Question

Two vertices of triangle ABC are B(5,-1) and C(2,3) .If orthocentre is at origin find the third vertex.

Answer 1

Dear student
The orthocenter of the triangle is the point where the altitudes of the triangle intersect.

Finding the slope of the altitude from vertex (5,-1) which passes through (0, 0).


m = (y2-y1)/(x2-x1)
m = (-1-0)/(5-0)
m = -1/5
Therefore if the altitude from vertex (5,-1) has a slope of -1/5, then the slope of the opposite side with which is it is perpendicular is: - 1/-1/5 or 5. Now this side with slope 5 passes through vertex (-2,3).
Solving for the equation of this line:

y = 5x + b
3 = 5(-2) + b
b = 3 + 10
b = 13
Therefore, the equation of this side is:
y = 5x + 13 or
5x - y + 13 = 0 eq. 1

Finding the slope of the altitude from vertex (-2,3) which passes through (0, 0).

m = (y2-y1)/(x2-x1)
m = (3 - 0)/(-2-0)
m = -3/2

Therefore if the altitude from vertex (-2,3) has a slope of -3/2, then the slope of the opposite side with which is it is perpendicular is: - 1/-3/2 or 2/3. Now this side with slope 2/3 passes through vertex (5,-1).
Solving for the equation of this line:

y = mx + b
-1 = (2/3)(5) + b
b = -1 - 10/3
b = -13/3

Therefore the equation of this side is:
y = (2/3)x - 13/3 eq. 2
simplifying:
3y = 2x - 13
2x - 3y - 13 = 0 eq.2
5x - y + 13 = 0 eq. 1: multiply by -3:

-15x + 3y - 39 = 0 eq.1
2x - 3y - 13 = 0 eq.2
____________________
- 13x - 52 = 0
- 13x = 52
x = 52/-13
x = -4

Solving for y by using eq. 2:
2(-4) - 3y - 13 = 0
-8 - 3y - 13 = 0
-21 - 3y = 0
y = 21/-3
y = -7

Therefore the coordinates of the third point are
(-4, - 7) ANS
Regards