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.
Solution and Explanation
According to the section formula, the coordinates of the point that divides the line segment joining the points (1, 0) and (2, 3) in the ratio 1:n is given by
\((\frac {n(1)+1(2)}{1+n}, \frac {n(0)+1(3)}{1+n})=(\frac {n+2}{n+1},\frac {3}{n+1})\)
The slope of the line joining the points (1, 0) and (2, 3) is
\(m=\frac {3-0}{2-1}=3\)
We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.
Therefore, slope of the line that is perpendicular to the line joining the points (1, 0) and (2, 3) = \(-\frac 1m = -\frac 13\)
Now, the equation of the line passing through \((\frac {n+1}{n+1},\frac {3}{n+1}) \) and whose slope is \(-\frac 13\) is given by
\((\frac {y-3}{n+1})=-\frac 13(x- \frac {n+2}{n+1})\)
⇒ \(3[(n+1)y-3]=-[x(n+1-(n+2)]\)
⇒ \(3(n+1)y-9=-(n+1)x+n+2\)
⇒ \((1+n)x+3(1+n)y=n+11\)
Top Questions on Various Forms of the Equation of a Line
- A line passes through (2, 2) and is perpendicular to the line 3x + y = 3. Its y-intercept is
- KCET - 2023
- Mathematics
- Various Forms of the Equation of a Line
- Suppose $ P(2,\,y,\,z) $ lies on the line through $ A(3,-1,4) $ and $ B(-4,2,1) $ . Then, the value of z is equal to
- KCET - 2011
- Mathematics
- Various Forms of the Equation of a Line
- $(0, - 1)$ and $(0, 3)$ are two opposite vertices of a square. The other two vertices are :
- BITSAT - 2006
- Mathematics
- Various Forms of the Equation of a Line
- If lines \(\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}\) and \(\frac{x-1}{3k}=\frac{y-5}{1}=\frac{z-6}{-5}\) are mutually perpendicular then k is equal to
- KCET
- Mathematics
- Various Forms of the Equation of a Line
- Find the equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.
- CBSE Class XI
- Mathematics
- Various Forms of the Equation of a Line
Questions Asked in CBSE Class XI exam
- Distinguish between the following pairs of sentences.
(i) He was visibly moved.
(ii) He was visually impaired.- CBSE Class XI
- The adventure
- How, according to you, can peace and liberty be maintained in a state?
- CBSE Class XI
- The tale of melon City
Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?
- CBSE Class XI
- Pressure
- Describe the change in hybridisation (if any) of the Al atom in the following reaction.
\(AlCl_3+Cl^- \rightarrow AlCl_4^-\)- CBSE Class XI
- Hybridisation
- Why did the narrator of the story want to forget the address?
- CBSE Class XI
- The address
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