The coordinates of the point of trisection of | vedclass

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Question

(A) Let the points be $A(-2, 3)$ and $B(3, -1)$.Let $C$ be the point of trisection nearer to $A$. Then $C$ divides the line segment $AB$ in the ratio

Vedclass · Accepted Answer

$\left( -\frac{1}{3}, \frac{5}{3} \right)$

Vedclass · Answer

(A) Let the points be $A(-2, 3)$ and $B(3, -1)$.Let $C$ be the point of trisection nearer to $A$. Then $C$ divides the line segment $AB$ in the ratio $1:2$.Using the section formula,the coordinates of $C$ are given by:$x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} = \frac{1(3) + 2(-2)}{1 + 2} = \frac{3 - 4}{3} = -\frac{1}{3}$$y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} = \frac{1(-1) + 2(3)}{1 + 2} = \frac{-1 + 6}{3} = \frac{5}{3}$Thus,the coordinates of $C$ are $\left( -\frac{1}{3}, \frac{5}{3} \right)$.

The coordinates of the point of trisection of the line segment joining the points $(-2, 3)$ and $(3, -1)$ that is nearer to $(-2, 3)$ are

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