Given:
coordinates of the orthocenter of the triangle whose vertices are ( - 1, 3), (2, - 1) and (0, 0).
Assuming:
A (0, 0), B (−1, 3) and C (2, −1) be the vertices of the triangle ABC. Let AD and BE be the altitudes.
To find:
Orthocenter of the triangle.
Explanation:
AD⊥BC and BE⊥AC
∴ The slope of AD × Slope of BC = −1
The slope of BE × Slope of AC = −1
Here, the slope of BC = \(\frac{-1-3}{2+1}=-\frac{4}{3}\)
and slope of AC = \(\frac{-1-0}{2-0}=-\frac{1}{2}\)
∴ slope of AD × ( - 4/3) = - 1 and slope of BE × ( - 1/2) = - 1
⇒ slope of AD = \(\frac{3}{4}\) and slope of BE = 2
The equation of the altitude AD passing through A (0, 0) and having slope is
y - 0 = \(\frac{3}{4}\) ( x - 0)
⇒ y = \(\frac{3}{4}\) x …..(1)
The equation of the altitude BE passing through B (−1, 3) and having slope 2 is
y - 3 = 2(x + 1)
⇒ 2x – y + 5 = 0 …….(2)
Solving (1) and (2):
x = − 4, y = − 3
Hence, the coordinates of the or the centre is (−4, −3).
