})}_{2}^{2} + {(x_{2} - x_{1})}_{2}^{2}}$ $d=sqrt((y_2-y_1)^2+(x_2-x_1)^2)$

$\Rightarrow d = \sqrt{{(- 3 - 0)}^{2} + {(6 - 0)}^{2}}$ $=>d=sqrt((-3-0)^2+(6-0)^2)$

$\Rightarrow d = \sqrt{{(- 3)}^{2} + {(6)}^{2}}$ $=>d=sqrt((-3)^2+(6)^2)$

$\Rightarrow d = \sqrt{9 + 36}$ $=>d=sqrt(9+36)$

$\Rightarrow d = \sqrt{45}$ $=>d=sqrt(45)$

$\Rightarrow d = 3 \sqrt{5}$ $=>d=3sqrt(5)$

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

$A = \frac{b h}{2}$ $A=(bh)/2$

$\Rightarrow 15 = \frac{3 \sqrt{5} \cdot h}{2}$ $=>15=(3sqrt5*h)/2$

$\Rightarrow 30 = 3 h \sqrt{5}$ $=>30=3hsqrt5$

$\Rightarrow 10 = h \sqrt{5}$ $=>10=hsqrt5$

$\Rightarrow \frac{10}{\sqrt{5}} = h$ $=>10/sqrt5=h$

$\Rightarrow 2 \sqrt{5} = h$ $=>2sqrt5=h$

This is the length of $O B$ $OB$

We can get two equations from this:

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

We can come up with the following:

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

$\Rightarrow 2 = \frac{Y}{X}$ $=>2=Y/X$

$\Rightarrow 2 X = Y$ $=>2X=Y$

We also use the distance formula using the fact that the length of segment $B O$ $BO$ equals $2 \sqrt{5}$ $2sqrt5$

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

We also know $d$ $d$ !

$\Rightarrow 2 \sqrt{5} = \sqrt{{(2 X)}^{2} + {(X)}^{2}}$ $=>2sqrt5=sqrt((2X)^2+(X)^2)$

$\Rightarrow 2 \sqrt{5} = \sqrt{4 X^{2} + X^{2}}$ $=>2sqrt5=sqrt(4X^2+X^2)$

$\Rightarrow 2 \sqrt{5} = \sqrt{5 X^{2}}$ $=>2sqrt5=sqrt(5X^2)$

$\Rightarrow 2 \sqrt{5} = X \sqrt{5}$ $=>2sqrt5=Xsqrt(5)$

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

Now, the important part:

We know the distance from $A$ $A$ to $O$ $O$ is $3 \sqrt{5}$ $3sqrt5$ .

Using our information from the problem, we know that the distance from $O$ $O$ to $C$ $C$ is $\frac{3 \sqrt{5}}{3} \Rightarrow \sqrt{5}$ $(3sqrt5)/3=>sqrt5$

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

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

Therefore,

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

$\Rightarrow - \frac{X}{2} = Y$ $=>-X/2=Y$

$\Rightarrow X = - 2 Y$ $=>X=-2Y$

Using our information, we have:

$\sqrt{5} = \sqrt{{(Y - 0)}^{2} + {(X)}^{2}}$ $sqrt5=sqrt((Y-0)^2+(X)^2)$ Substitute $X$ $X$ with $- 2 Y$ $-2Y$

$\sqrt{5} = \sqrt{{(Y - 0)}^{2} + {(- 2 Y)}^{2}}$ $sqrt5=sqrt((Y-0)^2+(-2Y)^2)$

$\sqrt{5} = \sqrt{Y^{2} + 4 Y^{2}}$ $sqrt5=sqrt(Y^2+4Y^2)$

$\sqrt{5} = \sqrt{5 Y^{2}}$ $sqrt5=sqrt(5Y^2)$

$\sqrt{5} = Y \sqrt{5}$ $sqrt5=Ysqrt(5)$

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

Therefore, $X = - 2$ $X=-2$

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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..

1 Answer

Explanation:

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