Question

find the centroid,the incentre,the circumcentre and the orthocentre of the triangle whose sides have the equations 3x-4y=0,12y+5x=0 andy-15=0

Answer 1

Unless there is a standard formula, this is a tedious calculation.

First I would start with finding the coordinates of the three vertices.

Take the first two equations and solve for x and y. Call that vertex A.
Take the second and third equation and solve for x and y. Call that vertex B.
Take the first and third equation and solve for x and y. Call that vertex C.

That is just the start ... but one step at a time.

Answer 2

yaa ryt we'll get the coordinates of all three vertices of triangle

Answer 3

Then go through the definition of each of the centers (American spelling) and see how you can find them.

Orthocenter is the intersection of the altitudes. Altitudes are lines drawn from vertex and is perpendicular to the opposite side. You are given the equations of the three sides of the triangle. So you know the slopes of each line. If two lines are perpendicular the product of their slopes is -1. So you can find the slopes of the altitude line. Knowing the slope and the vertex the altitude passes through you can find the equation of the altitude line. Find the equations of just two altitude lines and solve for x and y and that point, ehich will be the intersection of two altitudes, will be the orthocenter. You don't need to find the equations of all three altitude lines. Just any two will do as their intersection will be the orthocenter.

You can use similar logic to find all the other centers by starting with the definition of what each center is.

Answer 4

you know what that's what i thoght abt this question bt can't there be any simpler method to it.....it's a typical,tedious and long method

Answer 5

Are you allowed to graph the triangle on a graph paper, draw the altitudes, etc., and determine the coordinates of the various centers from the graph sheet?

I haven't touched geometry in a very long time. Others may have some ideas or formulas that you can simply plug in and get the answers. Or there may not be any such quick ways. If you leave the question open and bump it periodically you may get ideas from others.

Answer 6

One of the vertices is the origin (0, 0)
Two of the lines pass through the origin and the third one is parallel to the x axis (y=15). So the calculations may not be that bad if you get started.

Answer 7

i m getting 3 coordinates of vertices as (0,0) ,(3,-5/4) ,(5,15/4)

Answer 8

That should get you going.

Answer 9

hey im getting it as (36,-15)how are u getting (-36,15)

Answer 10

12y + 5x = 0
y - 15 = 0 Therefore, y = +15
(12)(15) + 5x = 0
divide by 5
(12)(3) + x = 0
36 + x = 0
x = -36

Answer 11

but how am i going to find out circumcentre of triangle

Answer 12

The circumcenter is the point of intersection of the three perpendicular bisectors.

A perpendicular bisectors of a triangle is each line drawn perpendicularly from the midpoint of the sides of the triangle.

Answer 13

The midpoint of (x1, y1) and (x2, y2) is { (x1 + x2)/2, (y1 + y2)/2 }

For example, A is (0, 0) and B is (-36, 15).
So the midpoint of AB is { (0 - 36)/2, (0 + 15)/2 }
= (-18, 7.5).
The slope of the line AB is -5/12.
Therefore, the slope of the line perpendicular to AB is +12/5 (negative reciprocal of -5/12)
Knowing the slope of a line perpendicular to AB and knowing the coordinates of the midpoint you can find the equation of the perpendicular bisector:
y = mx + b ; m = 12/5 ; so y = 12/5x + b.
This passes through the midpoint (-18, 7.5). Substitute in the equation above.
7.5 = (12/5)(-18) + b. So b = 50.7
The equation for the perpendicular bisector of AB is:
y = 12/5x + 50.7 = 2.4x + 50.7 ------- (1)

The perpendicular bisector of BC is simpler because it is going to be parallel to the y axis which means the equation will be: x = some constant. This bisector has to pass through the midpoint of BC which is: { (-36 +20)/2, (15 + 15)/2 } = (-8, 15)
So x = constant = -8. Or
x = -8 ---------------- (2)

From equations (1) and (2) you can find the intersection point and that will be your circumcenter.