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Question
Answer the following question:
A(1, 4), B(2, 3) and C(1, 6) are vertices of ∆ABC. Find the equation of the altitude through B and hence find the co-ordinates of the point where this altitude cuts the side AC of ∆ABC.
Solution
Vertices of the triangle are A(1, 4), B(2, 3), and C(1, 6).
Let BD be the altitude through vertex B.
Since both the points A and C have the same x co-ordinates i.e. 1, the given points lie on a line parallel to Y-axis.
∴ The equation of the line AC is x = 1 ....(i)
AC is parallel to Y-axis and therefore, altitude BD is parallel to X-axis.
Since the altitude BD passes through B(2, 3), the equation of altitude BD is y = 3 .......(ii)
From (i) and (ii),
Point of intersection of AC and altitude BD is (1, 3).
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