The orthocenter of the triangle formed by (0, 0), (8, 0) and (4, 6) is

4, \frac{8}{3}

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3, 4

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- 3, 4

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Solution

The correct option is A $4, \frac{8}{3}$
Sol: Let A ≡ (0, 0), B ≡ (8, 0) and C ≡ (4, 6).

Slope of BC = $\frac{6 - 0}{4 - 0} = \frac{3}{2}$

Equation of the line through A(0, 0) and perpendicular to BC is
(y – 0) = $\frac{2}{3}$ (x – 0) i.e. 2x – 3y = 0 …… (1)

Slope of CA = $\frac{6 - 0}{4 - 0} = \frac{3}{2}$

Equation of the line through B(8, 0) and perpendicular to CA is

(y – 0) = $- \frac{2}{3}$ (x – 8) i.e., 2x + 3y = 16 …… (2)

Solving (1) and (2), the orthocenter is $4, \frac{8}{3}$

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Similar questions

I. The centroid of the triangle formed by $(2, 3, - 1), (5, 6, 3), (2, - 3, 1)$ is	a) $(2, 2, 2)$
II. The circumcentre of the triangle formed by $(1, 2, 3), (2, 3, 1), (3, 1, 2)$ is	b) $(3, 1, 4)$
III. The orthocenter of the triangle formed by $(2, 1, 5), (3, 2, 3), (4, 0, 4)$ is	c) $(1, 1, 0)$
IV. The incentre of the triangle formed by $(0, 0, 0), (3, 0, 0), (0, 4, 0)$ is	d) $(3, 2, 1)$
	e) $(0, 0, 0)$

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