The point which divides externally the line j | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

If $P, Q, R$ are collinear points such that $P(7, 7)$,$Q(3, 4)$ and $PR = 10$,then what are the coordinates of $R$?

Find the distance between the points $P \left( 3, -\frac{\pi}{6} \right)$ and $Q \left( 4, \frac{\pi}{3} \right)$ in polar coordinates.

Find a point whose $x$-coordinate and $y$-coordinate are equal,and which is equidistant from the points $A(1, 0)$ and $B(0, 3)$.

In what ratio does the point $(8, 4)$ divide the line segment joining the points $(5, -2)$ and $(9, 6)$?

Find the points of trisection of the line segment joining the points $A(2, 1)$ and $B(5, 3)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module