Pair of Straight Lines Class 12: A straight line, often known as a line, is an infinite one-dimensional shape with no breadth but only length. A straight line comprises an infinite number of points. We represent a straight line with a linear equation in two variables. Under certain conditions, a second-degree equation in two variables represents a pair of straight lines.

A pair of straight lines can also be represented as a product of two linear equations in \(x\) and \(y,\) representing a straight line. The concept of pair of straight lines is very helpful in the mathematical world as it simplifies our complex problems more easily. In this article, we would learn how a pair of straight lines are created, pair of straight lines formulas, the general question related to pair of straight lines, and more.

Pair of Straight Lines Class 12: Overview

A pair of straight lines are formed when the product of two linear equations in \(x\) and \(y\) represents a straight line are multiplied together.
Let \({L_1} = 0,{L_2} = 0\) be the equations of two straight lines.
If \(P\left({{x_1},{y_1}} \right)\) is a point on \({L_1}\) then it satisfies the equation \({L_1} = 0.\) Similarly, if \(P\left({{x_1},{y_1}} \right)\) is a point on \({L_2} = 0,\) then it satisfies the equation.

If \(P\left({{x_1},{y_1}} \right)\) lies on \({L_1}\) or \({L_2},\) then \(P\left({{x_1},{y_1}} \right)\) satisfies the equation \({L_1}\,{L_2} = 0.\)
\(\therefore {L_1}\,{L_2} = 0\) represents the pair of straight lines \({L_1} = 0\) and \({L_2} = 0\) and the joint equation of \({L_1} = 0\) and \({L_2} = 0\) is given by \({L_1}.{L_2} = 0\)

On expanding the above equation, we get an equation of the form \(a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0\) which is a second degree (non – homogeneous) equation in \(x\) and \(y.\)
If \(a,b,h\) are not all zero, then \(a{x^2} + 2hxy + b{y^2} = 0\) represent the general equation of a second degree homogeneous equation in \(x\) and \(y.\) \(a{x^2} + 2hxy + b{y^2} = 0\) represent a pair of straight lines passing through the origin.

If \(a,b,h\) are not all zero, then \(a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0\) represent the general equation of a second-degree non – homogeneous equation in \(x\) and \(y.\)
For example: if \(5x + y = 0\) and \(x + y = 0\) are the two straight lines, then the pair of straight lines are represented by \(\left({5x + y} \right)\left({x + y}\right) = 0 \Rightarrow 5{x^2} + 6xy + {y^2} = 0\)
A pair of straight lines can be separated into two lines easily also, and after separating the lines, we can easily apply the basic formulas involved for straight lines.

In order to separate the equation of pair of straight lines, we can have either of the following steps:
a. Factorisation of the given second-degree equation of a pair of straight lines.
b. Or, we can make a quadratic in \(x\) from the given combined equation of a pair of straight lines, then the quadratic formula \(x = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\)
c. Or, we can take the general equation of two straight lines, multiply it together, and then compare its coefficient with the given combined equation of a pair of straight lines.

Pair of Straight Lines Formulas

The list of Pair of Straight Lines Formulas is included below:

1. \(a{x^2} + 2hxy + b{y^2} = 0\) is a homogenous equation of second degree, represents a pair of straight lines passing through the origin. So,
a. If \({h^2} > ab,\) the two straight lines are real and different.
b. If \({h^2} = ab,\) the two straight lines are coincident.
c. If \({h^2} < ab,\) the two straight lines are imaginary, having origin as the point of intersection.
2. \(a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0\) represent the general equation of a pair of straight lines if \(\Delta = \left|{\begin{array}{{c}} a \hfill & h \hfill & g \hfill \\ h \hfill & b \hfill & f \hfill \\ g \hfill & f \hfill & c \hfill \\ \end{array} } \right| = 0.\) So,
a. If \({h^2} > ab,\) the two straight lines are real and different.
b. If \({h^2} = ab,\) the two straight lines are coincident or parallel.
c. If \({h^2} < ab,\) the two straight lines are imaginary, having a fixed point of intersection.

Angle between Pair of Straight Lines

Consider the equation of a pair of straight lines passing through the origin as:
\(a{x^2} + 2hxy + b{y^2} = 0……\left( 1 \right)\)
Let \({m_1}\) and \({m_2}\) be the slopes of these two lines. By dividing \(\left( 1 \right)\) by \({x^2}\) and substituting \(\frac{y}{x} = m\) we get, \(b{m^2} + 2hm + a = 0\)
This quadratic in \(m\) will have its roots as \({m_1}\) and \({m_2}.\) Thus, \({m_1} + {m_2} = \frac{{ – 2h}}{b}\) and \({m_1}{m_2} = \frac{a}{b}\)
If the angle between the two lines is \(\theta \)
Then,
\(\tan \theta = \left|{\frac{{{m_2} – {m_1}}}{{1 + {m_2}{m_1}}}}\right|\)
\( = \left|{\frac{{\sqrt {{{\left({{m_1} + {m_2}} \right)}^2} – 4{m_1}{m_2}} }}{{1 + {m_2}{m_1}}}} \right|\)
\( = \left|{\frac{{\sqrt {{{\left({\frac{{ – 2h}}{b}} \right)}^2} – 4\frac{a}{b}} }}{{1 + \frac{a}{b}}}} \right|\)
\(\tan \theta = \left|{\frac{{2\sqrt {{h^2} – ab} }}{{a + b}}} \right|\)
This formula is also valid for the general equation of straight line \(a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0\)

Pair of Straight Lines JEE Questions: Condition for Pair of Straight Lines

Some important conditions or results are extremely helpful when dealing with problems based on pair of straight lines.
A general equation of second-degree equation \(a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0,\) which represent a pair of straight lines, will have the following results.
1. Two lines will be coincident or parallel if \(\tan \,\theta = 0\) i.e. if \({h^2} – ab = 0\)
2. Two lines will be perpendicular if tan⁡θ is not defined, i.e. if \(a + b = 0\)
3. Two lines will be equally inclined to axes, if the coefficient of \(xy = 0,\) i.e. if \(h = 0\)
4. The angle between a pair of straight lines is given by \(\tan \theta = \left|{\frac{{2\sqrt {{h^2} – ab} }}{{a + b}}} \right|\)
5. If \(a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0\) represent a pair of straight lines, then the sum of slopes of lines is \(\frac{{ – 2h}}{b}\) and product of the slopes is \(\frac{a}{b}\)

Solved Examples – Pair of Straight Lines Class 12

Q.1. Find the acute angle between the pair of the line represented by the equation \({x^2} – 7xy + 12{y^2} = 0\)
Ans: Given equation is \({x^2} – 7xy + 12{y^2} = 0.\)
Comparing with \(a{x^2} + 2hxy + b{y^2} = 0\)
\(a = 1,b = 12,h = \, – \frac{7}{2}\)
Let \(\theta \) be the angle between the lines, then \(\tan \theta = \frac{{2\sqrt {{h^2} – ab} }}{{a + b}} = \frac{{\sqrt[2]{{\frac{{49}}{4} – 12}}}}{{1 + 12}} = \frac{{\sqrt[2]{{\frac{1}{4}}}}}{{13}} = \frac{1}{{13}}\)
\(\tan \,\theta = \frac{1}{{13}} \Rightarrow \theta = {\tan ^{ – 1}}\left( {\frac{1}{{13}}} \right)\)

Q.2. If \(a{x^2} + 2hxy + b{y^2} = 0\) represents two straight lines such that the slope of one line is twice the slope of the other, prove that \(8{h^2} = 9ab.\)
Ans: The equation of the pair of lines is \(a{x^2} + 2hxy + b{y^2} = 0\)
Let \(y = {m_1}x\) and \(y = {m_2}x\) be the lines represented by \(a{x^2} + 2hxy + b{y^2} = 0\)
\(\therefore {m_1} + {m_2} = – \frac{{2h}}{b},{m_1}{m_2} = \frac{a}{b}\)
Given \({m_2} = 2{m_1}\,\therefore {m_1} + 2{m_2} = \, – \frac{{2h}}{b},{m_1}.2{m_1} = \frac{a}{b}\)
\(\therefore 3\,{m_1} = – \frac{{2h}}{b};\) and \(\,{m_1}^2 = \, – \frac{a}{{2\,b}}\)
\(\therefore {\left({ – \frac{{2h}}{{3b}}} \right)^2} = \frac{a}{{2b}} \Rightarrow \frac{{4{h^2}}}{{9{b^2}}} = \frac{a}{{2b}} \Rightarrow 8{h^2} = 9ab\)

Q.3. Show that \(2{x^2} + 3xy – 2{y^2} + 3x + y + 1 = 0\) represents a pair of perpendicular lines.
Ans: \(2{x^2} + 3xy – 2{y^2} + 3x + y + 1 = 0\)
On comparing the given equation with the general equation of pair of straight lines
\(a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0\)
We get, \(a = 2,b = \,- 2,2h = 3 \Rightarrow h = \frac{3}{2}\)
If two lines are perpendicular, then \(a + b = 0\)
\(2 + \left({ – 2} \right) = 0\)
Hence the given pair of the straight line is perpendicular.

Q.4. Show that the equation \(4{x^2} + 4xy +{y^2} – 6x – 3y – 4 = 0\) represents a pair of parallel lines. Find the distance between them
Ans: \(4{x^2} + 4xy +{y^2} – 6x – 3y – 4 = 0\)
\(a = 4,b = 1,h = \frac{4}{2} = 2\)
\({h^2} – ab = {2^2} – \left( 4 \right)\left( 1 \right) = 4 – 4 = 0\)
\( \Rightarrow \) The given equation represents a pair of parallel lines
To find the separate equations \(4{x^2} + 4xy + {y^2} = {\left({2x + y}\right)^2}\)
So, \(4{x^2} + 4xy + {y^2} – 6x – 3y – 4 = \left({2x + y + l} \right)\left({2x + y + m} \right)\)
Coefficient of \(x \Rightarrow 2m + 2l = \,- 6 \Rightarrow l + m = \, – 3 \ldots \left( 1 \right)\)
Coefficient of \(y \Rightarrow l + m = \, – 3 \ldots \left( 2 \right)\)
Constant term \( \Rightarrow lm = \, – 4 \ldots \left( 3 \right)\)
Now \(l + m = \,- 3\) and \(lm = \, – 4 \Rightarrow l = \, – 4,m = 1\)
So the separate equations are \(2x + y + 1 = 0\) and \(2x + y -4 = 0\) The distance between the parallel lines is \(\frac{{1 + 4}}{{\sqrt {4 + 1} }} = \frac{5}{{\sqrt 5 }} = \sqrt 5 \,{\text{units}}\)

Q.5. Find the different equations of lines represented by the equation \({x^2} – 6xy + 8{y^2} = 0\)
Ans: Given \({x^2} – 6xy + 8{y^2} = 0\)
\(x = \frac{{6y \pm \sqrt {{{\left({ – 6y} \right)}^2} – 4 \times 8{y^2}} }}{2}\)
\( \Rightarrow x = 3y \pm y\sqrt {9 – 8} \)
\(x = 3y \pm y\)
\(\therefore x = 4y\) or \(x = 2y\)
\( \Rightarrow x – 4y = 0\) or \(x – 2y = 0\)
Separate equation of lines represented by the equation \({x^2} – 6xy + 8{y^2} = 0\) will be \(\left({x – 4y} \right)\left({x – 2y} \right) = 0\)
\( \Rightarrow x – 4y = 0\) and \(x – 2y = 0\).

Summary on Pair of Straight Lines Theorems

In this article, we learned about a pair of straight lines, pair of straight lines formulas, and the conditions involved in it. We also saw some solved examples related to pairs of straight lines. We have learned how to pair of straight lines is formed and how they can be separated. A pair of straight lines are formed when the product of two linear equations in \(x\) and \(y\) represents a straight line are done together. A pair of straight lines can easily be separated; thereafter, we can apply the basic formulas for straight lines. We have also seen the formula for the angle between a pair of straight lines, which is very helpful to solve complex problems of pair of straight lines without separating the lines.

FAQs Pair of Straight Lines Important Questions

Q.1. How to solve pair of straight lines?
Ans: \(a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0\) represent the general equation of a pair of straight lines if \(\Delta = 0.\) Hence, In order to solve pair of straight lines, we just have to factorise into two different linear equations, which are basically straight lines. When the lines are separated, then we can apply the basics formulas related to the straight lines.

Q.2. How to find an angle between a pair of straight lines?
Ans: If \(a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0\) represent a pair of straight lines, then the angle between the straight lines is given by \(\tan \theta = \left|{\frac{{2\sqrt{{h^2} – ab} }}{{a + b}}} \right|\).

Q.3. How to separate the equation of pair of straight lines?
Ans: In order to separate the equation of pair of straight lines, we can have either of the following steps:
a. Factorization of the given second-degree equation of a pair of straight lines.
b. Or, we can make a quadratic in \(x\) from the given combined equation of a pair of straight lines, then the quadratic formula \(x = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\)
c. Or, we can take the general equation of two straight lines, multiply it together, and then compare its coefficient with the given combined equation of a pair of straight lines.

Q.4. How to find the slope of pair of straight lines?
Ans: If \(a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0\) represent a pair of straight lines, then \({m_1} + {m_2} = \, – \frac{{2h}}{b},{m_1}{m_2} = \frac{a}{b}\) where \({m_1}\) and \({m_2}\) are the slopes of the straight lines. From the given relations, we can easily find the slope of pair of straight lines.

Q.5. How do you find a pair of straight lines?
Ans: A pair of straight lines is created when the product of two linear equations in \(x\) and \(y\) represents a straight line are done together. For example, if \(5x + y = 0\) and \(x + y = 0\) are the two straight lines, then the pair of straight lines are represented by \(\left({5x + y} \right)\left({x + y} \right) = 0 \Rightarrow 5{x^2} + 6xy + {y^2} = 0\).