Question

Find the centroid, the in-centre, the circum-centre and the orthocentre of the triangle whose sides
have the equations 3x - 4y = 0, 12y +5x = 0 and y - 15 = 0.​

Answer 1

Answer with explanation:

The equation of sides of Triangle are:

  1. →3 x - 4 y = 0,

2.→ 12 y +5 x = 0

and 3.→ y - 15 = 0.​

To find the vertices of Triangle we will find their Point of Intersection.

⇒Line 1 and Line 2, have y intercept equal to 0.So,their Point of intersection will be at (0,0).

⇒Point of Intersection of Line 2 and Line 3

  y=15

Putting the value of y in equation 2

 $12 \times 15 +5 x=0\\\\5 x= -180\\\\x=\frac{-180}{5}\\\\x=-36$

So, point of Intersection of Line 2 and line 3 will be = (-36,15)

⇒Point of Intersection of Line 1 and Line 3

  y=15

Putting the value of y in equation 2

 $3 x -4 \times 15 =0\\\\3 x= 60\\\\x=\frac{60}{3}\\\\x=20$

So, point of Intersection of Line 1 and line 3 will be = (20,15)

So,Vertices of Triangle are =(0,0),(-36,15),(20,15)

Length of Sides of Triangle Using Distance formula

   $D={(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}\\\\AB=\sqrt{(0+36)^2+(0-15)^2}\\\\AB=\sqrt{1296+225}\\\\AB=\sqrt{1521`}\\\\AB=39\\\\BC=20+36\\\\BC=56\\\\CA=\sqrt{(20)^2+(15)^2}\\\\CA=\sqrt{625}\\\\CA=25$

Centroid

  $=(\frac{\text{Sum of x coordinates}}{3},\frac{\text{Sum of y coordinates}}{3})\\\\=(\frac{0-36+20}{3},\frac{0+15+15}{3})\\\\=(\frac{-16}{3},10)$

Incenter

$x=\frac{56 \times 0+25 \times (-36)+39 \times 20}{39+25+56}\\\\x=\frac{-120}{120}\\\\x= -1\\\\y=\frac{56 \times 0+25 \times 15+39 \times 15}{39+25+56}\\\\y=\frac{15 \times 64}{120}\\\\y=8$

= (-1,8)

Circumcenter

The Point of Intersection of Perpendicular bisectors of triangle is called Circumcenter.

Mid point of AB, BC and CA is

  $P=(\frac{0-36}{2},\frac{0+15}{2})\\\\=(-18,\frac{15}{2})\\\\Q=(\frac{0+20}{2},\frac{0+15}{2})=(10,\frac{15}{2})\\\\R=(\frac{20-36}{2},\frac{15+15}{2})\\\\=(-8,15)$

⇒ Equation of line Perpendicular to Ax+By +C=0,is equal to

Bx -A y+K=0

⇒Equation of line Perpendicular to , 3 x -4 y=0, is given by

     4 x +3 y+ m=0

The line will pass through ,$(-18,\frac{15}{2})$.

$4 \times -18+3 \times \frac{15}{2}+m=0\\\\ -72 +\frac{45}{2}+m=0\\\\m=\frac{99}{2}\\\\4 x+3 y+\frac{99}{2}=0\\\\8 x+6 y+99=0----(A)$

⇒Equation of Line perpendicular to, y-15=0 , is

    x+h=0

It will Pass through (-8,15).

-8 +h=0

h=8

So, equation of line perpendicular to , y-15=0, will be

→x+8=0------(B)

Point of Intersection of Line A and Line B will be

-64+ 6 y+99=0

6 y=35

$y=\frac{35}{6}$

So, Circumcenter of Triangle ABC is

         $(-8,\frac{35}{6})$

⇒Orthocentre of Triangle is Point of Intersection of Altitudes drawn from Opposite Vertex.

Line,4 x +3 y+ m=0, will pass through (20,15).

⇒4×20+3×15+m=0

⇒80+45+m=0

⇒m= -125

So, equation of line ⊥ to AB and Passing through vertex C will be

  4 x +3 y -125=0

Equation of Line perpendicular to, y-15=0 , is →x+h=0.It will Pass through (0,0).

x+0=0

So, equation of line ⊥ to, y-15=0, will be , x=0.

Point of Intersection of ,line , x=0 and 4 x+3 y-125=0 , will be

 ⇒4 × 0+ 3 y -125=0

3 y=125

$y=\frac{125}{3}\\\\=(0,\frac{125}{3})$ is Orthocentre of triangle.