The ordinate of incentre of the triangle formed by lines 3x+4y=0 and 5x-12y=0 and y=15 is ? Plz tell which formula has to be applied... Options are:
(A) 3 (B) 6 (C)8  (D) 12

Dear student

By solving the equation of the 3 sides of triangle simultaneously, we get three vertices of triangle.

$$\begin{gathered} 3\mathrm{x}+4\mathrm{y}=0 ....\left(1\right) \\ 5\mathrm{x}-12\mathrm{y}=0 ....\left(2\right) \\ \mathrm{y}=15 ....\left(3\right) \\ \mathrm{The} \mathrm{point} \mathrm{of} \mathrm{intersection} \mathrm{of} \left(1\right) \mathrm{and} \left(2\right) \mathrm{is} \left(0, 0\right). \\ \mathrm{Put} \mathrm{y} = 15 \mathrm{in} \left(1\right) \mathrm{we} \mathrm{get}, 3\mathrm{x}+4\times 15=0 \Rightarrow 3\mathrm{x}+60=0 \Rightarrow \mathrm{x}=-20 \\ \mathrm{so} \mathrm{point} \mathrm{of} \mathrm{intersection} \mathrm{of} \left(1\right) \mathrm{and} \left(3\right) \mathrm{is} \left(-20, 15\right) \\ \mathrm{Put} \mathrm{y} = 15 \mathrm{in} \left(2\right) \mathrm{we} \mathrm{get}, 5\mathrm{x}-12\times 15=0 \Rightarrow 5\mathrm{x}-180=0 \Rightarrow \mathrm{x}=36 \\ \mathrm{So}, \mathrm{point} \mathrm{of} \mathrm{intersection} \mathrm{of} \left(2\right) \mathrm{and} \left(3\right) \mathrm{is} \left(36, 15\right) \end{gathered}$$

Hence, the coordinate of vertices of triangle are (0,0) , (-20,15) and (36,15)

$$\begin{gathered} \mathrm{Now}, \mathrm{distance} \mathrm{between} \left(-20, 15\right) \mathrm{and} \left(36,15\right) =\mathrm{a} = \sqrt{{\left(36+20\right)}^2 + {\left(15-15\right)}^2} = 56 \mathrm{units} \\ \mathrm{Now}, \mathrm{distance} \mathrm{between} \left(0,0\right) \mathrm{and} \left(36,15\right) = \mathrm{b} = \sqrt{{\left(36-0\right)}^2 + {\left(15-0\right)}^2} = \sqrt{1296+225} = 39 \mathrm{units} \\ \mathrm{Now}, \mathrm{distance} \mathrm{between} \left(0,0\right) \mathrm{and} \left(-20,15\right)= \mathrm{c} = \sqrt{{\left(-20-0\right)}^2 + {\left(15-0\right)}^2} = \sqrt{400+225} = 25 \mathrm{units} \\ \mathrm{Now}, \mathrm{coordinates} \mathrm{of} \mathrm{incentre} = \left(\frac{{\mathrm{ax}}_1+{\mathrm{bx}}_2+{\mathrm{cx}}_3}{\mathrm{a}+\mathrm{b}+\mathrm{c}}, \frac{{\mathrm{ay}}_1+{\mathrm{by}}_2+{\mathrm{cy}}_3}{\mathrm{a}+\mathrm{b}+\mathrm{c}}\right) \\ \mathrm{Ordinate} \mathrm{of} \mathrm{incentre} = \frac{{\mathrm{ay}}_1+{\mathrm{by}}_2+{\mathrm{cy}}_3}{\mathrm{a}+\mathrm{b}+\mathrm{c}} = \frac{56\times 0 + 39\times 15+25\times 15}{56+39+25} = \frac{960}{120} = 8 \end{gathered}$$

 
Hence, option C is correct.

Regards