Is my teacher's final answer wrong?

Question

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..

Claire Akin · Accepted Answer

Your teacher is correct...

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

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#

Violet Aldrich · Answer

undefined To determine if your teacher's final answer is wrong, you need to provide the context and details of the problem or question. Without knowing the specifics, it's impossible to assess the accuracy of the answer. Please provide more information about the question or problem your teacher addressed.