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Standard VI

Mathematics

Perpendicular Bisector and Its Construction

In a Δ A B ...

Question

In a $Δ A B C,$ the sides $B C = 5, C A = 4, A B = 3.$ If $A (0, 0)$ and the internal bisector of angle $A$ meet $B C$ in $D (\frac{12}{7}, \frac{12}{7})$ then in-centre of $Δ A B C$

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Solution

Consider $△ A B C$

Vertice $A (0, 0), B (3, 0), C (0, 4)$

Sides $a = 5, b = 4, c = 3$

Incenter $= (\frac{a x_{1} + b x_{2} + c x_{3}}{a + b + c}, \frac{a y_{1} + b y_{2} + c y_{3}}{a + b + c}) ⟹ (\frac{12}{12}, \frac{12}{12}) ⟹ (1, 1)$

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In a ΔABC, the sides BC=5,CA=4,AB=3. If A(0,0) and the internal bisector of angle A meet BC in D(127,127) then in-centre of ΔABC

Consider △ABCVertice A(0,0),B(3,0),C(0,4)Sides a=5,b=4,c=3Incenter =(ax1+bx2+cx3a+b+c,ay1+by2+cy3a+b+c)⟹(1212,1212)⟹(1,1)

In a $Δ A B C,$ the sides $B C = 5, C A = 4, A B = 3.$ If $A (0, 0)$ and the internal bisector of angle $A$ meet $B C$ in $D (\frac{12}{7}, \frac{12}{7})$ then in-centre of $Δ A B C$

Consider $△ A B C$

Vertice $A (0, 0), B (3, 0), C (0, 4)$

Sides $a = 5, b = 4, c = 3$

Incenter $= (\frac{a x_{1} + b x_{2} + c x_{3}}{a + b + c}, \frac{a y_{1} + b y_{2} + c y_{3}}{a + b + c}) ⟹ (\frac{12}{12}, \frac{12}{12}) ⟹ (1, 1)$