The incentre of the triangle with vertices (1 | vedclass

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

Question

(D) Let the vertices be $A(1, \sqrt{3})$,$B(0, 0)$,and $C(2, 0)$.Calculate the side lengths:$AB = \sqrt{(1-0)^2 + (\sqrt{3}-0)^2} = \sqrt{1 + 3} = 2$$

Vedclass · Accepted Answer

$\left(1, \frac{1}{\sqrt{3}}\right)$

Vedclass · Answer

(D) Let the vertices be $A(1, \sqrt{3})$,$B(0, 0)$,and $C(2, 0)$.Calculate the side lengths:$AB = \sqrt{(1-0)^2 + (\sqrt{3}-0)^2} = \sqrt{1 + 3} = 2$$BC = \sqrt{(2-0)^2 + (0-0)^2} = \sqrt{4} = 2$$AC = \sqrt{(2-1)^2 + (0-\sqrt{3})^2} = \sqrt{1 + 3} = 2$Since all sides are equal,the triangle is an equilateral triangle.For an equilateral triangle,the incentre is the same as the centroid.The centroid $(G)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.$G = \left(\frac{1+0+2}{3}, \frac{\sqrt{3}+0+0}{3}\right) = \left(\frac{3}{3}, \frac{\sqrt{3}}{3}\right) = \left(1, \frac{1}{\sqrt{3}}\right)$.

The incentre of the triangle with vertices $(1, \sqrt{3}), (0, 0)$ and $(2, 0)$ is:

Similar Questions

In a triangle $PQR$,the coordinates of the points $P$ and $Q$ are $(-2, 4)$ and $(4, -2)$ respectively. If the equation of the perpendicular bisector of $PR$ is $2x - y + 2 = 0$,then the centre of the circumcircle of the $\Delta PQR$ is

Find the orthocenter of the triangle with vertices $A(0, 0)$,$B(3, 4)$,and $C(4, 0)$.

The orthocentre of the triangle formed by the vertices $(0, 0)$,$(8, 0)$,and $(4, 6)$ is

If the coordinates of the vertices of a triangle are $(1, a)$,$(2, b)$,and $(c^2, 3)$,then the centroid of the triangle:

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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module