I am trying to solve part iv of this question. The picture should explain the question. I know the answer is $8$,but have no clue how to arrive at this.
Find the coordinates of a point in a given triangle
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analytic-geometry
triangles
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0Have you tried using distance formula ? What work have you done so far ? – 2017-02-01
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0I have, to find the base length. – 2017-02-01
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0Did you try using the distance formula to find $RQ$ and $RP$ and equated those two together ? – 2017-02-01
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0That makes sense, I'll give it a shot, thanks for the hint – 2017-02-01
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0Indeed you are correct, I just need to get to grips with the maths behind it. Thank you. – 2017-02-01
2 Answers
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The point $(x_0,y_0)$ that lies in the middle of the 2 points $P$ and $Q$ is $$(x_0,y_0)=(4,4)$$ You are looking for an isosceles triangle, so that a line from $R$ to the base would be orthogonal to it.
Let the slope $m_1$ be the one corresponding to $QR$ and the slope $m_2$ be that of the middle point and $R$.
Demand $m_1\cdot m_2=-1$: $$m_1=\frac{6-0}{8-2}=\frac{1}{2}$$and $$m_2=\frac{4-r}{4-2} $$and so you can solve for $r$:$$\frac{1}{2}\cdot\frac{4-r}{4-2}=-1$$ to get $$r=8$$
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0Wouldn't that just give me coordinates that would make the triangle a right angled one? – 2017-02-01
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0Yes it would. My mistake, I thought you are asking for help in part c. of the exercise. I shall delete my answer, or edit it. – 2017-02-01
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0I appreciate the help though. – 2017-02-01
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0Edited my original answer for the correct part of the question. – 2017-02-01
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$$RQ^2=RP^2\to (2-0)^2+(2-r)^2=(8-2)^2+(6-r)^2\\ 8-4r=72-12r\to r=8$$