We know that there are infinite points in the coordinate plane. Consider an arbitrary point P(x,y)Â on the XYÂ plane and a line L. How will we confirm whether the point is lying on the line L? This is where the importance of equation of a straight line comes into the picture in two-dimensional geometry.
Equation of a straight line contains terms in x and y. If the point P(x,y) satisfies the equation of the line, then the point P lies on the line L.
Different forms of equations of a straight line
- Equations of horizontal and vertical lines
Equation of the lines which are horizontal or parallel to the X-axis is y = a, where a is the y – coordinate of the points on the line.
Similarly, equation of a straight line which is vertical or parallel to Y-axis is x = a, where a is the x-coordinate of the points on the line.
For example, the equation of the line which is parallel to X-axis and contains the point (2,3) is y= 3.
Similarly, the equation of the line which is parallel to Y-axis and contains the point (3,4) is x = 3.
2. Point-slope form equation of line
Consider a non-vertical line LÂ whose slope is m, A(x,y)Â be an arbitrary point on the line and \(\begin{array}{l}P(x_1,y_1)\end{array} \)
be the fixed point on the same line.
Slope of the line by the definition is,
\(\begin{array}{l}m\end{array} \)
= \(\begin{array}{l}\frac{y~-~y_1}{x~-~x_1}\end{array} \)
\(\begin{array}{l}y~-~y_1\end{array} \)
= \(\begin{array}{l}m(x~-~x_1)\end{array} \)
For example, equation of the straight line having a slope \(\begin{array}{l}m\end{array} \)
= \(\begin{array}{l}2\end{array} \)
and passes through the point \(\begin{array}{l}(2,3)\end{array} \)
is
y – 3 = 2(x – 2)
y= 2x-4+3
2x-y-1Â = 0
3. Two-point form equation of line
Let P(x,y)Â be the general point on the line LÂ which passes through the points \(\begin{array}{l}A(x_1,y_1)\end{array} \)
and \(\begin{array}{l}B(x_2,y_2)\end{array} \)
.
Since the three points are collinear,
slope of PAÂ = slope of AB
\(\begin{array}{l}\frac{y~-~y_1}{x~-~x_1}\end{array} \)
= \(\begin{array}{l}\frac{y_2~-~y_1}{x_2~-~x_1}\end{array} \)
\(\begin{array}{l}y – y_{1} = (y_{2} – y_{1}). \frac{x – x_{1} }{x_{2} – x_{1}}\end{array} \)
4. Slope-intercept form equation of line
Consider a line whose slope is \(\begin{array}{l}m\end{array} \)
which cuts the \(\begin{array}{l}Y\end{array} \)
-axis at a distance ‘a’ from the origin. Then the distance a is called the \(\begin{array}{l}y\end{array} \)
– intercept of the line. The point at which the line cuts \(\begin{array}{l}y\end{array} \)
-axis will be \(\begin{array}{l}(0,a)\end{array} \)
.
Then, equation of the line will be
y-a = m(x-0)
y = mx+a
Similarly, a straight line having slope m cuts the X-axis at a distance b from the origin will be at the point (b,0). The distance b is called x- intercept of the line.
Equation of the line will be:
y = m(x-b)
5. Intercept form
Consider a line L having x– intercept a and y– intercept b, then the line touches X– axis at (a,0) and Y– axis at (0,b).
By two-point form equation,
\(\begin{array}{l}y~-~0\end{array} \)
= \(\begin{array}{l}\frac{b-0}{0~-~a} (x~-~a)\end{array} \)
\(\begin{array}{l}y\end{array} \)
= \(\begin{array}{l}-\frac{b}{a} (x~-~a)\end{array} \)
\(\begin{array}{l}y\end{array} \)
= \(\begin{array}{l}\frac{b}{a} (a~-~x)\end{array} \)
\(\begin{array}{l}\frac{x}{a} ~+ ~\frac{y}{b} \end{array} \)
= \(\begin{array}{l}1\end{array} \)
For example, equation of the line which has \(\begin{array}{l}x\end{array} \)
– intercept \(\begin{array}{l}3\end{array} \)
and \(\begin{array}{l}y\end{array} \)
– intercept \(\begin{array}{l}4\end{array} \)
is,
\(\begin{array}{l}\frac{x}{3} ~+ ~\frac{y}{4}\end{array} \)
= \(\begin{array}{l}1\end{array} \)
\(\begin{array}{l}4x~ + ~3y\end{array} \)
= \(\begin{array}{l}12\end{array} \)
6. Normal form
Consider a perpendicular from the origin having length l to line L and it makes an angle β with the positive X-axis.
Let OP be the perpendicular from the origin to the line L.
Then,
\(\begin{array}{l}OQ\end{array} \)
= \(\begin{array}{l}l~ cosβ\end{array} \)
\(\begin{array}{l}PQ\end{array} \)
= \(\begin{array}{l}l~ sinβ\end{array} \)
Coordinates of the point \(\begin{array}{l}P\end{array} \)
are; \(\begin{array}{l}P(l ~cos~β,l ~sin~β)\end{array} \)
slope of the line \(\begin{array}{l}OP\end{array} \)
is \(\begin{array}{l}tan~β\end{array} \)
Therefore,
\(\begin{array}{l}Slope~ of ~the ~line ~L\end{array} \)
= \(\begin{array}{l}-\frac{1}{tan~β}\end{array} \)
= \(\begin{array}{l}-\frac{cos~β}{sin~β}\end{array} \)
Equation of the line \(\begin{array}{l}L\end{array} \)
having slope \(\begin{array}{l}-\frac{cos~β}{sin~β}\end{array} \)
and passing through the point \(\begin{array}{l}(l ~cos~β,l ~sin~β)\end{array} \)
is,
\(\begin{array}{l}y~-~l~ sin~β\end{array} \)
= \(\begin{array}{l}-\frac{cos~β}{sin~β} (x~-~l ~cosβ)\end{array} \)
\(\begin{array}{l}y ~sin~β~-~l~ sin^2~ β\end{array} \)
= \(\begin{array}{l}-x ~cos~β~+~l~cos^2~β\end{array} \)
\(\begin{array}{l}x~ cos~β~ + ~y ~sin~β\end{array} \)
= \(\begin{array}{l}l(sin^2~β ~+ ~cos^2~β)\end{array} \)
\(\begin{array}{l}x ~cos~β~ + ~y ~sin~β\end{array} \)
= \(\begin{array}{l}l\end{array} \)
You have learnt about the different forms of equation of a straight line. To know more about straight lines and its properties, log onto www.byjus.com.’
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