A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1: n. Find the equation of the line.

Asked by Abhisek | 11 months ago |  63

Solution :-

We know that the coordinates of a point dividing the line segment joining the points (x1, y1) and (x2, y2) 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 (x0, y0), if and only if, its coordinates satisfy the equation y – y0 = m (x – x0)

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

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

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 ago

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