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Question

(A) The formula for external division of a 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

Vedclass · Accepted Answer

$(3, -11)$

Vedclass · Answer

(A) The formula for external division of a line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m : n$ is given by $\left( \frac{mx_2 - nx_1}{m - n}, \frac{my_2 - ny_1}{m - n} \right)$.Given points are $(x_1, y_1) = (3, -1)$ and $(x_2, y_2) = (3, 4)$ with ratio $m : n = 2 : 3$.Substituting the values:$x = \frac{2(3) - 3(3)}{2 - 3} = \frac{6 - 9}{-1} = \frac{-3}{-1} = 3$.$y = \frac{2(4) - 3(-1)}{2 - 3} = \frac{8 + 3}{-1} = \frac{11}{-1} = -11$.Therefore,the coordinates of the point are $(3, -11)$.

Find the coordinates of the point that divides the line segment joining the points $(3, -1)$ and $(3, 4)$ externally in the ratio $2 : 3$.

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