Question

The coordinates of the incentre of the triangle having sides $3x - 4y = 0$, $5x + 12y = 0$ and $y - 15 = 0$ are
A) $\left( { - 1,8} \right)$
B) $\left( {1, - 8} \right)$
C) $\left( {2,6} \right)$
D) None of these

Answer 1

Hint: First we are to find the points of intersection of the three lines given in order to find the coordinates of the sides of the triangle. Then we can find the distance of the sides of the triangle. Further by using the formula of the incentre of a triangle, that is,
Incentre $ = \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)$
Where, $a$ is the length of the side opposite to $\left( {{x_1},{y_1}} \right)$
$b$ is the length of the side opposite to $\left( {{x_2},{y_2}} \right)$
$c$ is the length of the side opposite to $\left( {{x_3},{y_3}} \right)$.

Complete step by step answer:
The given lines are,
$3x - 4y = 0 - - - \left( 1 \right)$
$5x + 12y = 0 - - - \left( 2 \right)$
$y - 15 = 0 - - - \left( 3 \right)$
So, in triangle $ABC$, the equations $\left( 1 \right),\left( 2 \right),\left( 3 \right)$ represents the sides $AB,BC,CA$, respectively.
Now, solving $\left( 1 \right)$ and $\left( 2 \right)$, we get the point $B$.
So, $\left( 1 \right) \times 3 + \left( 2 \right)$ gives,
$3\left( {3x - 4y} \right) + \left( {5x + 12y} \right) = 0$
$ \Rightarrow 9x - 12y + 5x + 12y = 0$
Now, simplifying, we get,
$ \Rightarrow 14x = 0$
$ \Rightarrow x = 0$
Substituting this in $\left( 1 \right)$, we get,
$3\left( 0 \right) - 4y = 0$
$ \Rightarrow 0 - 4y = 0$
$ \Rightarrow 4y = 0$
$ \Rightarrow y = 0$
Therefore, $x = 0,y = 0$.
Hence, the point $B\left( {0,0} \right)$.
Now, solving $\left( 2 \right)$ and $\left( 3 \right)$, we get the point $C$.
From $\left( 3 \right)$, we get,
$y = 15$
Substituting this value in $\left( 2 \right)$, we get,
$5x + 12\left( {15} \right) = 0$
$ \Rightarrow 5x + 180 = 0$
Now, subtracting both sides by $180$, we get,
$ \Rightarrow 5x = - 180$
Dividing both sides by $5$, we get,
$ \Rightarrow x = - 36$
Therefore, we get, $x = - 36,y = 15$.
Hence, the point $C\left( { - 36,15} \right)$.
Now, solving $\left( 1 \right)$ and $\left( 3 \right)$, we get the point $A$.
From $\left( 3 \right)$, we get,
$y = 15$
Substituting this value in $\left( 1 \right)$, we get,
$3x - 4\left( {15} \right) = 0$
$ \Rightarrow 3x - 60 = 0$
Now, adding $60$ to both sides, we get,
$ \Rightarrow 3x = 60$
Now, dividing both sides by $3$, we get,
$ \Rightarrow x = 20$
Therefore, we get, $x = 20,y = 15$.
Hence, the point $A\left( {20,15} \right)$.
Therefore the points of the triangle $ABC$ are $A\left( {20,15} \right)$, $B\left( {0,0} \right)$, $C\left( { - 36,15} \right)$.
Now, by using the distance formula we can find the distance between the points.
Therefore, the distance formula is $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
Now, using this formula, we get,
$AB = c = \sqrt {{{\left( {0 - 20} \right)}^2} + {{\left( {0 - 15} \right)}^2}} $
$ \Rightarrow c = \sqrt {{{\left( { - 20} \right)}^2} + {{\left( { - 15} \right)}^2}} $
\[ \Rightarrow c = \sqrt {400 + 225} \]
Now, simplifying the square root, we get,
$ \Rightarrow AB = c = \sqrt {625} = 25$
And, $BC = a = \sqrt {{{\left( { - 36 - 0} \right)}^2} + {{\left( {15 - 0} \right)}^2}} $
$ \Rightarrow a = \sqrt {{{\left( { - 36} \right)}^2} + {{\left( {15} \right)}^2}} $
\[ \Rightarrow a = \sqrt {1296 + 225} \]
Now, simplifying the square root, we get,
$ \Rightarrow BC = a = \sqrt {1521} = 39$
Again, $AC = b = \sqrt {{{\left( { - 36 - 20} \right)}^2} + {{\left( {15 - 15} \right)}^2}} $
$ \Rightarrow b = \sqrt {{{\left( { - 56} \right)}^2} + {{\left( 0 \right)}^2}} $
\[ \Rightarrow b = \sqrt {3136 + 0} \]
Now, simplifying the square root, we get,
$ \Rightarrow AC = b = \sqrt {3136} = 56$
Therefore, $AB = c = 25,BC = a = 39,AC = b = 56$.
And, the points are, $A\left( {{x_1},{y_1}} \right) = \left( {20,15} \right)$, $B\left( {{x_2},{y_2}} \right) = \left( {0,0} \right)$, $C\left( {{x_3},{y_3}} \right) = \left( { - 36,15} \right)$.
Now, using the formula of the incentre of a triangle, we get,
Incentre $ = \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)$
Now, substituting the values, we get,
Incentre $ = \left( {\dfrac{{39\left( {20} \right) + 56\left( 0 \right) + 25\left( { - 36} \right)}}{{39 + 56 + 25}},\dfrac{{39\left( {15} \right) + 56\left( 0 \right) + 25\left( {15} \right)}}{{39 + 56 + 25}}} \right)$
Now, simplifying, we get,
$ = \left( {\dfrac{{780 + 0 - 900}}{{120}},\dfrac{{585 + 0 + 375}}{{120}}} \right)$
\[ = \left( {\dfrac{{ - 120}}{{120}},\dfrac{{960}}{{120}}} \right)\]
Now, diving the terms, we get,
Incentre $ = \left( { - 1,8} \right)$
Therefore, the coordinates of the incentre of the triangle are $\left( { - 1,8} \right)$, the correct option is (A).

Note:
The incentre of a triangle is the point inside the triangle where the angle bisectors of the three angles meet inside the triangle. Together with the centroid, circumcentre and orthocentre, it is one of the four triangle centres known to the ancient Greeks. Care should be taken while carrying out the calculations so as to be sure of the final answer.