Question

The orthocentre of the triangle with vertices
$(2,\frac{\sqrt{3}-1}{2}),(\frac{1}{2},-\frac{1}{2})and(2,-\frac{1}{2})$ is

• A. $(\frac{3}{2},\frac{\sqrt{3}-3}{6})$
• B. $(2,-\frac{1}{2})$
• C. $(\frac{5}{4},\frac{\sqrt{3}-2}{4})$
• D. $(\frac{1}{2},-\frac{1}{2})$

Answer 1

The correct option is B

$(2,-\frac{1}{2})$

Let A, B and C be $(2,\frac{\sqrt{3}-1}{2}),(\frac{1}{2},-\frac{1}{2})and(2,-\frac{1}{2})$
${D_{AB}}^2=(2-\frac{1}{2}{)}^2+(\frac{\sqrt{3}-1}{2}+\frac{1}{2}{)}^2$
${D_{AB}}^2=3$

${D_{BC}}^2=(\frac{1}{2}-2{)}^2+(-\frac{1}{2}+\frac{1}{2}{)}^2$
${D_{BC}}^2=\frac{9}{4}$

${D_{AC}}^2=(2-2{)}^2+(\frac{\sqrt{3}-1}{2}+\frac{1}{2}{)}^2$
${D_{AC}}^2=\frac{3}{4}$

Looking closely, we see that
${BC}^2+{AC}^2={AB}^2$
The points $(2,\frac{\sqrt{3}-1}{2}),(\frac{1}{2},-\frac{1}{2})and(2,-\frac{1}{2})$
are the vertices of a right triangle.
Since $(2,-\frac{1}{2})$ is the vertex where the right angle is formed. In a right angled triangle the vertex containing the right angle is the orthocentre. (orthocentre is the point of intersection of the altitudes of a triangle)
$\therefore$ orthocentre is $(2,-\frac{1}{2})$.