Question

Find the orthocenter& circumcenter of a triangle when their vertices are A(1, 2), B(2, 6), C(3,-4).

Answer 1

Note:

Ortho centre :a point of intersection of altitudes of a triangle meets the opposite angle.

Given:

For Orthocentre:.

A(1, 2), B(2, 6), C(3,-4). are vertices of a triangle:

Slope of AB[m1]=$\frac{6-2}{2-1}$=4

Since it is perpendicular to CX.

slope of CX=m2

we have for slope of perpendicular

m1m2=-1

m2=-¼

It passes through the point C(3,-4)

equation of line CX becomes;

(y-y1)=m(x-x1)

y+4=-¼(x-3)

4y+16=-x+3

x+4y+16-3=0

x+4y+13=0........[1]

again:

Slope of AC[m1]=$\frac{-4-2}{3-1}$=-3

Since it is perpendicular to BY

slope of BY=m2

we have for slope of perpendicular

m1m2=-1

m2=⅓

It passes through the point B(2,6)

equation of line BY becomes;

(y-y1)=m(x-x1)

y-6=⅓(x-2)

3y-18=x-2

x-3y+18-2=0

x-3y+16=0.........[2]

Subtracting equation 1&2.

x+4y+13=0

x-3y+16=0

-__________

7y-3=0

y=$\frac{3}{7}$

again

Substituting value of y in equation 1.

x+4*$\frac{3}{7}$+13=0

x=-13-$\frac{12}{7}$

x=$\frac{-103}{7}$=-14$\frac{5}{7}$

So

orthocenter is (-14$\frac{5}{7}$,$\frac{3}{7}$)

And for circumcenter.

Circumcentre: a point of intersection of perpendicular bisector of the triangle.

Now

X,Y and Z are the midpoint of AB,AC and BC respectively.

X(a,b)=($\frac{2+1}{2}$,$\frac{2+6}{2}$)=

($\frac{3}{2}$,4)

Slope of AB=4

Slope of OX=-¼

Equation of line OX passes through ($\frac{3}{2}$,4)is

y-4=-¼(x-$\frac{3}{2}$)

4y-16=-x+$\frac{3}{2}$

8y-16*2=-2x+3

2x+8y=3+32

2x+8y=35

x+4y=$\frac{35}{2}$........[1]

again

Y(c,d)=($\frac{3+1}{2}$,$\frac{-4+2}{2}$=(2,-1)

Slope of AC:-3

Slope of OY=⅓

Equation of line OY passes through (2,-1) is

y+1=⅓(x-2)

3y+3=x-2

x-3y=3+2

x-3y=5......[2]

Multiplying equation 2 by 3 and

Subtracting equation 1&2.

x+4y=35/2

x-3y=5

-_______

7y=$\frac{25}{2}$

y=$\frac{25}{14}$

Substituting value of y in equation 2.

x-3*$\frac{25}{14}$=5

x=5+$\frac{75}{14}$

x=$\frac{145}{14}$

x=10$\frac{5}{14}$

circumcenter of a triangle: (10$\frac{5}{14}$,1$\frac{11}{14}$)

Answer 2

Answer:

iven, the vertices of the triangle,  

A=(1,2)

B=(2,6)

C=(3,−4)

Slope of AB

=x2​−x1​y2​−y1​​=2–16−2​=14​=4

Slope of CF

= Perpendicular slope of AB

=  

Slope of AB

−1  =−14

The equation of CF is given as,

y–y  

1

​  

=m(x–x  

1

​  

)

y+4=−14(x–3)

4y+16=−x+3

x+4y=−13 ——————————– (1)

Slope of BC

=  

x  

2

​  

–x  

1

​  

 

y  

2

​  

–y  

1

​  

 

​  

 

=  

3–2

−4–6

​  

 

=  

1

−10

​  

=−10

Slope of AD

=Perpendicular slope of BC

=  

Slope of BC

−1

​  

 

=  

−10

−1

​  

 

=  

10

1

​  

 

The equation of AD is given as,

y–y  

1

​  

=m(x–x  

1

​  

)

y+2=  

10

1

​  

(x–1)

10y+20=x–1

x–10y=21 ——————————– (2)

Subtracting equation (1) and (2),

x+4y=−13

−x+10y=−21

——————

14y=−34

y=−2.429

Substituting the value of y in equation (1),

x+4y=−13

x+4(−2.429)=−13

x–9.714=−13

x=−3.286

Orthocenter =(−3.286,−2.429)

Step-by-step explanation: