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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1 Answer
Mar 15, 2018

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#