Question

Find the incentre of the triangle formed by the lines x+y-7=0,x-y+1=0,x-3y+5=0

Answer 1

Answer:

The in center point of triangle is (3,$\sqrt{5}+1$ )

Step-by-step explanation:

Given statement is:

x+y-7=0         eq-1

x+y+1=0         eq-2

x-3y+5=0       eq-3

Take eq-1 and eq-2 and find value of x and y.

we get x=3 and y=4. And x=3 is the is the internal bisector of these two lines

Take eq-2 and eq-3 and find value of x and y.

$x+\frac{y}{2}(\sqrt{5}-1)-\sqrt{5} =0\\x-\frac{y}{2}(\sqrt{5}+1)+\sqrt{5} =0$

now find value of internal bisector y using internal bisector x=3 and above equation:

we get,

$3-\frac{y}{2}(\sqrt{5}+1)+\sqrt{5} =0\\y=\frac{(3+\sqrt{5} )2 }{\sqrt{5}+1 }\\y=\frac{(6+2\sqrt{5)} }{\sqrt{5}+2 } \\y=\frac{\((\sqrt{5}+1)^{2} }{\sqrt{5}+1} \\\\y=\sqrt{5}+1$

hence the in center point of triangle is (3,$\sqrt{5}+1$ )

Answer 2

Answer:

The in center point of triangle is (3,\sqrt{5}+15+1 )

Step-by-step explanation:

Given statement is:

x+y-7=0         eq-1

x+y+1=0         eq-2

x-3y+5=0       eq-3

Take eq-1 and eq-2 and find value of x and y.

we get x=3 and y=4. And x=3 is the is the internal bisector of these two lines

Take eq-2 and eq-3 and find value of x and y.

\begin{gathered}x+\frac{y}{2}(\sqrt{5}-1)-\sqrt{5} =0\\x-\frac{y}{2}(\sqrt{5}+1)+\sqrt{5} =0\\\end{gathered}x+2y(5−1)−5=0x−2y(5+1)+5=0

now find value of internal bisector y using internal bisector x=3 and above equation:

we get,

\begin{gathered}3-\frac{y}{2}(\sqrt{5}+1)+\sqrt{5} =0\\y=\frac{(3+\sqrt{5} )2 }{\sqrt{5}+1 }\\y=\frac{(6+2\sqrt{5)} }{\sqrt{5}+2 } \\y=\frac{\((\sqrt{5}+1)^{2} }{\sqrt{5}+1} \\\\y=\sqrt{5}+1\end{gathered}

hence the in center point of triangle is (3,\sqrt{5}+15+1 )