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Question
Answer the following question:
The vertices of a triangle are A(1, 4), B(2, 3) and C(1, 6) Find equations of altitudes of ∆ABC
Solution
Let AX, BY and CZ be the altitudes through the vertices A, B and C respectively of ∆ABC.
Slope of BC = – 3
∴ Slope of AX = `1/3` ...[∵ AX ⊥ BC]
Since altitude AX passes through (1, 4) and has slope `1/3`,
equation of altitude AX is
y – 4 = `1/3(x - 1)`
∴ 3y – 12 = x – 1
∴ x – 3y + 11 = 0
Since both the points A and C have same x co-ordinates i.e. 1,
the points A and C lie on the line x = 1.
AC is parallel to Y-axis and therefore, altitude
BY is parallel to X-axis
Since the altitude BY passes through B(2, 3),
the equation of altitude BY is y = 3.
Also, slope of AB = – 1
∴ Slope of CZ = 1
Since altitude CZ passes through (1, 6) and has slope 1,
equation of altitude CZ is
y – 6 = 1(x – 1)
∴ x – y + 5 = 0
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