Let ABC be an equilateral triangle with ortho | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) In an equilateral triangle,the orthocenter coincides with the centroid. Let $O(0,0)$ be the orthocenter. The altitude $AD$ passes through $O$. The

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) In an equilateral triangle,the orthocenter coincides with the centroid. Let $O(0,0)$ be the orthocenter. The altitude $AD$ passes through $O$. The slope of $BC$ is $m_{BC} = -\frac{1}{2\sqrt{2}}$. Since $AD \perp BC$,the slope of $AD$ is $m_{AD} = 2\sqrt{2}$. The equation of $AD$ is $y = 2\sqrt{2}x$,so $\beta = 2\sqrt{2}\alpha$. The distance from $O(0,0)$ to $BC$ $(x+2\sqrt{2}y-4=0)$ is $OD = \frac{|0+0-4|}{\sqrt{1^2+(2\sqrt{2})^2}} = \frac{4}{\sqrt{9}} = \frac{4}{3}$. In an equilateral triangle,the centroid divides the altitude in the ratio $2:1$. Thus,$AO = 2OD = 2(\frac{4}{3}) = \frac{8}{3}$. The total length of the altitude $AD = AO + OD = \frac{8}{3} + \frac{4}{3} = 4$. The vertex $A$ lies on the line $y = 2\sqrt{2}x$ at a distance $4$ from the line $BC$. The coordinates of $A$ are $(\alpha, 2\sqrt{2}\alpha)$. The distance from $A$ to $x+2\sqrt{2}y-4=0$ is $\frac{|\alpha+2\sqrt{2}(2\sqrt{2}\alpha)-4|}{\sqrt{1+8}} = \frac{|9\alpha-4|}{3} = 4$. This gives $9\alpha-4 = 12$ or $9\alpha-4 = -12$. So $\alpha = \frac{16}{9}$ or $\alpha = -\frac{8}{9}$. Since $O(0,0)$ and $A$ must be on the same side of $BC$ (as $O$ is the centroid),we test the sign of the expression $x+2\sqrt{2}y-4$ at $(0,0)$,which is $-4$. For $A(\alpha, 2\sqrt{2}\alpha)$,the expression is $9\alpha-4$. Thus $9\alpha-4 We need to find the greatest integer less than or equal to $|\alpha+\sqrt{2}\beta| = |-\frac{8}{9} + \sqrt{2}(-\frac{16\sqrt{2}}{9})| = |-\frac{8}{9} - \frac{32}{9}| = |-\frac{40}{9}| = \frac{40}{9} \approx 4.44$. The greatest integer is $4$.

Let $ABC$ be an equilateral triangle with orthocenter at the origin and the side $BC$ on the line $x+2\sqrt{2}y=4$. If the coordinates of the vertex $A$ are $(\alpha, \beta)$,then the greatest integer less than or equal to $|\alpha+\sqrt{2}\beta|$ is

Similar Questions

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

The distance (in units) between the circumcentre and the centroid of the triangle formed by the vertices $(1, 2)$,$(3, -1)$ and $(4, 0)$ is

If in triangle $ABC$,$A \equiv (1, 10)$,circumcentre $\equiv \left( -\frac{1}{3}, \frac{2}{3} \right)$ and orthocentre $\equiv \left( \frac{11}{3}, \frac{4}{3} \right)$,then the coordinates of the mid-point of the side opposite to $A$ are:

Assertion $(A)$: If the centroid and circumcenter of a triangle are known,its orthocenter can be found.
Reason $(R)$: The centroid,orthocenter,and circumcenter of a triangle are collinear.

The coordinates of the orthocentre of the triangle whose sides are given by the lines $x = 3$,$y = 4$,and $3x + 4y = 6$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module