Question:

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

Updated On: Oct 10, 2024
  • $2+\sqrt{2}$
  • $2-\sqrt{2}$
  • $1+\sqrt{2}$
  • $1-\sqrt{2}$
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The Correct Option is B

Solution and Explanation

Given m id-points of a triangle are (0, 1), (1, 1) and (1, 0). P lotting these points on a 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$=\frac{2\times0+2\sqrt{2}.0+2.2}{2+2+\sqrt{2}}$
$=\frac{2}{2+\sqrt{2}}\times\frac{2-\sqrt{2}}{2-\sqrt{2}}=2-\sqrt{2}$
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Questions Asked in JEE Main exam

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Concepts Used:

Straight lines

A straight line is a line having the shortest distance between two points. 

A straight line can be represented as an equation in various forms,  as show in the image below:

 

The following are the many forms of the equation of the line that are presented in straight line-

1. Slope – Point Form

Assume P0(x0, y0) is a fixed point on a non-vertical line L with m as its slope. If P (x, y) is an arbitrary point on L, then the point (x, y) lies on the line with slope m through the fixed point (x0, y0) if and only if its coordinates fulfil the equation below.

y – y0 = m (x – x0)

2. Two – Point Form

Let's look at the line. L crosses between two places. P1(x1, y1) and P2(x2, y2)  are general points on L, while P (x, y) is a general point on L. As a result, the three points P1, P2, and P are collinear, and it becomes

The slope of P2P = The slope of P1P2 , i.e.

\(\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}\)

Hence, the equation becomes:

y - y1 =\( \frac{y_2-y_1}{x_2-x_1} (x-x1)\)

3. Slope-Intercept Form

Assume that a line L with slope m intersects the y-axis at a distance c from the origin, and that the distance c is referred to as the line L's y-intercept. As a result, the coordinates of the spot on the y-axis where the line intersects are (0, c). As a result, the slope of the line L is m, and it passes through a fixed point (0, c). The equation of the line L thus obtained from the slope – point form is given by

y – c =m( x - 0 )

As a result, the point (x, y) on the line with slope m and y-intercept c lies on the line, if and only if

y = m x +c