Question

# If the vertices P,Q,R of a triangle PQR are rational points, which of the following points of the triangle PQR is (are) always rational point(s) ?(a) Centroid (b) Incentre(c) Circumcentre(d) Orthocentre

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Hint: Orthocentre is the point of concurrence of the altitudes of a triangle.

Circumcentre is the point of concurrency of perpendicular bisectors of a triangle.
Incentre is the point of concurrency of angle bisectors of a triangle.
Centroid is the point of concurrence of medians of a triangle.
Let the coordinates of the vertices be $P\left( {{x}_{1}},{{y}_{1}} \right),Q\left( {{x}_{2}},{{y}_{2}} \right)\ and R\left( {{x}_{3}},{{y}_{3}} \right)$.

Now, we know the centroid of a triangle with vertices $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\ and \left( {{x}_{3}},{{y}_{3}} \right)$ is given as :
$\left(\frac{{x}_{1}+{x}_{2}+{x}_{3}}{3},\frac{{y}_{1}+{y}_{2}+{y}_{3}}{3}\right)$$\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)$
So , the centroid of $\mathrm{\Delta }PQR$$\Delta PQR$ is :
$G\left(\frac{{x}_{1}+{x}_{2}+{x}_{3}}{3},\frac{{y}_{1}+{y}_{2}+{y}_{3}}{3}\right)$$G\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)$
Now , in the equation it is given that the coordinates of the vertices are rational.
We know , the closure property of rational numbers says that the sum, difference, product or quotient of two rational numbers is always a rational number.
So , by closure property of rational number we can say that,
$\left(\frac{{x}_{1}+{x}_{2}+{x}_{3}}{3}\right)$$\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3} \right)$ is rational and $\left(\frac{{y}_{1}+{y}_{2}+{y}_{3}}{3}\right)$$\left( \dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)$ is rational.
Hence, centroid is a rational point.
Now, we know the slope of line joining two point $\left( {{a}_{1}},{{b}_{1}} \right)\ and \left( {{a}_{2}},{{b}_{2}} \right)$ is given as :
$m=\left(\frac{{b}_{2}-{b}_{1}}{{a}_{2}-{a}_{1}}\right)$$m=\left( \dfrac{{{b}_{2}}-{{b}_{1}}}{{{a}_{2}}-{{a}_{1}}} \right)$
So, the slope of $PQ$$PQ$ is given as :
${m}_{PQ}=\left(\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\right)$${{m}_{PQ}}=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)$
Slope of $PR$$PR$ is given as :
${m}_{PR}=\left(\frac{{y}_{3}-{y}_{1}}{{x}_{3}-{x}_{1}}\right)$${{m}_{PR}}=\left( \dfrac{{{y}_{3}}-{{y}_{1}}}{{{x}_{3}}-{{x}_{1}}} \right)$
And the slope of $QR$$QR$ is given as :
${m}_{QR}=\left(\frac{{y}_{2}-{y}_{3}}{{x}_{2}-{x}_{3}}\right)$${{m}_{QR}}=\left( \dfrac{{{y}_{2}}-{{y}_{3}}}{{{x}_{2}}-{{x}_{3}}} \right)$
Now, we know if two lines are perpendicular then product of their slopes is $\left(-1\right)$$\left( -1 \right)$
So,
$\begin{array}{rl}& {m}_{PQ}×{m}_{\mathrm{\perp }PQ}=-1\\ & ⇒{m}_{\mathrm{\perp }PQ}=\frac{-1}{{m}_{PQ}}\\ & ⇒{m}_{\mathrm{\perp }PQ}=\frac{{x}_{1}-{x}_{2}}{{y}_{2}-{y}_{1}}............\left(i\right)\end{array}$\begin{align} &amp; {{m}_{PQ}}\times {{m}_{\bot PQ}}=-1 \\ &amp; \Rightarrow {{m}_{\bot PQ}}=\dfrac{-1}{{{m}_{PQ}}} \\ &amp; \Rightarrow {{m}_{\bot PQ}}=\dfrac{{{x}_{1}}-{{x}_{2}}}{{{y}_{2}}-{{y}_{1}}}............\left( i \right) \\ \end{align}
And,
$\begin{array}{rl}& {m}_{\mathrm{\perp }QR}×{m}_{QR}=-1\\ & ⇒{m}_{\mathrm{\perp }QR}=\frac{-1}{{m}_{QR}}\\ & ⇒{m}_{\mathrm{\perp }QR}=\frac{{x}_{3}-{x}_{2}}{{y}_{2}-{y}_{3}}............\left(ii\right)\end{array}$\begin{align} &amp; {{m}_{\bot QR}}\times {{m}_{QR}}=-1 \\ &amp; \Rightarrow {{m}_{\bot QR}}=\dfrac{-1}{{{m}_{QR}}} \\ &amp; \Rightarrow {{m}_{\bot QR}}=\dfrac{{{x}_{3}}-{{x}_{2}}}{{{y}_{2}}-{{y}_{3}}}............\left( ii \right) \\ \end{align}
And,
$\begin{array}{rl}& {m}_{\mathrm{\perp }PR}×{m}_{PR}=-1\\ & ⇒{m}_{\mathrm{\perp }PR}=\frac{-1}{{m}_{PR}}\\ & ⇒{m}_{\mathrm{\perp }PR}=\frac{{x}_{1}-{x}_{3}}{{y}_{3}-{y}_{1}}............\left(iii\right)\end{array}$\begin{align} &amp; {{m}_{\bot PR}}\times {{m}_{PR}}=-1 \\ &amp; \Rightarrow {{m}_{\bot PR}}=\dfrac{-1}{{{m}_{PR}}} \\ &amp; \Rightarrow {{m}_{\bot PR}}=\dfrac{{{x}_{1}}-{{x}_{3}}}{{{y}_{3}}-{{y}_{1}}}............\left( iii \right) \\ \end{align}
Now, since $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\ and \left( {{x}_{3}},{{y}_{3}} \right)$ are rational then ${{m}_{\bot PQ}},{{m}_{\bot QR}}\ and {{m}_{\bot PR}}$, given by equation $\left( i \right),\left( ii \right)\ and \left( iii \right)$ must be rational.
Now, let’s consider the line perpendicular to sides and passing from the opposite vertex.
We know equation of line passing from $\left(a,b\right)$$\left( a,b \right)$ with slope $m$$m$ is given as :
$\left(y-b\right)=m\left(x-a\right)$$\left( y-b \right)=m\left( x-a \right)$
So, equation of line perpendicular to $PQ$$PQ$ and passing through $R$$R$ is given as :
$\left(y-{y}_{3}\right)={m}_{\mathrm{\perp }QR}\left(x-{x}_{3}\right)$$\left( y-{{y}_{3}} \right)={{m}_{\bot QR}}\left( x-{{x}_{3}} \right)$
$\left(y-{y}_{3}\right)=\frac{{x}_{1}-{x}_{2}}{{y}_{2}-{y}_{1}}\left(x-{x}_{3}\right)..........\left(iv\right)$$\left( y-{{y}_{3}} \right)=\dfrac{{{x}_{1}}-{{x}_{2}}}{{{y}_{2}}-{{y}_{1}}}\left( x-{{x}_{3}} \right)..........\left( iv \right)$
And equation of line perpendicular to $QR$$QR$ and passing through $P$$P$ is given as :
$\left(y-{y}_{1}\right)={m}_{\mathrm{\perp }QR}\left(x-{x}_{1}\right)$$\left( y-{{y}_{1}} \right)={{m}_{\bot QR}}\left( x-{{x}_{1}} \right)$
$\left(y-{y}_{1}\right)=\frac{{x}_{3}-{x}_{2}}{{y}_{2}-{y}_{3}}\left(x-{x}_{1}\right)..........\left(v\right)$$\left( y-{{y}_{1}} \right)=\dfrac{{{x}_{3}}-{{x}_{2}}}{{{y}_{2}}-{{y}_{3}}}\left( x-{{x}_{1}} \right)..........\left( v \right)$
And equation of line perpendicular to $PR$$PR$ and passing through $Q$$Q$ is given as :
$\left(y-{y}_{2}\right)={m}_{\mathrm{\perp }PR}\left(x-{x}_{2}\right)$$\left( y-{{y}_{2}} \right)={{m}_{\bot PR}}\left( x-{{x}_{2}} \right)$
$\left(y-{y}_{2}\right)=\frac{{x}_{1}-{x}_{3}}{{y}_{3}-{y}_{1}}\left(x-{x}_{2}\right)..........\left(vi\right)$$\left( y-{{y}_{2}} \right)=\dfrac{{{x}_{1}}-{{x}_{3}}}{{{y}_{3}}-{{y}_{1}}}\left( x-{{x}_{2}} \right)..........\left( vi \right)$
Now we know that ${{x}_{1}},{{x}_{2}},{{x}_{3}},{{y}_{1}},{{y}_{2}}\ and {{y}_{3}}$ all are rational.
Hence, the coordinates of the point satisfying all three equations will also be rational $......\left(vii\right)$$......\left( vii \right)$
Now we know the point of intersection of the perpendiculars from the vertices to the opposite sides is the orthocentre $......\left(viii\right)$$......\left( viii \right)$
Hence, from statements $\left(vii\right)\mathrm{&}\left(viii\right)$$\left( vii \right)\And \left( viii \right)$ we can say that ortho centres will be rational points.
Now, let’s consider the midpoint of the sides $PQ,QR\ and PR$ .

We know, the coordinates of the midpoint of the line joining $\left( {{a}_{1}},{{b}_{1}} \right)\ and \left( {{a}_{2}},{{b}_{2}} \right)$ is given as :
$\left(\frac{{a}_{1}+{a}_{2}}{2},\frac{{b}_{1}+{b}_{2}}{2}\right)$$\left( \dfrac{{{a}_{1}}+{{a}_{2}}}{2},\dfrac{{{b}_{1}}+{{b}_{2}}}{2} \right)$ .
So, the coordinates of midpoint of side $PQ$$PQ$ is given as $D\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)$$D\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$ .
The coordinates of midpoint of side $QR$$QR$ is given as $E\left(\frac{{x}_{2}+{x}_{3}}{2},\frac{{y}_{2}+{y}_{3}}{2}\right)$$E\left( \dfrac{{{x}_{2}}+{{x}_{3}}}{2},\dfrac{{{y}_{2}}+{{y}_{3}}}{2} \right)$ .
And the coordinates of midpoint of side $PR$$PR$ is given as $F\left(\frac{{x}_{1}+{x}_{3}}{2},\frac{{y}_{1}+{y}_{3}}{2}\right)$$F\left( \dfrac{{{x}_{1}}+{{x}_{3}}}{2},\dfrac{{{y}_{1}}+{{y}_{3}}}{2} \right)$ .
Now let’s find the equation of line perpendicular to the sides and passing through the midpoint of the sides i.e. perpendicular bisector of sides. So, the equation of perpendicular bisector of $PQ$$PQ$ is given as:
$y-\left(\frac{{y}_{1}+{y}_{2}}{2}\right)=\left(\frac{{x}_{1}-{x}_{2}}{{y}_{2}-{y}_{1}}\right)\left(x-\left(\frac{{x}_{1}+{x}_{2}}{2}\right)\right)......\left(ix\right)$$y-\left( \dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)=\left( \dfrac{{{x}_{1}}-{{x}_{2}}}{{{y}_{2}}-{{y}_{1}}} \right)\left( x-\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2} \right) \right)......\left( ix \right)$
The equation of perpendicular bisector of $QR$$QR$ is given as:
$y-\left(\frac{{y}_{2}+{y}_{3}}{2}\right)=\left(\frac{{x}_{3}-{x}_{2}}{{y}_{2}-{y}_{3}}\right)\left(x-\left(\frac{{x}_{2}+{x}_{3}}{2}\right)\right)......\left(x\right)$$y-\left( \dfrac{{{y}_{2}}+{{y}_{3}}}{2} \right)=\left( \dfrac{{{x}_{3}}-{{x}_{2}}}{{{y}_{2}}-{{y}_{3}}} \right)\left( x-\left( \dfrac{{{x}_{2}}+{{x}_{3}}}{2} \right) \right)......\left( x \right)$
The equation of perpendicular bisector of $PR$$PR$ is given as:
$y-\left(\frac{{y}_{1}+{y}_{3}}{2}\right)=\left(\frac{{x}_{1}-{x}_{3}}{{y}_{3}-{y}_{1}}\right)\left(x-\left(\frac{{x}_{1}+{x}_{3}}{2}\right)\right)......\left(xi\right)$$y-\left( \dfrac{{{y}_{1}}+{{y}_{3}}}{2} \right)=\left( \dfrac{{{x}_{1}}-{{x}_{3}}}{{{y}_{3}}-{{y}_{1}}} \right)\left( x-\left( \dfrac{{{x}_{1}}+{{x}_{3}}}{2} \right) \right)......\left( xi \right)$
Now we know, $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\ and \left( {{x}_{3}},{{y}_{3}} \right)$ are rational coordinates. Hence, any point satisfying all three equations i.e. $\left( ix \right),\left( x \right)\ and \left( xi \right)$ will have rational coordinates. $.............\left(xii\right)$$.............\left( xii \right)$
Now, we know the circumcentre is the point of intersection of the perpendicular bisectors of the sides of the triangle$..........\left(xiii\right)$$..........\left( xiii \right)$
So, from statements $\left(xii\right)\mathrm{&}\left(xiii\right)$$\left( xii \right)\And \left( xiii \right)$ we can conclude that the coordinates of the circumcentre are rational.
Now, we know the incentre of a triangle with vertices $\left({x}_{1},{y}_{1}\right),\left({x}_{2},{y}_{2}\right)\mathrm{&}\left({x}_{3},{y}_{3}\right)$$\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\And \left( {{x}_{3}},{{y}_{3}} \right)$ is given as :
$\left(\frac{a{x}_{1}+b{x}_{2}+c{x}_{3}}{a+b+c},\frac{a{y}_{1}+b{y}_{2}+c{y}_{3}}{a+b+c}\right)$$\left( \dfrac{a{{x}_{1}}+b{{x}_{2}}+c{{x}_{3}}}{a+b+c},\dfrac{a{{y}_{1}}+b{{y}_{2}}+c{{y}_{3}}}{a+b+c} \right)$ where $a,b,c$$a,b,c$ are the sides of the triangle opposite to the corresponding vertices .
Now, let $a$$a$ be the length of side $QR,b$$QR,b$ be the length of side $PR$$PR$ and $c$$c$ be the length of side $PQ$$PQ$ .
Now, we know the distance between the points $\left({a}_{1},{b}_{1}\right)\mathrm{&}\left({a}_{2},{b}_{2}\right)$$\left( {{a}_{1}},{{b}_{1}} \right)\And \left( {{a}_{2}},{{b}_{2}} \right)$ is given as :
\begin{align} &amp; \sqrt{{{\left( {{a}_{1}}-{{a}_{2}} \right)}^{2}}+{{\left( {{b}_{1}}-{{b}_{2}} \right)}^{2}}} \\ &amp; \text{So, }a=\sqrt{{{\left( {{x}_{2}}-{{x}_{3}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{3}} \right)}^{2}}}, \\ &amp; b=\sqrt{{{\left( {{x}_{1}}-{{x}_{3}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{3}} \right)}^{2}}}, \\ &amp; \text{and }c=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}} \\ \end{align}
Clearly , we can see that $a,b$$a,b$ and $c$$c$ may or may not be rational.
So, the coordinates of incentre may or may not be rational.
Hence , if the vertices $P,Q,R$$P,Q,R$ of a triangle $PQR$$PQR$ are rational points , then the centroid , the orthocentre and the circumcentre will always be rational points.
Options (a) Centroid (c) Circumcentre (d) Orthocentre are correct.

Note: While simplifying the equations , please make sure that sign mistakes do not occur. These mistakes are very common and can cause confusions while solving. Ultimately the answer becomes wrong. So, sign conventions should be carefully taken .