Question

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..

Answer 1

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` 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=(y_2-y_1)/(x_2-x_1)`

`=>m=(-3-0)/(6-0)`

`=>m=-3/6`

`=>m=-1/2`

Since line `OB` is perpendicular to `AC`, we find the negative reciprocal of our slope.

`=>M=-1*(1divide-1/2)`

`=>M=-1*(1*-2/1)`

`=>M=-1*(-2)`

`=>M=2`

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

`d=sqrt((y_2-y_1)^2+(x_2-x_1)^2)`

`=>d=sqrt((-3-0)^2+(6-0)^2)`

`=>d=sqrt((-3)^2+(6)^2)`

`=>d=sqrt(9+36)`

`=>d=sqrt(45)`

`=>d=3sqrt(5)`

If we set this as our base, then our hight is `OB`.

`A=(bh)/2`

`=>15=(3sqrt5*h)/2`

`=>30=3hsqrt5`

`=>10=hsqrt5`

`=>10/sqrt5=h`

`=>2sqrt5=h`

This is the length of `OB`

We can get two equations from this:

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

We can come up with the following:

`m=(Y-0)/(X-0)` We found `m` long ago!

`=>2=Y/X`

`=>2X=Y`

We also use the distance formula using the fact that the length of segment `BO` equals `2sqrt5`

`d=sqrt((Y-0)^2+(X-0)^2)` Substitute `Y` with `2X`.

We also know `d`!

`=>2sqrt5=sqrt((2X)^2+(X)^2)`

`=>2sqrt5=sqrt(4X^2+X^2)`

`=>2sqrt5=sqrt(5X^2)`

`=>2sqrt5=Xsqrt(5)`

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

Now, the important part:

We know the distance from `A` to `O` is `3sqrt5`.

Using our information from the problem, we know that the distance from `O` to `C` is `(3sqrt5)/3=>sqrt5`

We also know that the slope from `A` to `C` is `-1/2`.

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

Therefore,

`-1/2=Y/X`

`=>-X/2=Y`

`=>X=-2Y`

Using our information, we have:

`sqrt5=sqrt((Y-0)^2+(X)^2)` Substitute `X` with `-2Y`

`sqrt5=sqrt((Y-0)^2+(-2Y)^2)`

`sqrt5=sqrt(Y^2+4Y^2)`

`sqrt5=sqrt(5Y^2)`

`sqrt5=Ysqrt(5)`

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

Therefore, `X=-2`