We know that the coordinates of a point dividing the line segment joining the points (x_{1}, y_{1}) and (x_{2}, y_{2}) internally in the ratio m: n are

\( \dfrac{1(2)+n(1)}{1+n}, \dfrac{1(3)+n(0)}{1+n}\)

\( = (\dfrac{2+n}{1+n}, \dfrac{3}{1+n})\)

We know that slope, m =\( \dfrac{y_2+y_1}{x_2+x_1}\)

= \( \dfrac{3-0}{2-1}\)

= \( \dfrac{3}{1}\)

= 3

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

Then, m = (\( - \dfrac{1}{m}\)) = \( -\dfrac{1}{3}\)

We know that the point (x, y) lies on the line with slope m through the fixed point (x_{0}, y_{0}), if and only if, its coordinates satisfy the equation y – y_{0} = m (x – x_{0})

\( = (\dfrac{2+n}{1+n}, \dfrac{3}{1+n})\)

\((y-\dfrac{3}{n+1})=\dfrac{-1}{3}(x-\dfrac{2+n}{1+n})\)

3((1 + n) y – 3) = (-(1 + n) x + 2 + n)

3(1 + n) y – 9 = – (1 + n) x + 2 + n

(1 + n) x + 3(1 + n) y – n – 9 – 2 = 0

(1 + n) x + 3(1 + n) y – n – 11 = 0

The equation of the line is (1 + n) x + 3(1 + n) y – n – 11 = 0.

Answered by Pragya Singh | 11 months agoFind the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.

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