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the line joining B to A is sort of the vector (2,-4). Find the vector orthogonal to this: y=1/2x. You have to add a constant to return it back to its position (before they were all translated down to the origin). Use the fact that BC goes through (1,6) to get y = 1/2x + 11/2. That's the equation of BC. Now you also have the equation of AC. Where these two lines meet is the point C. AC = x-1. BC = 1/2x + 11/2 x-1 = 1/2x + 11/2 this gives you x = 13. Throw that into the equation: c = (13,12) Calculate some lengths in R2 to get the perimeter.
the line joining B to A is sort of the vector (2,-4). Find the vector orthogonal to this: y=1/2x. You have to add a constant to return it back to its position (before they were all translated down to the origin). Use the fact that BC goes through (1,6) to get y = 1/2x + 11/2. That's the equation of BC. Can you explain this more. I understand how you gor 2, -4... but what's orthogonal? Nad how did that equate to y=1/2x?
in two dimensions, orthogonal means that it's at right angle to the other vector. You don't have a lot of info about that other line, BC, so you have to use that I think. Two vectors are orthogonal iff the dot product is equal to zero. So, if you have two vectors (x1,y1) (x2,y2) the dot product is calculated like this: x1*x2 + y1 * y2. If that number is zero then the two vectors are orthogonal. Use that to find the vector BC. But then you need to add a constant so that it's translated to the right position, because at first it will be centred at the origin.
Ok... So how do you get the perimeter?
Would it be 13-1 +21-6 and 3-1 +6-2?
a^2 + b^2 = c^2. pythagoras. Do the same for the other side. Add them, double it.
Would the first be c^2= 12^2+6^2?
and the second a^2=2^2 + 4^2?
That is right. perimeter = 2* [ sqrt(2^2 + 4^2) + sqrt(12^2 + 6^2) ]
Since it is a rectangle, you know the 4 included angles all will be 90 degrees. That being the case you know the lince BC will be perpendicular to the line BA. the line BA has a slope of:(6-2)/(1-3)=4/-2=-2 The line perpendicular will have a slope which is the negative reciprocal or 1/2. The equation of BC is 1/2=(6-y)/(1-x) cross multiplying the line equation results in: y=(1/2)x + 11/2 The same as scarydoor got but not using vectors.
You are given the equation for line AC, as y=x-1 now solve the system of two equations: y=x-1 and y=(1/2) x +11/2 obtaining values of x=13 and y=12 The coordinates of C, again no vectors.
It is now possible to obtain the length of BC (and AD) as you know the terminating coordinates of BC using the distance formula.
The widths BA (CD) you were given the terminating coordinates for BA, using the distance formula you can calculate the length of that segment, now you can calculate the perimeter. I have nothing against vectors, but you sounded kind of wary of them, so I am just pointing out an alternative method.