Pair of Straight Lines in Two Dimensional Geometry

Pair of Straight Lines in Two Dimensional Geometry

 

Some Formulae on Pair of Straight Lines

Here are some important formulas on Pair of Straight Lines.

1. A second-degree homogeneous equation of the form ax2 + 2hxy + by2 = 0 represents a pair of straight lines through the origin if h2 ≥ ab.

2. The angle between the pair of straight lines ax2 + 2hxy + by2 = 0 are \tan \theta =\frac{2\sqrt{{{h}^{2}}-ab}}{a+b}

3. If the two straight lines be coincident, then θ = 0 and hence {{h}^{2}}-ab=0

4. If the lines be at right angles, then θ = 90° and hence a + b =0

5. The equation of bisectors of the angles between the lines ax2 + 2hxy + by2 = 0 are

\[\frac{{{x}^{2}}-{{y}^{2}}}{a-b}=\frac{xy}{h}\]

6. The equation a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0 represents a pair of straight lines if

m1 1
 Example 01

Find the value of λ, so that the equation 6{{x}^{2}}+xy+\lambda {{y}^{2}}+2x-31y-20=0 may represents a pair of straight lines.

Solution:

We have 6{{x}^{2}}+xy+\lambda {{y}^{2}}+2x-31y-20=0
Comparing it with a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0 we get,
a = 6, h = \frac{1}{2}, b = λ, g = 1, f = -\frac{31}{2}, c = –20
If the curve represents a pair of straight lines then,

m2 1

\[\Rightarrow 6\left( -20\times \lambda -\frac{31}{2}\times \frac{31}{2} \right)-\frac{1}{2}\left( -20\times \frac{1}{2}-\frac{31}{2}\times 1 \right)+1\left( -\frac{31}{2}\times \frac{1}{2}-1\times \lambda  \right)=0\]

\[\Rightarrow \lambda +12=0\]

\[\therefore \,\,\lambda =-12\]

 Example 02

Show that the equation {{x}^{2}}+3xy+2{{y}^{2}}=0 represents a pair of straight lines. Find these straight lines and the angle between them.

Solution:

The given equation is a homogeneous equation of second degree. So, it represents a pair of straight lines through the origin.
The given equation {{x}^{2}}+3xy+2{{y}^{2}}=0 can be written as \left( x+y \right)\left( x+2y \right)=0.
Hence the two straight lines represented by the given equation are

\[x+y=0\]

\[x+2y=0\]

Comparing {{x}^{2}}+3xy+2{{y}^{2}}=0 with a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0 we get,

\[a=1, h= \frac{3}{2}, b=2, g=0, f=0, c=0\]

The angle between them is

\[{{\tan }^{-1}}\left\{ \frac{2\sqrt{{{h}^{2}}-ab}}{a+b} \right\}={{\tan }^{-1}}\left\{ \frac{2\sqrt{{{\left( \frac{3}{2} \right)}^{2}}-1\times 2}}{1+2} \right\}={{\tan }^{-1}}\left( \frac{1}{3} \right)\]

 Example 03

Show that the equation to the pair of straight lines through the origin perpendicular to the pair of straight lines a{{x}^{2}}+2hxy+b{{y}^{2}}=0 is b{{x}^{2}}-2hxy+a{{y}^{2}}=0.

Solution:

Let y={{m}_{1}}x and y={{m}_{2}}x be the straight lines represented by the equation a{{x}^{2}}+2hxy+b{{y}^{2}}=0, so that

\[{{m}_{1}}+{{m}_{2}}=-\frac{2h}{b},\,\,{{m}_{1}}{{m}_{2}}=\frac{a}{b}………\left( 1 \right)\]

The straight lines perpendicular to y={{m}_{1}}x and y={{m}_{2}}x passing through the origin are y=-\frac{1}{{{m}_{1}}}x and y=-\frac{1}{{{m}_{2}}}x respectively.
Hence the required pair of straight lines is

\[\left( y+\frac{1}{{{m}_{1}}}x \right)\left( y+\frac{1}{{{m}_{2}}}x \right)=0\]

\[\Rightarrow \left( {{m}_{1}}y+x \right)\left( {{m}_{2}}y+x \right)=0\]

\[\Rightarrow {{m}_{1}}{{m}_{2}}{{y}^{2}}+\left( {{m}_{1}}+{{m}_{2}} \right)xy+{{x}^{2}}=0\]

\[\Rightarrow \frac{a}{b}{{y}^{2}}-\frac{2h}{b}xy+{{x}^{2}}=0\,\,\left[ By\,\,\left( 1 \right) \right]\]

\[\Rightarrow b{{x}^{2}}-2hxy+a{{y}^{2}}=0\]

Hence the equation to the pair of straight lines through the origin perpendicular to the pair of straight lines a{{x}^{2}}+2hxy+b{{y}^{2}}=0 is b{{x}^{2}}-2hxy+a{{y}^{2}}=0.

 Example 04

Prove that the equation to the straight lines through the origin each of which makes an angle α with the straight lines y = x is {{x}^{2}}-2xy\sec 2\alpha +{{y}^{2}}=0.

Solution:

The slope of the straight line y = x is 1=\tan \frac{\pi }{4}.
So, the straight line makes an angle \frac{\pi }{4} with x-axis.
Hence the required straight lines makes angle \left( \frac{\pi }{4}-\alpha \right) and \left( \frac{\pi }{4}+\alpha \right) with x-axis.
As they pass through the origin, their equations are

\[y=x\tan \left( \frac{\pi }{4}-\alpha  \right);\,\,y=x\tan \left( \frac{\pi }{4}+\alpha  \right)\]

\[i.e.,\,y-\frac{1-\tan \alpha }{1+\tan \alpha }x=0\,\,and\,\,y-\frac{1+\tan \alpha }{1-\tan \alpha }x=0\]

Therefore the required equation of the pair of straight lines is

\[\left( y-\frac{1-\tan \alpha }{1+\tan \alpha }x \right)\left( y-\frac{1+\tan \alpha }{1-\tan \alpha }x \right)=0\]

\[\Rightarrow {{y}^{2}}-\left( \frac{1-\tan \alpha }{1+\tan \alpha }+\frac{1+\tan \alpha }{1-\tan \alpha } \right)xy+{{x}^{2}}=0\]

\[\Rightarrow {{y}^{2}}-\frac{2\left( 1+{{\tan }^{2}}\alpha  \right)}{1-{{\tan }^{2}}\alpha }xy+{{x}^{2}}=0\]

\[\Rightarrow {{x}^{2}}-\frac{2}{\cos 2\alpha }xy+{{y}^{2}}=0\,\,\left[ \because \frac{1-{{\tan }^{2}}\alpha }{1-{{\tan }^{2}}\alpha }=\cos 2\alpha  \right]\]

\[\Rightarrow {{x}^{2}}-2xy\sec 2\alpha +{{y}^{2}}=0\]

 Example 05

If the pair of straight lines {{x}^{2}}-2pxy-{{y}^{2}}=0 and {{x}^{2}}-2qxy-{{y}^{2}}=0 be such that each pair bisects the angles between the other pair, then prove that pq = –1.

Solution:

The equation of the bisectors of the angles between the pair of straight lines {{x}^{2}}-2pxy-{{y}^{2}}=0 is

\[\frac{{{x}^{2}}-{{y}^{2}}}{1-\left( -1 \right)}=\frac{xy}{-p}\]

\[\Rightarrow \frac{{{x}^{2}}-{{y}^{2}}}{2}=\frac{xy}{-p}\]

\[\Rightarrow p{{x}^{2}}-p{{y}^{2}}=-2xy\]

\[\Rightarrow p{{x}^{2}}+2xy-p{{y}^{2}}=0………\left( 1 \right)\]

But, by the condition of the problem, (1) is identical with {{x}^{2}}-2qxy-{{y}^{2}}=0. Comparing, we have

\[\frac{p}{1}=\frac{1}{-q}=\frac{p}{1}\]

\[\therefore \,\,pq=-1\,\,\left[ \text{Proved} \right]\]

 Example 06

Show that the angle between one of the straight lines given by a{{x}^{2}}+2hxy+b{{y}^{2}}=0 and one of the straight lines given by a{{x}^{2}}+2hxy+b{{y}^{2}}+\lambda \left( {{x}^{2}}+{{y}^{2}} \right)=0 is equal to the angle between the other two straight lines of the system.

Solution:

\[We have, a{{x}^{2}}+2hxy+b{{y}^{2}}+\lambda \left( {{x}^{2}}+{{y}^{2}} \right)=0\]

\[i.e.,\left( a+\lambda  \right){{x}^{2}}+2hxy+\left( b+\lambda  \right){{y}^{2}}=0\]

The equation of the bisectors of the angles between the straight lines a{{x}^{2}}+2hxy+b{{y}^{2}}=0 is

\[\frac{{{x}^{2}}-{{y}^{2}}}{a-b}=\frac{xy}{h}\]

Also, the equation of the bisectors of the angles between the straight lines i.e.,\left( a+\lambda  \right){{x}^{2}}+2hxy+\left( b+\lambda  \right){{y}^{2}}=0 is

\[\frac{{{x}^{2}}-{{y}^{2}}}{\left( a+\lambda  \right)-\left( b+\lambda  \right)}=\frac{xy}{h}\]

\[\Rightarrow \frac{{{x}^{2}}-{{y}^{2}}}{a-b}=\frac{xy}{h}\]

Thus the two pairs have the same bisectors.

 Example 07

Find the condition that one of the straight lines given by a{{x}^{2}}+2hxy+b{{y}^{2}}=0 may coincide with one of the straight lines given by a'{{x}^{2}}+2h'xy+b'{{y}^{2}}=0.

Solution:

Let the two pairs have a common straight line y=mx.
Then substituting mx of y in the two equations and cancelling x2, we have

\[a+2hm+b+{{m}^{2}}=0\]

\[a’+2h’m+b'{{m}^{2}}=0\]

Therefore by cross-multiplication, we get

\[\frac{{{m}^{2}}}{ah’-a’h}=\frac{2m}{ba’-b’a}=\frac{1}{hb’-h’b}\]

\[\Rightarrow \frac{{{m}^{2}}}{m}=\frac{2\left( ah’-a’h \right)}{ba’-b’a};\,m=\frac{ba’-b’a}{2\left( hb’-h’b \right)}\]

\[\Rightarrow m=\frac{2\left( ah’-a’h \right)}{ba’-b’a};\,m=\frac{ba’-b’a}{2\left( hb’-h’b \right)}\]

\[\therefore \frac{2\left( ah’-a’h \right)}{ba’-b’a}=\frac{ba’-b’a}{2\left( hb’-h’b \right)}\]

\[\Rightarrow 4\left( ah’-a’h \right)\left( hb’-h’b \right)={{\left( ba’-b’a \right)}^{2}}\]

This is the required condition.

 Example 08

The gradient of one of the straight lines of  a{{x}^{2}}+2hxy+b{{y}^{2}}=0 is twice that of the other. Show that 8{{h}^{2}}=9ab.

Solution:

Let the two straight lines be y={{m}_{1}}x and y={{m}_{2}}x
Then substituting mx of y in the two equations and canceling x2, we have

\[\therefore \,2{{m}_{2}}+{{m}_{2}}=-\frac{2h}{b}\,\,and\,2{{m}_{2}}\times {{m}_{2}}=\frac{a}{b}\,\]

\[\Rightarrow 3{{m}_{2}}=-\frac{2h}{b}\,\,and\,2m_{2}^{2}=\frac{a}{b}\,\]

\[\Rightarrow {{m}_{2}}=-\frac{2h}{3b}\,\,and\,m_{2}^{2}=\frac{a}{2b}\,\]

\[\therefore \,{{\left( -\frac{2h}{3b} \right)}^{2}}=\frac{a}{2b}\,\]

\[\Rightarrow \frac{4{{h}^{2}}}{9{{b}^{2}}}=\frac{a}{2b}\,\]

\[\therefore \,\,8{{h}^{2}}=9ab\,\,\left[ \text{Proved} \right]\]

 Example 09

Prove that the product of the perpendiculars from the point (x1, y1) on the straight lines a{{x}^{2}}+2hxy+b{{y}^{2}}=0 is \frac{ax_{1}^{2}+2h{{x}_{1}}{{y}_{1}}+by_{1}^{2}}{\sqrt{{{\left( a-b \right)}^{2}}+4{{h}^{2}}}}.

Solution:

Let the equation a{{x}^{2}}+2hxy+b{{y}^{2}}=0 represent two straight lines OA and OB passing through the origin, whose equations are

\[y-{{m}_{1}}x=0………\left( 1 \right)\]

\[y-{{m}_{2}}x=0………\left( 2 \right)\]

So that,

\[{{m}_{1}}+{{m}_{2}}=-\frac{2h}{b},\,\,{{m}_{1}}{{m}_{2}}=\frac{a}{b}………\left( 3 \right)\]

PL = p1 = Length of the perpendicular from P(x1, y1) upon OA = \left| \frac{{{y}_{1}}-{{m}_{1}}{{x}_{1}}}{\sqrt{1+m_{1}^{2}}} \right|
And PM = p2 = Length of the perpendicular from P(x1, y1) upon OB = \left| \frac{{{y}_{1}}-{{m}_{2}}{{x}_{1}}}{\sqrt{1+m_{2}^{2}}} \right|

Pair of Straight Lines Q9

Hence the product of perpendiculars = p1 × p2 

\[\left| \frac{\left( {{y}_{1}}-{{m}_{1}}{{x}_{1}} \right)\left( {{y}_{1}}-{{m}_{2}}{{x}_{1}} \right)}{\sqrt{1+m_{1}^{2}}\sqrt{1+m_{2}^{2}}} \right|\]

\[=\left| \frac{y_{1}^{2}-\left( {{m}_{1}}+{{m}_{2}} \right){{x}_{1}}{{y}_{1}}+{{m}_{1}}{{m}_{2}}x_{1}^{2}}{\sqrt{1+m_{1}^{2}+m_{2}^{2}+{{\left( {{m}_{1}}{{m}_{2}} \right)}^{2}}}} \right|\]

\[=\left| \frac{y_{1}^{2}-\left( {{m}_{1}}+{{m}_{2}} \right){{x}_{1}}{{y}_{1}}+{{m}_{1}}{{m}_{2}}x_{1}^{2}}{\sqrt{1+{{\left( {{m}_{1}}+{{m}_{2}} \right)}^{2}}-2{{m}_{1}}{{m}_{2}}+{{\left( {{m}_{1}}{{m}_{2}} \right)}^{2}}}} \right|\]

\[=\left| \frac{y_{1}^{2}-\left( -\frac{2h}{b} \right){{x}_{1}}{{y}_{1}}+\left( \frac{a}{b} \right)x_{1}^{2}}{\sqrt{1+{{\left( -\frac{2h}{b} \right)}^{2}}-2\frac{a}{b}+{{\left( \frac{a}{b} \right)}^{2}}}} \right|\]

\[=\frac{ax_{1}^{2}+2h{{x}_{1}}{{y}_{1}}+by_{1}^{2}}{\sqrt{{{\left( a-b \right)}^{2}}+4{{h}^{2}}}}\,\,\left[ \text{Proved} \right]\]

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