Find the co-ordinates of the in-centre of the triangle whose vertices

1.	Find the co-ordinates of the in-centre of the triangle whose vertices are (–36, 7), (20, 7) and (0, –8).
Answer» Let A(–36, 7), B(20, 7) and C(0, –8) be the vertices of the given triangle. Then, a = BC = \(\sqrt{(0-20)^2+(-8-7)^2}\) = \(\sqrt{400+225}\) = \(\sqrt{625}\) = 25 b = AC =\(\sqrt{(0-36)^2+(-8-7)^2}\) = \(\sqrt{1296+225}\) = \(\sqrt{1521}\) = 39 c = AB = \(\sqrt{(20+36)^2+(7-7)^2}\) = \(\sqrt{56^2}\) = 56. The co-ordinates of the incentre of the ΔABC are \(\bigg[\frac{ax_1+bx_2+cx_3}{a+b+c},\frac{ay_1+by_2+cy_3}{a+b+c}\bigg]\) = \(\bigg(\frac{25(-36)+39(20)+56\times0}{25+39+56},\frac{25(7)+39(7)+56(-8)}{25+39+56}\bigg)\) = \(\bigg(\frac{-900+780}{120},\frac{175+273-448}{120}\bigg)\) = \(\bigg(\frac{-120}{120},\frac{0}{120}\bigg)\) = (–1, 0)

Find the co-ordinates of the in-centre of the triangle whose vertices are (–36, 7), (20, 7) and (0, –8).

Answer»

Let A(–36, 7), B(20, 7) and C(0, –8) be the vertices of the given triangle.

Then, a = BC = \(\sqrt{(0-20)^2+(-8-7)^2}\) = \(\sqrt{400+225}\) = \(\sqrt{625}\) = 25

b = AC =\(\sqrt{(0-36)^2+(-8-7)^2}\) = \(\sqrt{1296+225}\) = \(\sqrt{1521}\) = 39

c = AB = \(\sqrt{(20+36)^2+(7-7)^2}\) = \(\sqrt{56^2}\) = 56.

The co-ordinates of the incentre of the ΔABC are

\(\bigg[\frac{ax_1+bx_2+cx_3}{a+b+c},\frac{ay_1+by_2+cy_3}{a+b+c}\bigg]\) = \(\bigg(\frac{25(-36)+39(20)+56\times0}{25+39+56},\frac{25(7)+39(7)+56(-8)}{25+39+56}\bigg)\)

= \(\bigg(\frac{-900+780}{120},\frac{175+273-448}{120}\bigg)\) = \(\bigg(\frac{-120}{120},\frac{0}{120}\bigg)\) = (–1, 0)

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