Question

If coordinates of the vertices of a triangle are $(2,0),(6,0)$ and $(1,5)$, then distance between its orthocentre and circumcentre is

• A. $4$
• B. $6$
• C. $5$
• D. $Noneofthese$

Answer 1

The correct option is A $4$

Let the vertices of triangle be $A(2,0),B(6,0),C(1,5)$

Let the cir-cum center be $P(h,k)$

Then,

$AP=BP$

$A{P}^2=B{P}^2$

$(h-2{)}^2+k^2=(h-6{)}^2+k^2$

$h=4$

Also, $AP=CP$

$A{P}^2=C{P}^2$

$(h-2{)}^2+k^2=(h-1{)}^2+(k-5{)}^2$

$h^2+4-4h+k^2=h^2+1-2h+k^2+25-10k$

$k=3$

So, cir-cum center of the triangle $ABC$ is $(4,3)$

Centroid $\left(G\right)$ of the $\Delta ABC=(\frac{2+6+1}{3},\frac{0+0+5}{3})=(3,\frac{5}{3})$

We know that,

Distance between the centroid and orthocenter is equal to the twice of the distance between the centroid and the circumcentre

So, Distance between the orthocentre and circumcentre $=3\times GC=3\times \sqrt{(4-3{)}^2+(3-\frac{5}{3}{)}^2}=5units$