Question

The coordinates of the incentre of the triangle having sides $3x-4y=0,5x+12y=0$ and $y-15=0$ are

• A. $(-1,8)$
• B. $(1,–8)$
• C. $(2,6)$
• D. None of these

Answer 1

The correct option is A

$(-1,8)$

Explanation for the correct option:

Step-1 : Find the vertices of traingle :

Given the sides of triangle are

$$\begin{gathered} AC\equiv 3x-4y=0...\left(i\right) \\ AB\equiv 5x+12y=0...\left(ii\right) \\ BC\equiv y-15=0...\left(iii\right) \end{gathered}$$

Finding vertices of the triangle by solving equations $\left(i\right),\left(ii\right)\&\left(iii\right)$

From $\left(ii\right)\&\left(ii\right)$ we get

$$\begin{gathered} x=0,y=0 \\ \therefore A(0,0) \end{gathered}$$

From $\left(ii\right)\&\left(iii\right)$ we get

$$\begin{gathered} x=-36 \\ \Rightarrow y=15 \\ \therefore B(-36,15) \end{gathered}$$

From $\left(i\right)\&\left(iii\right)$ we get

$$\begin{gathered} y=15 \\ \Rightarrow x=20 \\ \therefore C(20,15) \end{gathered}$$

Therefore, the vertices of the triangle are $A(0,0),B(-36,15)\&C(20,15)$

Step-2 :The length of sides:

The distance formula is given by

$\begin{array}{rcl}D& =& \sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\end{array}$

Now calculating the sides of triangle by using the distance formula.

$\begin{array}{rcl}AB& =& \sqrt{{\left(-36-0\right)}^2+{\left(15-0\right)}^2}\\ & =& 39\\ & =& a\\ BC& =& \sqrt{{\left(20+36\right)}^2+{\left(15-15\right)}^2}\\ & =& 56\\ & =& b\\ AC& =& \sqrt{{\left(20-0\right)}^2+{\left(15+0\right)}^2}\\ & =& 25\\ & =& c\end{array}$

Step-3 : Find the incentre of triangle:

Now, the incentre of the triangle is

$$\begin{gathered} =\left(\frac{x_{1}a+x_{2}b+x_{3}c}{a+b+c},\frac{y_{1}a+y_{2}b+y_{3}c}{a+b+c}\right) \\ =\left(\frac{0\times 56-36\times 25+20\times 39}{120},\frac{0\times 56+15\times 25+15\times 39}{120}\right) \\ =\left(\frac{-900+780}{120},\frac{375+585}{120}\right) \\ =\left(\frac{-120}{120},\frac{960}{120}\right) \\ =\left(-1,8\right) \end{gathered}$$

Hence, option (A) is the correct answer.