# Technique of Homogenizing

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We now discuss a very useful application of the concept of pair of straight lines.

Consider a second degree curve \(S(x,\;y)\) with the equation

\[S(x,\;y) \equiv a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0\]

and a straight line

\[L \equiv px + qy + r = 0\]

intersecting \(S = 0\) in *A* and *B*. Let *O* be the origin.

What is the joint equation of *OA* and *OB*?

The insight that we use here is that since both *OA* and *OB* pass through the origin, their joint equation will be homogenous. We now construct a homogenous equation and show that both *A* and *B* satisfy it; that equation is then guaranteed to jointly represent *OA* and *OB*.

First of all, observe that since *A* and *B* satisfy the equation of *L*, i.e. \(px + qy + r = 0,\) they will also satisfy the relation

\[\begin{align}\frac{{px + qy}}{{ - r}} = 1 \qquad \quad : \qquad \quad {\text{Both A and B will satisfy this relation.}}\end{align}\]

Now, we homogenize the equation of the second degree curve \(S(x,\;y)\) using the relation above; consider this equation :

\[\begin{align}a{x^2} + 2hxy + b{y^2} + 2gx\left( {\frac{{px + qy}}{{ - r}}} \right) + 2fy\left( {\frac{{px + qy}}{{ - r}}} \right) + c{\left( {\frac{{px + qy}}{{ - r}}} \right)^2} = 0 \qquad \qquad \qquad ...(1)\end{align}\]

Can you understand why we’ve done this? The equation we obtain above is a second degree homogenous equation, and so it must represent two straight lines passing through the origin. Which two straight lines? Since *A* and *B* satisfy the equation of the original curve as well as the relation \(\begin{align}\frac{{px + qy}}{{ - r}} = 1,\end{align}\) *A* and *B* both satisfy the homogenized equation in (1).

What does this imply ? That (1) is the joint equation of *OA* and *OB*!

Go over this discussion again if you find this confusing. You must fully understand the described technique which will find very wide usage in subsequent chapters.

**Example - 24**

Find the joint equation of the straight lines passing through the origin *O* and the points of intersection of the line \(3x + 4y - 5 = 0\) and the curve \(2{x^2} + 3{y^2} = 5.\)

**Solution:** One approach is of course to explicitly determine the two points of intersection, say *A* and *B*, writing the equations of *OA* and *OB*, thereby obtaining the required joint equation. You are urged to do this as an exercise.

However, we’ll use the homogenizing technique just described:

The required equation can be obtained by first writing the line as

\[\left( {\frac{{3x + 4y}}{5}} \right) = 1\]

and then using this relation to homogenize the equation of the curve:

\[\begin{align} &2{x^2} + 3{y^2} = 5{\left( {\frac{{3x + 4y}}{5}} \right)^2}\\ \Rightarrow \qquad & 10{x^2} + 15{y^2} = 9{x^2} + 16{y^2} + 24xy\\ \Rightarrow \qquad & {x^2} - 24xy - {y^2} = 0 \end{align}\]

This is the required joint equation of *OA* and *OB*.

**Example - 25**

Find the value of \(m,\) if the lines joining the origin *O* to the points of intersection *A*, *B* of \(y = 1 + mx{\text{ and }}{x^2} + {y^2} = 1\) are perpendicular.

**Solution:** The joint equation of *OA* and *OB* is

\[\begin{align} &{x^2} + {y^2} = {(y - mx)^2}\\ \Rightarrow \qquad & (1 - {m^2}){x^2} + 2mxy = 0 \qquad \qquad \qquad ...(1) \end{align}\]

The condition for perpendicularity is

\[a+b=0\]

which when applied to (1) yields

\[\begin{align} &1 - {m^2} = 0\\ &m = \pm 1 \end{align}\]

This example was more or less trivial and a little knowledge of circles would have enabled you to solve this question without resorting to the homogenizing approach; however, the fact that this approach is very powerful in many cases will become apparent in later examples.

# TRY YOURSELF - II

**Q1. **Find the values(s) of *m* for which the following equation(s) represents a pair of straight lines:

(a) \(x^2 + \lambda xy - 2y^2 + 3y - 1 = 0\)

(b) \(4x^2 + 10xy + \lambda y^2 + 5x + 10y = 0\)

**Q2. **Find the angle of intersection of the straight lines given by the equation

\(3x^2 - 7xy + 2y^2 + 9x + 2y - 12 = 0\)

**Q3. **Show that the lines joining the origin to the points common to \(x^2 + hxy - y^2 + gx + fy = 0\) and \(fx - gy = \lambda \) are at right angles for all values of \(\lambda. \)

**Q4. **Find the angle between the lines given by \(x^2 - 2pxy + y^2 = 0\).

**Q5. **Prove that the lines joining the origin to the points of intersection of the line \(\begin{align}\frac{x}{h} + \frac{y}{k} \ne 2\end{align}\) with the curve \((x - h)^2 + (y - k)^2 = c^2 \), are perpendicular if \(h^2 + k^2 = c^2 \).

**Q6. **Find the joint equations of the straight lines passing through (1, 1) and parallel to the lines given by \(x^2 - 5xy + 4y^2 + x+2y - 2 = 0\).

**Q7. **Evaluate the point of intersection for the lines represented by the general equation \(ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\).

**Q8. **Find the joint equation of the images of the pair of lines \(ax^2 + 2hxy + by^2 = 0\) in the mirror \(y = 0\).

**Q9. **Find the joint equation of the angle bisectors of the lines given by \(x^2 + 2xy\sec \theta + y^2 = 0\).

**Q10. **If the pairs of straight lines

\(\begin{align} & ax^2 + 2hxy + by^2 = 0 \\ &a'x^2 + 2h'xy + b'y^2 = 0 \end{align}\)

have a line in common, show that

\(\begin{align}\left( {\frac{{ab'-a'b}}{2}}\right)^2=(ha'-h'a)(h'b-hb')\end{align}\)

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- Practice worksheets in and after class for conceptual clarity
- Personalized curriculum to keep up with school