Question

A(0,6), B(8,12), C(8,0) are the Co-ordinate of vertices of triangle ABC. Then...... $\begin{array}{cc}1.Co-ordinateofcentroid& p.(20,6)\\ 2.Co-ordinateofIncentre& q.(0,16)\\ 3.Co-ordinateofExcentre& r.\left(\frac{16}{3}\right),6)\\ & s.(0,-4)\\ & t.(5,6)\end{array}$

• A. 1-r 2 - t 3-p
• B. 1-r 2 - t 3-q
• C. 1-r 2 - t 3-s
• D. 1-r 2 - q 3-t

Answer 1

Answer: (A), (B), (C)

The correct options are
A

1-r 2 - t 3-p

B

1-r 2 - t 3-q

C

1-r 2 - t 3-s

1. Co-ordinate of centroid for $△$ABC

$X=\frac{x_1+x_2+x_3}{3}$
=$\frac{0+8+8}{3}=\frac{16}{3}$
$Y=\frac{y_1+y_2+y_3}{3}=\frac{6+12+0}{3}=6$
Co-ordinate of centroid $G(\frac{16}{3},6)$

2. In $△$ ABC

Side AB=C=$\sqrt{(0-8{)}^2+(6-12{)}^2}$

$=\sqrt{(64+36)}$
=10
Side BC=a=$\sqrt{(8-8{)}^2+(12-0{)}^2}=12$
Side CA=b=$\sqrt{(8-0{)}^2+(6-0{)}^2}=10$
Co-ordinate of incentre I(x,y)
$X=\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c}$
$X=\frac{12\times 0+10\times 8+10\times 3}{12+10+10}=\frac{160}{32}=5$
$X=\frac{12\times 6+10\times 12+10\times 0}{12+10+10}=\frac{192}{32}=6$
Co-ordinates of incentre (5,6)

3. Co-ordinate of Ex-centre
There will be three excentre for O triangle
Excentre touching the side BC

$z_1(x,y),x=\frac{a{x}_1+x_2+c{x}_3}{-a+b+c}=\frac{-12\times 0+10\times 0+10\times 8}{-12+10+10}=\frac{160}{8}=20$

$y=\frac{-a{y}_1+b{y}_2+c{y}_3}{-a+b+c}=\frac{-12\times 6+10\times 12+10\times 0}{-12+10+10}=\frac{48}{8}=6$
$z_1(20,6)$
Excentre touching the side AC
$z_2(x,y),x=\frac{a{x}_1-b{x}_2+c{x}_3}{a-b+c}=\frac{12\times 0-10\times 3+10\times 0}{12-10+10}=0$
$Y=\frac{a{y}_1-b{y}_2+c{y}_3}{a-b+c}=\frac{12\times 6-10\times 12+10\times 8}{12-10+10}=\frac{-48}{12}=-4$

$z_2(0,-4)$
Excentre touching the side AB
$z_3(x,y),x=\frac{a{x}_1+b{x}_2-c{x}_3}{a+b-c}=\frac{12\times 0+10\times 3-10\times 8}{12+10-10}=0$
$Y=\frac{a{y}_1-b{y}_2+c{y}_3}{a+b-c}=\frac{12\times 6-10\times 12-10\times 0}{12+10-10}=16$
$z_3(0,16)$
Three excentre are $z_1(20,6)$ touching the side BC
$z_2(0,-4)$ touching the side CA
$z_3(0,16)$ touching the side AB