Question

The coordinates of the in-centre of the triangle formed by the lines 3x + 4y = 12 and the coordinate axes are _____.

Answer 1

Hint: In this question, we will be using the formula for the coordinates of the in-centre of a triangle when the length of the sides a, b and c are given, at the coordinates of the vertices are also given. The in-centre is given by-
$I\left( {\dfrac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}},\;\dfrac{{a{y_1} + b{y_2} + c{y_3}}}{{a + b + c}}} \right)$

Complete step by step answer:

The three sides of the triangle are x=0, y=0 and 3x + 4y = 12. Solving the equations, coordinates are obtained easily as A(4, 0), B(0, 3) and C(0, 0).
Now, from the graph, we can see clearly that a = 3 units and b = 4 units. The triangle formed is a a right angled triangle, so c can be calculated by Pythagoras’ theorem-
c2 = a2 + b2
c2 = 32 + 42
c2 = 25
c = 5 units
So, the coordinates of incentre can be calculated as-
${\text{I}}\left( {\dfrac{{3\left( 4 \right) + 4\left( 0 \right) + 5\left( 0 \right)}}{{3 + 4 + 5}},\dfrac{{3\left( 0 \right) + 4\left( 3 \right) + 5\left( 0 \right)\;}}{{3 + 4 + 5}}} \right)$
${\text{I}}\left( {\dfrac{{12}}{{12}},\dfrac{{\;12}}{{12}}} \right)$
${\text{I}}\left( {1,1} \right)$
This is the required answer. The coordinates of in-centre are (1, 1).
Note: If we don’t remember the correct formula, we can find the coordinates by finding the angle bisectors of any two pairs of lines. This can be done by-
By geometry, the angle bisector of x = 0 and y = 0 is x = y.
The angle bisector of x = 0 and 3x + 4y = 12 is-
$\dfrac{{3{\text{x}} + 4{\text{y}} - 12}}{5} = \dfrac{{\text{x}}}{1}$
3x + 4y -12 = -5x
8x + 4y - 12 = 0
4x + 2y - 6 =0
The in-centre is at the point of intersection of these two lines.
4x + 2y - 6 = 0 and x = y
4x + 2x - 6 = 0
x = 1 and y = 1
These are the coordinates- I(1, 1)