The points (a, b),(c, d),and left( frac{kc + | vedclass

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(D) The given points are $(a, b)$,$(c, d)$,and $P = \left( \frac{kc + la}{k + l}, \frac{kd + lb}{k + l} \right)$.By the section formula,a point dividi

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Collinear

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(D) The given points are $(a, b)$,$(c, d)$,and $P = \left( \frac{kc + la}{k + l}, \frac{kd + lb}{k + l} \right)$.By the section formula,a point dividing the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m:n$ is given by $\left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right)$.Here,the point $P$ divides the line segment joining $(a, b)$ and $(c, d)$ in the ratio $k:l$.Since the point $P$ lies on the line segment connecting $(a, b)$ and $(c, d)$,the three points are collinear.

The points $(a, b)$,$(c, d)$,and $\left( \frac{kc + la}{k + l}, \frac{kd + lb}{k + l} \right)$ are:

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