Question

The incentre of the triangle formed by the lines $y = \left| x \right|$ and $y = 1$ is:
A) $\left( {0,2 - \sqrt 2 } \right)$
B) $\left( {2 - \sqrt 2 ,0} \right)$
C) $\left( {2 + \sqrt 2 ,0} \right)$
D) $\left( {0,2 + \sqrt 2 } \right)$

Answer 1

Hint: We will first find the lengths of sides of triangle using the distance formula:
$d = \sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}} $
Then we will use the following formula for finding the incentre of the triangle: \[\left( {\dfrac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}}} \right),\left( {\dfrac{{a{y_1} + b{y_2} + c{y_3}}}{{a + b + c}}} \right)\]

Complete step by step solution: The incentre of a triangle is the point of intersection of the angle bisectors of the angles of the triangle. Also, an incentre is the centre of the circle inscribing the triangle.
An incentre is also a point that is equidistant from the sides of the triangle.
Now, consider a triangle $\Delta ABC$ as in the following figure, formed by the given lines $y = \left| x \right|$ and $y = 1$

Vertex A is \[\left( { - 1,1} \right)\]
Vertex B is \[\left( {1,1} \right)\]
Vertex C is \[\left( {0,0} \right)\]
For a triangle, having vertices $\left( {{x_1},{y_1}} \right);\left( {{x_2},{y_2}} \right);\left( {{x_3},{y_3}} \right)$ and a, b, c be the length of sides opposite to vertices $\left( {{x_1},{y_1}} \right);\left( {{x_2},{y_2}} \right);\left( {{x_3},{y_3}} \right)$respectively, then the coordinates of incentre of the triangle are given by:
\[\left( {\dfrac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}}} \right),\left( {\dfrac{{a{y_1} + b{y_2} + c{y_3}}}{{a + b + c}}} \right)\]
So, for the triangle $\Delta ABC$, we have:
$\begin{gathered}
  a = BC \\
  b = AC \\
  c = AB \\
\end{gathered} $
Let’s the coordinates of incentre are $\left( {h,k} \right)$
Now, by applying distance formula, between points $\left( {{x_1},{y_1}} \right)$and $\left( {{x_2},{y_2}} \right)$ we have
$d = \sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}} $
Thus, finding the values of lengths of sides of triangle by distance formula
\[\begin{gathered}
  a = \sqrt {{{\left( {1 - 0} \right)}^2} + {{\left( {1 - 0} \right)}^2}} = \sqrt 2 \\
  b = \sqrt {{{\left( { - 1 - 0} \right)}^2} + {{\left( {1 - 0} \right)}^2}} = \sqrt 2 \\
  c = \sqrt {{{\left( { - 1 - 1} \right)}^2} + {{\left( {1 - 1} \right)}^2}} = 2 \\
\end{gathered} \]
Thus, coordinates of incentre are,
\[\begin{gathered}
  h = \left( {\dfrac{{ - \sqrt 2 + \sqrt 2 }}{{\sqrt 2 + \sqrt 2 + 2\sqrt 2 }}} \right) = 0 \\
  k = \left( {\dfrac{{\sqrt 2 + \sqrt 2 }}{{\sqrt 2 + \sqrt 2 + 2\sqrt 2 }}} \right) = \dfrac{1}{2} \\
\end{gathered} \]
Hence, incentre of the triangle formed by lines $y = \left| x \right|$ and $y = 1$ is \[\left( {0,\dfrac{1}{2}} \right)\]

Note: Students must remember the definition of incentre carefully, and not get confused with other terms like orthocenter, circumcentre and centroid of a triangle. If the student finds difficulty in memorizing the formula for coordinates of incentre, he/she can find it by finding the intersection point of angle bisectors of the angles of the triangle.
For a triangle, having vertices $\left( {{x_1},{y_1}} \right);\left( {{x_2},{y_2}} \right);\left( {{x_3},{y_3}} \right)$ and a, b, c be the length of sides opposite to vertices $\left( {{x_1},{y_1}} \right);\left( {{x_2},{y_2}} \right);\left( {{x_3},{y_3}} \right)$respectively, then the coordinates of
Centroid: \[\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)\]
Circumcentre: For circumcentre, we can find the midpoints of each side. Then find the equations of the perpendicular bisector of the sides. Their point of intersection is the circumcentre.
Orthocenter: It is the point of intersection of altitudes. An altitude is perpendicular to the opposite side. So, the product of slopes of altitude and the opposite side will be -1, from here slope of altitude can be calculated. Thus, the equation of each altitude can be calculated with one vertex and slope, and their point of intersection as Orthocenter.