Question

The coordinates of the vertices of a triangle are $(0,1),(2,3)$ and $(3,5)$:
(a) Find centroid of the triangle.
(b) Find circumcentre and the circumradius.
(c) Find Orthocentre of the triangle.

Answer 1

Lets coordinate of a triangle are $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$

Here $A(0,1),B(2,3),C(3,5)$

$\left(a\right)$ Centroid of the triangle$=(x_c,y_c)$

$x_c=\frac{x_1+x_2+x_3}{3},y_c=\frac{y_1+y_2+y_3}{3}$

$x_c=\frac{0+2+3}{3}=\frac{5}{3},y_c=\frac{1+3+5}{3}=3$

So the centroid is $(\frac{5}{3},3)$.

$\left(b\right)$ To find out the circumcenter we have to solve any two bisector equations and find out the intersection points.

So, mid point of AB $=(\frac{0+2}{2},\frac{1+3}{2})=F(1,2)$

Slope of $AB=\frac{2-0}{3-1}=1$

Slope of the bisector is the negative reciprocal of the given slope.

So, the slope of the perpendicular bisector $=-1$

Equation of line with slope $-1$ and the coordinates $F(1,2)$ is,

$(y-2)=-1(x-1)$

$x+y=3$………………$\left(1\right)$

Similarly, for AC

Mid point of AC $=(\frac{0+3}{2},\frac{1+5}{2})=E(\frac{3}{2},3)$

Slope of $AC=\frac{5-1}{3-0}=\frac{4}{3}$

Slope of the bisector is the negative reciprocal of the given slope.

So, the slope of the perpendicular bisector $=-\frac{3}{4}$

Equation of line with slope $-\frac{3}{4}$ and the coordinates $E(\frac{3}{2},3)$ is,

$(y-3)=-\frac{3}{4}(x-\frac{3}{2})$

$y-3=-\frac{3}{4}x+\frac{9}{8}$

$\frac{3}{4}x+y=\frac{33}{8}$………………$\left(2\right)$

By solving equation (1) and (2),

$\left(1\right)-\left(2\right)\Rightarrow \frac{x}{4}=3-\frac{33}{8};x=-\frac{9}{2}$

Substitute the value of $x$ in to $\left(1\right)$

$-\frac{9}{2}-y=3$

$y=-\frac{15}{2}$

So the circumcenter is $(-\frac{9}{2},-\frac{15}{2})$

$\left(c\right)$ Let perpendicular bisectors from A,B,C of triangle be $F$ (on side AB), $E$ (on side AC),$D$(on side BC)

We know, slope of $AB=1$

Slope of $CF$ $=$ Perpendicular slope of $AB$

$=-\frac{1}{SlopeofAB}$

$=-1$

The equation of CF is given as,

$y-5=-1(x-3)$

$x+y=8$ ………………$\left(1\right)$

Slope of $AC=\frac{4}{3}$

Slope of AD $=$ Perpendicular slope of BC

$=-\frac{1}{SlopeofBC}$

$=-\frac{3}{4}$

The equation of $BE$ is given as,

$y-3=-\frac{3}{4}(x-2)$

$y-3=-\frac{3}{4}x+\frac{3}{2}$

$\frac{3}{4}x+y=\frac{9}{2}$ ………………$\left(2\right)$

By solving equation (1) and (2),

$\left(1\right)-\left(2\right)\Rightarrow \frac{x}{4}=8-\frac{9}{2};x=14$

Substitute the value of $x$ in to $\left(1\right)$

$14+y=8$

$y=-6$

So the Orthocenter is $(14,-6)$