The point which divides externally the line j | vedclass

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Question

(A) The section 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_

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The section 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 $(x_1, y_1) = (a + b, a - b)$ and $(x_2, y_2) = (a - b, a + b)$ with ratio $m:n = a:b$.Applying the formula for the $x$-coordinate:$x = \frac{a(a - b) - b(a + b)}{a - b} = \frac{a^2 - ab - ab - b^2}{a - b} = \frac{a^2 - 2ab - b^2}{a - b}$.Applying the formula 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 point is $\left( \frac{a^2 - 2ab - b^2}{a - b}, \frac{a^2 + b^2}{a - b} \right)$.

The point which divides externally the line joining the points $(a + b, a - b)$ and $(a - b, a + b)$ in the ratio $a:b$ is

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