Find the coordinates of the point which divid | vedclass

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(A) The section formula for external division in the ratio $m : n$ for points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\left( \frac{mx_2 - nx_1}{m -

Vedclass · Accepted Answer

$\left( \frac{a^2 - 2ab - b^2}{a - b}, \frac{a^2 + b^2}{a - b} \right)$

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(A) The section formula for external division in the ratio $m : n$ for points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\left( \frac{mx_2 - nx_1}{m - n}, \frac{my_2 - ny_1}{m - n} \right)$.Here,$m = a$,$n = b$,$(x_1, y_1) = (a + b, a - b)$,and $(x_2, y_2) = (a - b, a + b)$.For the $x$-coordinate:$x = \frac{a(a - b) - b(a + b)}{a - b} = \frac{a^2 - ab - ab - b^2}{a - b} = \frac{a^2 - 2ab - b^2}{a - b}$.For the $y$-coordinate:$y = \frac{a(a + b) - b(a - b)}{a - b} = \frac{a^2 + ab - ab + b^2}{a - b} = \frac{a^2 + b^2}{a - b}$.Thus,the coordinates are $\left( \frac{a^2 - 2ab - b^2}{a - b}, \frac{a^2 + b^2}{a - b} \right)$.

Find the coordinates of the point which divides the line segment joining the points $(a + b, a - b)$ and $(a - b, a + b)$ externally in the ratio $a : b$.

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