Question

Statement-$1:$ Let orthocentre of an acute triangle $ABC$ is at the origin and its circumcentre has the coordinates $(\frac{1}{2},\frac{-1}{2})$. If the base $BC$ has the equation $4x-2y=5,$ then the radius of the circle circumscribing the triangle $ABC$ is $\sqrt{\frac{5}{2}}.$

Statement-$2:$ In any acute triangle, the image of orthocentre in any side of the triangle lies on the circumcircle of the triangle.

• A. Statement-$1$ is true, statement-$2$ is true and statement-$2$ is correct explanation for statement-$1$.
• B. Statement-$1$ is true, statement-$2$ is true and statement-$2$ is NOT the correct explanation for statement-$1$.
• C. Statement-$1$ is true, statement-$2$ is false
• D. Statement-$1$ is false, statement-$2$ is true

Answer 1

The correct option is A Statement-$1$ is true, statement-$2$ is true and statement-$2$ is correct explanation for statement-$1$.

We know that, image of orthocentre about any side of a triangle lies on circumcircle.
Now, image of the origin $(0,0)$ about the lines $4x–2y=5$ is,
$\frac{x-0}{4}=\frac{y-0}{-2}=-2\left(\frac{-5}{16+4}\right)$
$\Rightarrow x=2,y=-1$
$\Rightarrow P\equiv (2,-1)$
$(2,–1)$ which must lie on the circumcircle of the $\Delta ABC$.
Hence, radius = distance between $(2,–1)$ and $(\frac{1}{2},-\frac{1}{2})$
$=\sqrt{\frac{9}{4}+\frac{1}{4}}=\sqrt{\frac{10}{4}}=\sqrt{\frac{5}{2}}$