Find the incentre of the triangle
formed by the straight lines x+y-7=0,x-y+1=0 and x-3y+5=0
Answers
Answer:
The in center point of triangle is (3,\sqrt{5}+1
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.
\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+
2
y
(
5
−1)−
5
=0
x−
2
y
(
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}+1
5
+1 )
Step-by-step explanation:
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