The orthocentre of the triangle having vertices at (2, 3)(2, 5)(4, 3) is
We have,
Given point are
(2, 3), (2, 5), (4, 3)
So,
Let
A (x1, y1)=(2, 3)
B (x2, y2)=(2, 5)
C (x3, y3)=(4, 3)
Slope of AC =y3−y1x3−x1=3−34−2=0
Let the slope of line
BD = m
So, BD⊥AC
m×0=1
m=∞
So, equation of BD⇒y−5=∞(x−2)
⇒x−2=0 ...... (1)
x=2
Slope of line
AB⇒y2−y1n2−n1=5−32−2=∞
CE⊥AB
Let the slope of CE be n.
n×∞=−1, n=0
Now,
Equation of CE is
y−3=n(x−4)
y−3=0(n−4)
y−3=0 ...... (2)
From equation (1) and (2) to, and
We get.
(n, y)=(2, 3) (othocentre.)
Hence, this is the answer.