Question

If in a triangle ABC, A = (1, 10), circumcentre $=\left( -\dfrac{1}{3},\dfrac{2}{3} \right)$ and orthocenter$=\left( \dfrac{11}{3},\dfrac{4}{3} \right)$, then the coordinates of the midpoint of the side opposite to A is.
(a)$\left( 1,-\dfrac{11}{3} \right)$
(b)(1, 5)
(c)(1, −3)
(d)(1, 6)

Answer 1

As we know that the orthocenter, centroid and circumcentre of any triangle are collinear and the centroid divides the distance from orthocenter to circumcentre in the ratio 2:1 and also, the centroid divides the median in the ratio 2:1. Then we will use section formula for calculation which states that, the section formula tells us the coordinates of the point which divides a given line segment into two parts such that their lengths are in the ratio m: n, then
If (a, b) is the point that divides a line between points $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ in the ratio m: n , then we can write as follows
$(a,b)=\left[ \left( \dfrac{m{{x}_{2}}+m{{x}_{1}}}{m+n} \right),\left( \dfrac{m{{y}_{2}}+m{{y}_{1}}}{m+n} \right) \right]$ .
Now, in this question, we will first find the coordinates of the centroid using the section formula and the fact that centroid divides the distance from orthocenter to circumcentre in the ratio 2:1. . Then, using the fact that the centroid divides the median in the ratio 2:1, we can find the midpoint of the side opposite to the vertex A.

Complete step-by-step answer:
As mentioned in this question, we have to find the coordinates of the midpoint of the side opposite to vertex A.
Now, let the centroid have the coordinates as (x, y) can be written as follows using the section formula
$=(x,y)$
Using section formula, we get value of $x=\left( \dfrac{m{{x}_{2}}+m{{x}_{1}}}{m+n} \right)$ and $y=\left( \dfrac{m{{y}_{2}}+m{{y}_{1}}}{m+n} \right)$
$=\left[ \left( \dfrac{m{{x}_{2}}+m{{x}_{1}}}{m+n} \right),\left( \dfrac{m{{y}_{2}}+m{{y}_{1}}}{m+n} \right) \right]$
Now, as centroid divides the distance from orthocenter to circumcentre in the ratio 2:1, so m = 2 and n = 1

So, we get
$=\left( \dfrac{\dfrac{-2}{3}+\dfrac{11}{3}}{3},\dfrac{\dfrac{4}{3}+\dfrac{4}{3}}{3} \right)$
On solving, we get
$=\left( 1,\dfrac{8}{9} \right)$
Hence, the coordinates of the centroid are $\left( 1,\dfrac{8}{9} \right)$ .

Now, the centroid divides the distance from orthocenter to circumcentre in the ratio 2:1.
Let the coordinates of the midpoint of the side opposite to vertex A be (a, b).
Therefore, we can write as follows using the section formula

& 1=\left( \dfrac{m{{x}_{2}}+m{{x}_{1}}}{m+n} \right)=\dfrac{2a+1}{3}and\ \dfrac{8}{9}=\left( \dfrac{m{{y}_{2}}+m{{y}_{1}}}{m+n} \right)=\dfrac{2b+10}{3} \\\ & a=1\ and\ b=\dfrac{-11}{3} \\\ \end{aligned}$$ so, the required point is $$\left( 1,\dfrac{-11}{3} \right)$$ . hence, option ( a ) is correct. **So, the correct answer is “Option (a)”.** **Note:** The students can make an error if they don’t know about the section formula so always remember that section formula is If (a, b) is the point that divides a line between points $$({{x}_{1}},{{y}_{1}})$$ and $$({{x}_{2}},{{y}_{2}})$$ in the ratio m: n , then we can write as follows$$(a,b)=\left[ \left( \dfrac{m{{x}_{2}}+m{{x}_{1}}}{m+n} \right),\left( \dfrac{m{{y}_{2}}+m{{y}_{1}}}{m+n} \right) \right]$$ and and the property that the orthocenter, centroid and circumcentre of any triangle are collinear and the centroid divides the distance from orthocenter to circumcentre in the ratio 2:1.