Let A(1,1), B(1,-1), C(-1,1) be the vertices | vedclass

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

Question

(B) The vertices are $A(1,1), B(1,-1), C(-1,1)$.Since $AB$ is vertical (along $x=1$) and $AC$ is horizontal (along $y=1$),$	riangle ABC$ is a right-a

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The vertices are $A(1,1), B(1,-1), C(-1,1)$.Since $AB$ is vertical (along $x=1$) and $AC$ is horizontal (along $y=1$),$	riangle ABC$ is a right-angled triangle at $A(1,1)$.$1$. Circumcentre $S$: For a right-angled triangle,the circumcentre is the midpoint of the hypotenuse $BC$.$S = \left(\frac{1+(-1)}{2}, \frac{-1+1}{2}ight) = (0,0)$.$2$. Orthocentre $O$: For a right-angled triangle,the orthocentre is the vertex at the right angle.$O = A = (1,1)$.$3$. Incentre $I$: The side lengths are $c = AB = \sqrt{(1-1)^2 + (1-(-1))^2} = 2$,$b = AC = \sqrt{(1-(-1))^2 + (1-1)^2} = 2$,and $a = BC = \sqrt{(1-(-1))^2 + (-1-1)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}$.The coordinates of the incentre $I$ are given by $\left(\frac{ax_A + bx_B + cx_C}{a+b+c}, \frac{ay_A + by_B + cy_C}{a+b+c}ight)$.$I = \left(\frac{2\sqrt{2}(1) + 2(1) + 2(-1)}{2\sqrt{2}+2+2}, \frac{2\sqrt{2}(1) + 2(-1) + 2(1)}{2\sqrt{2}+2+2}ight) = \left(\frac{2\sqrt{2}}{2\sqrt{2}+4}, \frac{2\sqrt{2}}{2\sqrt{2}+4}ight) = \left(\frac{\sqrt{2}}{\sqrt{2}+2}, \frac{\sqrt{2}}{\sqrt{2}+2}ight) = \left(\frac{1}{1+\sqrt{2}}, \frac{1}{1+\sqrt{2}}ight) = (\sqrt{2}-1, \sqrt{2}-1)$.Now,calculate $IS + OS$:$OS = \sqrt{(1-0)^2 + (1-0)^2} = \sqrt{1+1} = \sqrt{2}$.$IS = \sqrt{(\sqrt{2}-1-0)^2 + (\sqrt{2}-1-0)^2} = \sqrt{2(\sqrt{2}-1)^2} = \sqrt{2}(\sqrt{2}-1) = 2 - \sqrt{2}$.$IS + OS = (2 - \sqrt{2}) + \sqrt{2} = 2$.

Let $A(1,1), B(1,-1), C(-1,1)$ be the vertices of $\triangle ABC$. Let $S$ be the circumcentre,$O$ be the orthocentre,and $I$ be the incentre of the $\triangle ABC$. Then $IS + OS =$ ?

Similar Questions

Assertion: If $(0, 3), (1, 1)$ and $(-1, 2)$ are the midpoints of the sides of a triangle,then the centroid of the original triangle is $(0, 2)$.
Reason: The centroid of a triangle and the centroid of the triangle formed by joining the midpoints of the sides of the original triangle are the same.

The $x-$ coordinate of the incentre of the triangle that has the coordinates of midpoints of its sides as $(0,1), (1,1)$ and $(1,0)$ is

What is the circumcenter of the triangle with vertices $(0, 0)$,$(3, 0)$,and $(0, 4)$?

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

If the vertices of a triangle are $(a, b - c)$,$(b, c - a)$,and $(c, a - b)$,then the centroid of the triangle lies

Vedclass Test Series

Exam Paper Generator

Online Exam Module