Main 2013 | JEE 2013 | Discussion Page

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

$2+\sqrt{2}$

$2-\sqrt{2}$

$1+\sqrt{2}$

$1-+\sqrt{2}$

Option B

Given midpoints of a triangle are (0,1),(1,1) and (1,0) . Plotting these points on graph paper and make a triangle.

So the sides of a triangle will be 2,2 and $\sqrt{2^{2}+2^{2}}$ $i.e 2\sqrt{2}$

x- coordinate of incentre

= $\frac{2\times0+2\sqrt{2}.0+2.2}{2+2+2\sqrt{2}}$

$=\frac{2}{2+\sqrt{2}}\times\frac{2-\sqrt{2}}{2-\sqrt{2}}=2-\sqrt{2}$

Discussion Forum