A line perpendicular to the line segment joining the points $ (1,0) $ and $ (2,3) $ divides it in the ratio $ 1 : n $ . The equation of the line is
- $ 3y + x = \frac{n+11}{n+1} $
- $ 3y - x = \frac{n+11}{n+1} $
- $ 3y + x = \frac{n-11}{n+1} $
- $ 3y - x = \frac{n+11}{n-1} $
The Correct Option is A
Solution and Explanation
$\therefore $ Slope of $MN = - \frac{1}{3}$
Now, coordinates of
$P\equiv\left(\frac{2\times1+1\times n}{n+1}, \frac{3\times1+0\times n}{n+1}\right)$
$[$ By section formulaa$]$
$ \equiv\left(\frac{2+n}{n+1}, \frac{3}{n+1}\right) $
$\therefore$ Required equation is given as
$ \left(y - \frac{3}{ n+1}\right) = -\frac{1}{3}\left(x - \frac{2+n}{n+1}\right) $
$\Rightarrow 3y - \frac{9}{n+1} = -x + \frac{2+n}{n+1} $
$ \Rightarrow 3y +x = \frac{11 + n}{n+1} $
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Concepts Used:
Straight lines
A straight line is a line having the shortest distance between two points.
A straight line can be represented as an equation in various forms, as show in the image below:
The following are the many forms of the equation of the line that are presented in straight line-
1. Slope – Point Form
Assume P0(x0, y0) is a fixed point on a non-vertical line L with m as its slope. If P (x, y) is an arbitrary point on L, then the point (x, y) lies on the line with slope m through the fixed point (x0, y0) if and only if its coordinates fulfil the equation below.
y – y0 = m (x – x0)
2. Two – Point Form
Let's look at the line. L crosses between two places. P1(x1, y1) and P2(x2, y2) are general points on L, while P (x, y) is a general point on L. As a result, the three points P1, P2, and P are collinear, and it becomes
The slope of P2P = The slope of P1P2 , i.e.
\(\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}\)
Hence, the equation becomes:
y - y1 =\( \frac{y_2-y_1}{x_2-x_1} (x-x1)\)
3. Slope-Intercept Form
Assume that a line L with slope m intersects the y-axis at a distance c from the origin, and that the distance c is referred to as the line L's y-intercept. As a result, the coordinates of the spot on the y-axis where the line intersects are (0, c). As a result, the slope of the line L is m, and it passes through a fixed point (0, c). The equation of the line L thus obtained from the slope – point form is given by
y – c =m( x - 0 )
As a result, the point (x, y) on the line with slope m and y-intercept c lies on the line, if and only if
y = m x +c