Various Forms of the Equation of a Line: An infinite number of points come together in two dimensions and make a coordinate plane. Consider the point \(P(x, y)\) in a \(2D\) plane and the line \(l.\) To check if the point we are looking at is on the line \(l,\) or if it is above or below the line \(l,\) the equation of line comes into play.

We can use the equation of any line to determine whether or not a given point is on the line. If the point \(P(x, y)\) satisfies the equation of the line, then the point \(P\) lies on the line \(l.\) In this article, let us see the various types of line equations.

Different Forms of Equation of a Line

Based on the characteristics known of the straight line, there are five ways to write the equation of a line. The following are a few methods for determining and representing the equation of a line.

• Point-Slope Form
• Two point Form
• Slope-intercept form
• Intercept form
• Normal form

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Equation of a Line Parallel to y-axis

Let \(AB\) be a line parallel to the \(y-\)axis, then the abscissa of every point on the line \(AB\) is constant, a

Let \(P(x,y)\) be any point on \(AB.\)

From \(P\) draw \(NP\) perpendicular to \(y−\)axis. Then \(NP=x.\)

\(∴x=a,\) which is the required equation of the line \(AB.\)

When \(a=0,\) the line \(AB\) coincides with the \(y−\)axis, and the equation of the line becomes \(x=0.\) Hence, the equation of \(y−\)axis is given by \(x=0.\)

Equation of a Line Parallel to x−axis

Let \(CD\) be a line parallel to the \(x−\)axis. Observe that the ordinate of every point on the line \(CD\) is constant, \(b.\)

Let \(P(x,y)\) be any point on the line \(CD.\) From \(P,\) draw \(MP\) perpendicular to the \(x−\)axis. Then \(MP=y.\)

\(∴y=b,\) which is the required equation of the line \(CD.\)

When \(b=0,\) the line \(CD\) coincides with the \(x−\)axis. Thus, the equation of \(x−\)axis is given by \(y=0.\)

Point-Slope Form:

Let \(l\) be the given line passing through the fixed point \(A\left( {{x_1},{y_1}} \right)\) and having slope m. Let \(P\left( {x,y} \right)\) be any point on \(l\) then slope of the line \(l = \frac{{y – {y_1}}}{{x – {x_1}}}.\)

Slope of the line \(l=m\) (given)

Therefore, the required equation of a straight line passing through the fixed point \(A\left( {{x_1},{y_1}} \right)\) and having slope \(m\) is \(y – {y_1} = m\left( {x – {x_1}} \right).\)

This equation is known as the point-slope form.

Two-Point Form:

Let the line pass through the fixed points \(A\left( {{x_1},{y_1}} \right),\) and \(A\left( {{x_2},{y_2}} \right).\)

The slope of the line \(AB = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}}.\)

Therefore, the equation of the line through \(A\left( {{x_1},{y_1}} \right)\) and having slope  \(\frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}}\) is given by

\(y – {y_1} = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}}\left( {x – {x_1}} \right)\)

This is known as two-point form.

Slope-Intercept Form:

We know that a line that passes through the origin makes zero intercepts on axes.

A horizontal line has no \(x−\)intercept and a vertical line has no \(y−\)intercept. 

Case I:

Let \(l\) be the line having slope \(m\) and \(y−\)intercept, \(c.\) Then, \(l\) intersects the \(y−\)axis at a point \((0,c).\)

Now, using the Point-Slope form of the line, we have,

\((y – 0) = m(x – c)\)

\(∴y=mx+c\) is the equation of the straight line in the slope-intercept form. This equation is used when the slope of the line, and its intercept on the \(y−\)axis are known to us.

Case II:

Let \(l\) be a line having slope \(m\) and \(x−\)intercept \(d.\) Then the line \(l\) intersects the \(x−\)axis at a point \((d,0).\)

Now, using the point-slope form of the line, we have,

\(y – 0 = m(x – d)\)

\(∴y=m(x−d)\) is the required equation of the straight line in slope-intercept form whose slope is \(m,\) and \(x−\)intercept \(d.\)

Intercept Form:

This form is used to find the equation of a straight line which cuts off given intercepts on the axes. Let the two points \(A(a,0)\) and \(B(0,b)\) lie on the line \(l.\) Hence, \(l\) makes intercepts \(a\) and \(b\) on \(x−\)axis, and \(y−\)axis respectively.

Now, using the two point form of a line, we have

\(y – 0 = \frac{{b – 0}}{{0 – a}}(x – a)\)

\( \Rightarrow y = \frac{b}{a}(x – a)\)

\( \Rightarrow bx + ay = ab\)

Dividing both sides of the equation by \(ab,\) we have,

\(\frac{x}{a} + \frac{y}{b} = 1\)

Thus, \(\frac{x}{a} + \frac{y}{b} = 1\) is the required equation of the line in intercept-form which is making an intercepts \(a\) and \(b\) on the \(x−\)axis, and \(y−\)axis respectively.

Normal (or Perpendicular) Form:

Let \(l\) be the given line, and let it meet the \(x−\)axis and \(y−\)axis in points \(A\) and \(B\) respectively.

From the origin, \(O,\) draw \(ON\) perpendicular to \(l.\) Here, \(ON=p\) and \(\angle AON = \alpha \)

\(\therefore \angle NOB = {90^\circ } – \alpha \Rightarrow \angle OBN = \alpha \)

From \(∆OAN,\)

\(\cos \alpha = \frac{{ON}}{{OA}} = \frac{p}{{OA}}\)

\( \Rightarrow p = OA\cos \alpha \)

Similarly, from \(∆OBN,\)

\(\sin \alpha = \frac{{ON}}{{OB}} = \frac{p}{{OB}}\)

\( \Rightarrow p = OB\sin \alpha \)

Thus, the line \(l\) makes intercepts pcos, and psin on \(x−\)axis and \(y−\)axis respectively.

Using the intercept form of a line, the equation of the required line is:

\(x\cos \alpha + y\sin \alpha = p\)

Symmetrical (or Distance) Form:

Let \(l\) be the given line through \(\left( {{x_1},{y_1}} \right)\) and be the angle which the line makes with the positive \(x−\)axis.Let \(P(x,y)\) be any point on the line \(l\) and let \(AP=r.\) Draw \(PM\) and \(AN\) perpendiculars to the \(x−\)axis and \(AL\) perpendicular to \(MP\) then \(\angle LAP = \theta .\)

From the figure,

\(AL = NM = OM – ON = x – {x_1}\)

\(LP = MP – ML = MP – AN = y – {y_1}\)

From \(∆ALP,\)

\(\cos \theta = \frac{{x – {x_1}}}{r}\) \( \Rightarrow \frac{{x – {x_1}}}{{cos\theta }} = r…{\rm{(i)}}\)

\(\sin \theta = \frac{{y – {y_1}}}{r}\) \( \Rightarrow \frac{{y – {y_1}}}{{\sin \theta }} = r \ldots {\rm{(ii)}}\)

From \(\left( {\rm{i}} \right)\) and \(\left( {{\rm{ii}}} \right),\) we have \(\frac{{x – {x_1}}}{{{\mathop{\rm cos}\nolimits} \theta }} = \frac{{y – {y_1}}}{{\sin \theta }} = r,\) which is the required equation of the line in symmetrical (or distance) form.

Solved Examples – Various Forms of the Equation of a Line

Q.1. Find the equation of a straight line parallel to \(x−\)axis at a distance \(5\) units below it.
Ans: We know that the equation of a straight line parallel to the \(x−\)axis is given by \(y=b,\) where \(b\) is the distance of the line from the \(x−\)axis.
Since the line is at a distance \(5\) units below the \(x−\)axis. So, we have \(b=−5.\)
Thus, the required equation of the line is \(y=−5.\)

Q.2. Find the equation of a straight line which passes through the point \((−3,7)\) and makes intercepts on the axes equal in magnitude but opposite in sign.
Ans: Given that a line passes through the point \((−3,7).\)
Since the intercepts made by the line on the axes are equal in magnitude but opposite in sign.
Let the \(x−\)intercept be a and \(y−\)axis intercept be \(–a.\)
Now, using the intercept form of a line the equation of the line is given by:
\(\frac{x}{a} – \frac{y}{a} = 1\)
Coordinates of \((−3,7)\) satisfies the equation of the line.
\(\therefore – \frac{3}{a} – \frac{7}{a} = 1\)
\( \Rightarrow a = – 10\)
Hence, the required equation is: \( – \frac{x}{{10}} + \frac{y}{{10}} = 1\) i.e.,\(x – y + 10 = 0\)

Q.3. If the point \((−4,3)\) lies on a line intercepted between the axes and divides it internally in the ratio \(5:3\) then find the equation of the line.
Ans: Let the required equation of the line be \(\frac{x}{a} + \frac{y}{b} = 1,\) where a is the \(x−\)intercept, and \(b\) is the \(y−\)intercept.
Consider the following figure

Using the section formula for internal division, we have
\(( – 4,3) = \left( {\frac{{5 \times 0 + 3a}}{{5 + 3}},\frac{{5 \times b + 3 \times 0}}{{5 + 3}}} \right)\)
\( \Rightarrow – 4 = \frac{{3a}}{8},3 = \frac{{5b}}{8}\)
\(\therefore a = – \frac{{32}}{3},\) and \(b = \frac{{24}}{5}\)
Hence, the required equation is,
\(\frac{x}{{ – \frac{{32}}{3}}} + \frac{y}{{\frac{{24}}{5}}} = 1\)
\( \Rightarrow 9x – 20y + 96 = 0\)

Q.4. What line is parallel to \(y=2x−1\) through \((1,−3)\)?
Ans: Given:
\(y=2x−1,\)
Slope of this line\(=2\)
Since parallel lines have equal slopes, the slope of required line\(=2,\) and it passes through the point \((1,−3).\)
Let the required equation of the line be \(y=2x+c,\) where \(c=y−\)intercept of the line
Since the coordinates of the point \((1,−3)\) satisfies the equation \(y=2x+c,\)
\( \Rightarrow – 3 = 2(1) + c\)
\( \Rightarrow c = – 5\)
Hence, the required equation is,
\(y=2x−5\)

Q.5. Write the equation of a straight line on which the perpendicular from the origin is of length \(2\) units and this perpendicular makes an angle of \(240\) with the \(x−\)axis.
Ans: Perpendicular form of a line is given by,
\(x\cos \alpha + y\sin \alpha = p\)
Here,
\(\alpha = {240^\circ }\)
p=2 units
Hence, the equation of the required line is given by
\(x\cos {240^\circ } + y\sin {240^\circ } = 2\)
\( \Rightarrow x\cos \left( {{{180}^\circ } + {{60}^\circ }} \right) + y\sin \left( {{{180}^\circ } + {{60}^\circ }} \right) = 2\)
\( \Rightarrow – x\cos \left( {{{60}^\circ }} \right) – y\sin \left( {{{60}^\circ }} \right) = 2\)
\( \Rightarrow x\left( { – \frac{1}{2}} \right) + y\left( { – \frac{{\sqrt 3 }}{2}} \right) = 2\)
\( \Rightarrow x + \sqrt 3 y + 4 = 0\)

Summary

In this article, the different ways to write the equation of straight lines depending on the known factors is discussed. We can find the equation of a line given the slope and a point using the Point-Slope form i.e., \(y – {y_1} = m\left( {x – {x_1}} \right).\) We can also find the equation of a line given two points using the two-point form i.e., \(y – {y_1} = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}}\left( {x – {x_1}} \right)\)  If the slope of line is given, and the intercept of the line is given with the axes then we can find the equation of line using the slope-intercept form i.e., \(y=mx+c,\) where \(m = {\rm{slope}}\) of line, and \(c=y−\)intercept. Similarly, the Intercept form, and Normal form of the line are also explained.

Frequently Asked Questions (FAQs)

Q.1. What are the five forms of equations of the line?
Ans: There are five basic forms of writing the equation of a line based on the parameters known for the straight line.
1. Point-slope form
2. Two point form
3. Slope-intercept form
4. Intercept form
5. Normal form

Q.2. What is the equation of a line that passes through the origin?
Ans: Let the slope be m.
Using the point slope form, \(y – {y_1} = m\left( {x – {x_1}} \right),\) the equation of the line passing through the origin \(O(0,0)\) is given by,
\(y – 0 = m(x – 0)\) i.e.,
\(\therefore y = mx\)

Q.3. How do you convert general form to slope-intercept form?
Ans: The general form of the linear equation with two variables is given by
\(ax+by+c=0\)
Rearranging,
\(by=−ax−c\)
\( \Rightarrow y = – \frac{a}{b}x – \frac{c}{b},\) is the equation of the line in slope-intercept form where,
slope\( = – \frac{a}{b}\)
\(y−\) intercept\( = – \frac{c}{b}\)

Q.4. Write the formula for slope-intercept, and point-slope forms of a line.
Ans: The formula for slope-intercept form is \(y=mx+c,\) and point-slope form is \(y – {y_1} = m\left( {x – {x_1}} \right)\) where m is the slope of the line, and \(c\) is the \(y−\)intercept, and, \(\left( {{x_1},{y_1}} \right)\) is the point that lies on the line.

Q.5. Why is it important to know the various forms of equations of a line?
Ans: Knowing the different forms of equations of a line can help us in identifying the slope of the line, and the intercepts made by that line on the axes.
Example: Let the equation of line is \(y=mx+c\)
Here, slope of the line\(=m,\) and \(y−\)intercept of line\(=c\)
We can also check whether an arbitrary point \(P(x,y)\) lies on that line or not by satisfying the coordinates of the point in an equation.

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