Question

The vertices of a variable triangle are $(3,4),(5\cos \theta ,5\sin \theta ),and(5\sin \theta ,-5\cos \theta ),where\theta ϵR.$ then the locus of its orthocenter is

• A. ${(x-3)}^2+{(y-4)}^2=50$
• B. $(x+y-7{)}^2+(x-y-7{)}^2=100$
• C. $(x+y-7{)}^2+(x-y+7{)}^2=100$
• D. $(x-y-7{)}^2+(x-y-7{)}^2=100$

Answer 1

The correct option is A ${(x-3)}^2+{(y-4)}^2=50$

We have,

$A(x_1,y_1)=(3,4)$

$B(x_2,y_2)=(5cos\theta ,5sin\theta )$

$C(x_3,y_3)=(5sin\theta ,-5\cos \theta )$

Where $\theta \in R$

Let $O$ be the origin then it is equidistant from the vertices of $\Delta ABC$

Now,

$OA=\sqrt{{(3-0)}^2+{(4-0)}^2}$

$OA=5$

$OB=\sqrt{{(5\cos \theta -0)}^2+{(5\sin \theta -0)}^2}$

$=\sqrt{25{\cos }^2\theta +25{\sin }^2\theta }$

$=\sqrt{25({\sin }^2\theta +{\cos }^2\theta )}$

$OB=5$

$OC=\sqrt{{(5\sin \theta -0)}^2+{(-5\cos \theta -0)}^2}$

$=\sqrt{25{\sin }^2\theta +25{\cos }^2\theta }$

$=\sqrt{25}$

$OC=5$

So, We can conclude that, $O(0,0)$ is the circumcenter of $\Delta ABC$

Let $H(h,k)$ be the orthocenter of $\Delta ABC$.

The centroid $G$ is given by,

Coordinate of $G=(\frac{3+5\cos \theta +5\sin \theta }{3},\frac{4+5sin\theta -5\cos \theta }{3}).......\left(1\right)$

We note that, Centroid $G$ divides the line joining the orthocenter $H(h,k)$ and circumcentre $O(0,0)$ in the ratio $2:1$.

Using section formula, we get,

Coordinate of $G=(\frac{h}{3},\frac{k}{3})$……. (2)

By equation (1) and (2) to and we get,

$h=3+5\cos \theta +5\sin \theta$

$k=4+5\sin \theta -5\cos \theta$

This further implies that,

$h-3=5\cos \theta +5sin\theta$

$k-4=5sin\theta -5cos\theta$

Squaring both side and we get,

${(h-3)}^2+{(k-4)}^2={(5\cos \theta +5\sin \theta )}^2+{(5\sin \theta -5\cos \theta )}^2$

${(h-3)}^2+{(k-4)}^2=25{\cos }^2\theta +25{\sin }^2\theta +50\sin \theta \cos \theta +25{\cos }^2\theta +25{\sin }^2\theta -50\sin \theta \cos \theta$

${(h-3)}^2+{(k-4)}^2=50({\sin }^2\theta +{\cos }^2\theta )$

${(h-3)}^2+{(k-4)}^2=50.$

So the locus of orthocenter is,

${(x-3)}^2+{(y-4)}^2=50$

Hence, this is the answer.