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**How to Find Coordinates of Circumcentre of a Triangle IB Test**. Following along are instructions in the video below:

let us now understand how to determine coordinates of the circle center of a triangle here is a question from IV maths test paper coordinate geometry great tip the question is determine the coordinates of the circle center of triangle ABC with vertices a 3 1 B is minus 1/3 and C is minus 3 minus 3 now what is circumcenter circumcenter is a point of intersection of right bisectors of the sides of the triangle so that point will be equidistant from the vertices of the triangle so that is the significance of circumcenter so from that point you can draw a circle touching all the three vertices right so thats how it case gets his name circumcenter to begin with this approximately sketch our triangle we have three points a B and C so let me write minus 3 minus 3 which is probably the leftmost point here so lets say this point is C for us and let me say this is -3 minus 3 and then we have a point minus 1 plus 3 so lets say minus 1 plus 3 will be kind of here so you call this as B which is minus 1 plus 3 and then we have 3 & 1 so 3 & 1 should be somewhere here so lets say this point is a for us 3 & 1 so the approximately sketch it will really help us to solve the problem getting these points on a grid it is even better right so you approximately get the idea of you know how the things are before us I think this will work now when we say circumcenter what are we trying to find we are trying to find midpoint of a segment slope of the segment and then the right bisector so lets begin with a B so

lets find midpoint of a B first so midpoint of a B is what add these 2 divided by uses average value of x and y-coordinates thats how we see it right so we are calling this point as midpoint of a B so the x value will be minus 1 plus 3 divided by 2 and the Y value will be 3 plus 1 divided by 2 and that gives us 2 over 2 s plus 1 and 3 plus 1 is 4 1 & 2 so that is midpoint now what is the slope of a B the slope of a B will be let me call it small m a B slope will be rise over run so we can say 3 minus 1 difference of y-values minus 1 minus 3 3 minus 1 is 2 this is minus 4 so get minus 1/2 as the slope of a B now the right bisector is something which is perpendicular to this current so we are looking for line which is bisecting this goes through the midpoint so slope of this will be what so let me call this as right bisector R of a B right so this right bisector R of a B will have slope of what negative reciprocal of a piece low current so the slope is so lets first figure out what we are trying to use so we are trying to use slope of this perpendicular which will be negative reciprocals so negative reciprocal of 1/2 is 2 right and the point to be considered is ma b right so we are considering the point M a B which is 1 comma 2 so this is what we are using and now we will find equation of this line right so clearly from slope and a point we can write the

question as y equals to MX plus B now here Y is 2 X is 1 slope is 2 right so substituting these values we get 2 equals 2 slope of 2 X is 1 plus B and that gives us value of P as equal to 0 and then for the equation of this line is y equals 2 2 X so well call this as our equation one we got one of the right bisectors equation which is y equals to 2 X now well follow all these steps again and then find the right bisector of another section let us let us take this has to be PC okay so once midpoint of BC midpoint of BC will be average values of X minus 1 minus 3 divided by 2 3 and minus 3 3 I should write I should actually add so Im adding 3 and minus 3 okay that that was okay so I should act like this right and that gives me the values as this is minus 2 and that is U 3 so the x value will be minus 1 minus 3 when you add you get minus 4 and average minus 2 3 – you will give it zero similarly the slope of so this point which way of God is MBC right this is eliminated M BC and the slope of this line BC is difference of Y values so we write 3 minus of minus will be 6 divided by minus 1 minus this which is – so the slope is 3 for us fosters 3 now lets find the equation of right bisector of this one so it will be kind of like this is it okay yeah let me call this right bisector BC so for right bisector BC what do we have we have a

point which is M BC and the point is minus to 0 plus we know what the slope should be right slope of this bisector should be negative reciprocal this is minus 1 over 3 now taking these two things into consideration equation should be y equals two sofas minus 1/3 you can write minus 1/3 X plus P we need to find the value of B so well plug in minus 2 for X and 0 for y so we get 0 equals 2 minus 1 over 3 minus 2 plus B and that gives us a value of B as equal to this becomes plus when you take it to the other side 2 over 3 right or the equation in our case will be y equals 2 minus 1 over 3 X plus 2 over 3 right so well call this as a second right bisector now using equations of the two right bisectors we can find point of intersection right so now let us calculate the point of intersection great that is the circumcenter so the circumcenter of well the point of intersection of our a B first right bisector and the other right our PC so to find that what we can do is we are using these two equations now have a look y equals to two x and y equals 2 minus 1 over 3x plus 2 over 3 so if I substitute Y as 2 X then I get one equation in one variable which Im going to do so from equation 1 & 2 we get 2x equals 2 minus 1 over 3x plus 2 over 3 so everything is in X lets multiply by 3 so we get 6x equals 2 minus X plus 2 and then we can bring X to the sides we have 7x equals to

2 or X is equal to 2 over 7 so we get the value of X since we know so let this be equation 3 for us since we know Y is twice X so Y will be equal to let me further divide this page I like to work in just half piece since this video coverage is so less so y equals to 2 times X is just 2 times 2 over 7 so that is 4 over 7 right so the center of this point which let me call this as C circumcenter the coordinates of C are 2 over 7 and 4 over 7 you see that so that is our answer 2 over 7 4 over 7 let me write down our answer here clearly so answer is that the circumcenter is 2 over 7 4 over 7 so that as that is a coordinator of circumcenter right I hope is clearly visible to you now lets summarize this circumcenter is the point with right bisectors of the sides of a triangle intersect you could have also UCE as we do that youll find that line to be going like this it will also pass through the same point so the idea is find the midpoint of a side find the slope of the side negative reciprocal of the slope will give us slope of the right bisector find the equation of the right bisector using the midpoint and the negative reciprocal slope do it for two sides find the point of intersection and that point of intersection is circumcenter for you in triangle so thats how we do it I hope is absolutely clear – this is one of the most complicated questions which youll get and a time-consuming question in the test be thorough with the calculations thank you and all the best

tags:

Circumscribed Circle, Anil kumar, anil kumar math, aan learning centre, grade 11, grade 12, algebra 2, nelson, grade 10, unit 2, explore, AP Math, IB Math, e…

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