Question

Triangle ABC has vertices at A(2,3),B(-4,-3) and C(2,-3) find the coordinates of each point of concurrency.

Circumcenter ________
Orthocenter ________

Can someone please help me with this? and if you can explain how to do it?

Answer 1

Answer:

Circumcenter =(-1,0)

Orthocenter =(2,-3)

Step-by-step explanation:  

Given : Points A = (2,3), B = (-4,-3), C = (2,-3)  

Formula used :  

→Mid point of two points- $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

→Slope of two points - $\frac{y_2-y_1}{x_2-x_1})$

→Perpendicular of a line = $\frac{-1}{slope of line})$

Circumcenter- The point where the perpendicular bisectors of a triangle meets.

Orthocenter-The intersecting point for all the altitudes of the triangle.

To find out the circumcenter we have to solve any two bisector equations.

We solve for line AB and AC

So, mid point of AB =$(\frac{2-4}{2},\frac{3-3}{2})=(-1,0)$

Slope of AB =$\frac{-3-3}{-4-2}=1$

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

So, the slope of the perpendicular bisector = -1  

Equation of AB with slope -1 and the coordinates (-1,0) is,  

(y – 0) = -1(x – (-1))  

y+x=-1………………(1)  

Similarly, for AC  

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

Slope of AC = $\frac{-3-3}{2-2}=\frac{-6}{0}$  

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

So, the slope of the perpendicular bisector = 0  

Equation of AC with slope 0 and the coordinates (2,0) is,  

(y – 0) = 0(x – 2)  

y=0 ………………(2)  

By solving equation (1) and (2),  

put y=0 in equation (1)

y+x=-1

0+x=-1

⇒x=-1  

So the circumcenter(P)= (-1,0)

To find the orthocenter we solve the intersections of altitudes.

We solve for line AB and BC

So, mid point of AB =$(\frac{2-4}{2},\frac{3-3}{2})=(-1,0)$

Slope of AB =$\frac{-3-3}{-4-2}=1$

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

So, the slope of CF = -1  

Equation of AB with slope -1 and the coordinates (-1,0) gives equation CF  

(y – 0) = -1(x – (-1))  

y+x=-1………………(3)  

Similarly, mid point of BC =$(\frac{-4+2}{2},\frac{-3-3}{2})=(-1,-3)$

Slope of AB =$\frac{-3+3}{-4-2}=0$

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

So, the slope of AD = 0

Equation of AB with slope 0 and the coordinates (-1,-3) gives equation AD

(y-(-3)) = 0(x – (-1))  

y+3=0

y=-3………………(4)  

Solve equation (3) and (4),

Put y=-3 in equation (3)

y+x=-1

-3+x=-1

x=2

Therefore, orthocenter(O)= (2,-3)