Pair of Straight Lines: Equations, Notes & Formulas

Earlier, we studied the mathematical representation and properties of a single line. Now, it is time to advance this knowledge for a pair of straight lines.

In this post, we will study a pair of lines and develop mechanism to represent them using a single equation. Further, we will see the types of problems that come from this topic.

After going through this post, you will realize that most equations and formulas are simple extensions of the basic formulas you are already aware of.

Table of Contents

  1. Combined equation of a pair of straight lines
  2. Type of Problems
  3. Examples

Combined Equation of a Pair of Straight Lines

Let’s we have two lines $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$. For brevity, we denote them as $u \equiv a_1 x + b_1 y + c_1 = 0$ and $v \equiv a_2 x + b_2 y + c_2 = 0$, respectively.

pair of straight lines

We can describe this pair of straight lines by writing two equations $u = 0$ and $v=0$. However, can’t we have a single equation to describe this pair? There comes the concept of combined equation of a pair of lines.

It is to note that an equation of a line denotes all the points which lie on this line. In a similar fashion, the combined equation should represent all points that lie on either of the lines.

The equation $uv = 0$ is the combined equation of a pair of lines. It is easy to verify this proposition. Let $(x’, y’)$ be a point on the line $u=0$. Then,

\begin{aligned} uv &= (a_1 x’ + b_1 y’ + c_1) (a_2 x’ + b_2 y’ + c_2), \\
&= 0 (a_2 x’ + b_2 y’ + c_2), \\
&= 0.\end{aligned}

In a similar fashion, if $(x^{\prime \prime}, y^{\prime \prime})$ is a point on the line $v=0$, then

\begin{aligned} uv &= (a_1 x^{\prime \prime} + b_1 y^{\prime \prime} + c_1) (a_2 x^{\prime \prime} + b_2 y^{\prime \prime} + c_2), \\
&= (a_1 x^{\prime \prime} + b_1 y^{\prime \prime} + c_1) 0, \\
&= 0.\end{aligned}

Thus, all the points lying on either of the lines satisfy the equation. Now comes the reverse part. Let $(x_0, y_0)$ be a point which satisfy the equation $uv = 0$, i.e.,

$uv = (a_1 x_0 + b_1 y_0 + c_1) (a_2 x_0 + b_2 y_0 + c_2) = 0.$

This implies that either $u = 0$ or $v = 0$ because the product of two non-zero numbers can never be equal to zero. Thus the point $(x_0, y_0)$ lies on either of the lines. This completes the proof of the proposition.

Type of Problems

An important question may have come to your mind that what type of problems come from the topic of the combined equation of a pair of lines. Be happy that only two major types of problems come from this topic. These are

  1. Obtaining the combined equation of a given pair of lines.
  2. Obtaining the separate equations of lines for a given combined equation.

As we will see, the first type of problem is relatively easier to solve compared to the second type of problem.

Examples

Ex. 1: Find the combined equation of the following pair of lines

\begin{aligned}x + 2y – 1 = 0 \quad \text{and} \quad x-3y = 2\end{aligned}

Solution: In such problems, first of all, bring all line equations in the form $ax + by + c = 0$. Post that, the combined equation of the lines can be obtained as

\begin{aligned} &(x + 2y-1) (x – 3y -2) = 0, \\
&x (x – 3y -2) + 2y (x – 3y -2)-(x – 3y -2)= 0, \\
& x^2 – 3 xy – 2x + 2xy – 6 y^2 – 4y -x + 3y + 2 = 0, \\
& x^2 – xy – 6y^2 – 3x -y+2 =0.\end{aligned}

Thus, the combined equation of the given pair is given by $x^2 – xy – 6y^2 – 3x -y+2 =0$.


Ex. 2: Find the separate equation of lines represented by $x^2 – y^2 + x – y = 0$.

Solution: For such problems, we have to figure out a common factor in the equation and take it out.

\begin{aligned} (x-y)(x+y) + x-y &= 0, \\
(x-y)(x+y+1) & = 0.
\end{aligned}

Hence, the separate equations are $x-y = 0$ and $x+y+1 = 0$.


Ex. 3: Find the separate equation of lines represented by $3x^2 – 10xy-8y^2 = 0$.

Solution: Sometimes, the factors are not obvious in the equation. In that case, we have to artificially create a factor.

\begin{aligned}3x^2 – 10xy-8y^2&= 0, \\
3x^2-12xy+2xy-8y^2 &= 0, \\
3x(x-4y)+2y(x-4y) &=0,\\
(3x+2y)(x-4y) &=0.
\end{aligned}

Thus, the separate equations are $3x+2y =0$ and $x-4y =0$.


Ex. 4: Find the separate equation of lines represented by $x^2 +2 \text{cosec} \theta xy+ y^2 = 0$.

Solution: It is a very good problem Here, we have to use a trigonometric relation to create factors which is as follows

\begin{aligned} \text{cosec}^2 \theta-1&=\cot^2\theta,\\
\text{cosec}^2 \theta-\cot^2\theta&=1,\\
(\text{cosec} \theta-\cot\theta) (\text{cosec} \theta+\cot\theta) &=1.
\end{aligned}

Using the above relation, we can have the following innovation

\begin{aligned} x^2 +2 \text{cosec} \theta xy+ y^2&=0,\\
x^2 +2 \text{cosec} \theta xy + (\text{cosec} \theta-\cot\theta) (\text{cosec} \theta+\cot\theta)y^2 &=0, \\
(x+(\text{cosec} \theta-\cot\theta) y) (x+(\text{cosec} \theta+\cot\theta) y) &=0.
\end{aligned}

So, the separate equations are $x+(\text{cosec} \theta-\cot\theta) y = 0$ and $x+(\text{cosec} \theta+\cot\theta) y = 0$.

Summary

In this lesson, we obtained a combined equation of a pair of lines. We also looked at types of problems asked from this topic along with few examples.

If you have any doubt related with this topic or any question, then please drop a comment. I will be more than happy to answer your question.

Subscribe
Notify of
guest
0 Comments
Inline Feedbacks
View all comments