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Question
The points A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of ∆ABC. Find the coordinates of the point P on AD such that AP : PD = 2 : 1
Solution
According to the question,
The vertices of ΔABC = A, B and C
Coordinates of A, B and C = A(x1, y1), B(x2, y2), C(x3, y3)
Let the coordinates of a point P be (x, y)
Given,
The ratio in which the point P(x, y), divide the line joining,
`"A"(x_1, y_1)` and `"D"((x_2 + x_3)/2, (y_2 + y_3)/2)` = 2 : 1
Then,
Coordinates of P = `[(2 xx ((x_2 + x_3)/2) + 1 xx x_1)/(2 + 1), (2 xx ((y_2 + y_3)/2) + 1 xx y_1)/(2 + 1)]`
By using internal section formula;
= `((m_1x_2 + m_2x_1)/(m_1 + m_2), (m_1y_2 + m_2y_1)/(m_1 + m_2))`
= `((x_2 + x_3 + x_1)/3, (y_2 + y_3 + y_1)/3)`
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