We know that there are infinite points in the coordinate plane. Consider an arbitrary point \(P(x,y)\) on the \(XY\) 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 are discussed below.
1. 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, equation of the line which is parallel to \(X\)–Â axis and contains the point \((2,3)\) is \(y\) = \(3\).
Similarly, 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 \(L\) whose slope is \(m, ~A(x,y)\) be an arbitrary point on the line and \(P(x_1,y_1)\) be the fixed point on the same line.
Slope of the line by the definition is,
\(m\) = \(\frac{y~-~y_1}{x~-~x_1}\)
\(y~-~y_1\) = \(m(x~-~x_1)\)
For example, equation of the straight line having a slope \(m\) = \(2\) and passes through the point \((2,3)\) is
\(y~-~3\) = \(2(x~-~2)\)
\(y\) = \(2x~-~4~+~3\)
\(2x~-~y~-~1\) = \(0\)
3. Two-point form equation of line
Let \(P(x,y)\) be the general point on the line \(L\) which passes through the points \(A(x_1,y_1)\) and \(B(x_2,y_2)\).
Since the three points are collinear,
\(slope ~of ~PA\) = \(slope~ of~AB\)
\(\frac{y~-~y_1}{x~-~x_1}\) = \(\frac{y_2~-~y_1}{x_2~-~x_1}\)
\(y – y_{1} = (y_{2} – y_{1}). \frac{x – x_{1} }{x_{2} – x_{1}}\)
4. Slope-intercept form equation of line
Consider a line whose slope is \(m\) which cuts the \(Y\)-axis at a distance ‘a’ from the origin. Then the distance a is called the \(y\)– intercept of the line. The point at which the line cuts \(y\)-axis will be \((0,a)\).
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 as \(x\)– 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,
\(y~-~0\) = \(\frac{b-0}{0~-~a} (x~-~a)\)
\(y\) = \(-\frac{b}{a} (x~-~a)\)
\(y\) = \(\frac{b}{a} (a~-~x)\)
\(\frac{x}{a} ~+ ~\frac{y}{b} \) = \(1\)
For example, equation of the line which has \(x\)– intercept \(3\) and \(y\)– intercept \(4\) is,
\(\frac{x}{3} ~+ ~\frac{y}{4}\) = \(1\)
\(4x~ + ~3y\) = \(12\)
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,
\(OQ\) = \(l~ cosβ\)
\(PQ\) = \(l~ sinβ\)
Coordinates of the point \(P\) are; \(P(l ~cos~β,l ~sin~β)\)
slope of the line \(OP\) is \(tan~β\)
Therefore,
\(Slope~ of ~the ~line ~L\) = \(-\frac{1}{tan~β}\) = \(-\frac{cos~β}{sin~β}\)
Equation of the line \(L\) having slope \(-\frac{cos~β}{sin~β}\) and passing through the point \((l ~cos~β,l ~sin~β)\) is,
\(y~-~l~ sin~β\) = \(-\frac{cos~β}{sin~β} (x~-~l ~cosβ)\)
\(y ~sin~β~-~l~ sin^2~ β\) = \(-x ~cos~β~+~l~cos^2~β\)
\(x~ cos~β~ + ~y ~sin~β\) = \(l(sin^2~β ~+ ~cos^2~β)\)
\(x ~cos~β~ + ~y ~sin~β\) = \(l\)<
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.’
