# Current Affairs 11th Class

## Notes - Mathematics Olympiads - Pair Sraight Line

Category : 11th Class

Pair and Straight Line

Key Points to Remember

A hemogenous of equation of second degree of the form $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ represents the pair of straight lines which passes through the origin.

(a)        If the lines are real and distinct then

${{h}^{2}}>ab$

(b)        If the lines are real and coincidents if

${{h}^{2}}=ab.$

(c)        If the lines are imaginary then ${{h}^{2}}<ab.$

Let $y={{m}_{1}}x\to (1)$ and $y={{m}_{2}}x\to (2)$ are two lines which are passing through the origin.

Then $(y-{{m}_{1}}x)(y-{{m}_{2}}x)\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}$

${{y}^{2}}-({{m}_{1}}+{{m}_{2}})xy+{{m}_{1}}{{m}_{2}}{{x}^{2}}={{y}^{2}}+\frac{2h}{b}xy+\frac{a}{b}{{x}^{2}}$

Equation the coefficient of same variable of the$\frac{2h}{b}=({{m}_{1}}+{{m}_{2}})$we have, $({{m}_{1}}+{{m}_{2}})=\frac{2h}{b}$ & ${{m}_{1}}.{{m}_{2}}=\frac{a}{b}$

• Angle between the pair of straight lines

Let q be the angle between the two given pair of straight line $a{{x}^{2}}+2hxy+b{{y}^{2}}=0,$ which are passing through origin is written as

$\tan \theta =\pm \frac{2\sqrt{{{h}^{2}}}-ab}{a+b}$

for acute angle,

$\tan \theta =\left| \frac{2\sqrt{{{h}^{2}}}-ab}{a+b} \right|$

In cosine form

$\cos \theta =\frac{a+b}{\sqrt{{{(a+b)}^{2}}+4{{h}^{2}}}}$

Note:    If two lines are coincident

i.e.       $\theta =0{}^\circ$ or $180{}^\circ$

then $\frac{2\sqrt{{{h}^{2}}-ab}}{a+b}=0$

Hence ${{h}^{2}}=ab\Rightarrow {{h}^{2}}=a$

(b)        If two lines are perpendicular

i.e.        $\theta =90{}^\circ ,$ i.e. $\tan 90{}^\circ =\infty$

then Loosly, it can be written

$\frac{2\sqrt{{{h}^{2}}-ab}}{a+b}=\infty =\frac{1}{0}$

$a+b=0$ i.e.

$\therefore$Sum of the coefficient of ${{x}^{2}}$ and ${{y}^{2}}$ respectively is zero.

(c)        The equation of pair of straight lines passing through the origin and perpendicular to the given equation of pair of straight lines$/a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ is written as $b{{x}^{2}}-2hxy+a{{x}^{2}}=0$

The general equation of second degree in x and y be

$a{{x}^{2}}+b{{y}^{2}}+2hxy+2gx+2fy+c=0$ …….(i)

represents a pair of straight lines iff

$abc+2fgx-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$ (a)        If $\Delta \ne 0$ & ${{h}^{2}}-ab\ge 0$ then this general equation of second degree in x & y represents the equation of hyperbola.

(b)        If $\Delta \ne 0$ & ${{h}^{2}}-ab\le 0$ then this pair of lines represent the equation ellipse.

(c)        If $\Delta \ne 0$ & ${{h}^{2}}-ab=0$  then this pair of lines represent the equation of parabola

(d)        If $a=b=1$ & $h=0$ then this represents the equation of circle.

• Some points to remember

(a)        Angle between the lines:-

If $\theta$ is the angle between the two lines:

$a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$  …… (1)

then $\tan \theta =\pm \left| \frac{2\sqrt{{{h}^{2}}-ab}}{a+b} \right|$

$\Rightarrow \,\,\,\theta ={{\tan }^{-1}}\left| \frac{2\sqrt{{{h}^{2}}-ab}}{a+b} \right|$

Note:    It is the same form which is obtained by the pair of straight lines passing through the origin.

(b)        Point of intersection of lines:

The point of intersection of line (1) is obtained by the partially differentiation of

$\text{f}\equiv a{{x}^{\text{2}}}+b{{y}^{2}}+2hxy+2gx+2\text{f}y+c=0$

w.r.t x and y respective & making

$\frac{\partial \text{f}}{\partial x}=0$                            ….. (i)

&  $\frac{\partial \text{f}}{\partial y}=0$                        ….. (ii)

Then solve these equations and hence, we will obtain the value ofx and y, which are written as

$(x,y)=\left( \frac{bg-h}{{{h}^{2}}-ab},\frac{a\text{f-gh}}{{{h}^{2}}-ab} \right)$

Here,

$\frac{\partial \text{f}}{\partial x}=2ax+2hy+2g=0$ & $\frac{\partial \text{f}}{\partial x}=2by+2hx+2\text{f}=0$

If the two lines represented by equation (1) be parallel if ${{h}^{2}}=ab$ & $b{{g}^{2}}=a{{\text{f}}^{\text{2}}}$

• Distance between the parallel lines: If the two lines represented by (1) are parallel then the distance between these two lines are written as

$d=2\sqrt{\frac{{{g}^{2}}-ac}{a(a+b)}}$

The equation to the pair of lines through the origin and forming an equiletral triangle with line

$ax+by+c=0$is given by ${{(ax+by)}^{2}}-3{{(ay-bx)}^{2}}=0$

Also, the area of equiletral triangle is

$=\frac{{{c}^{2}}}{\sqrt{3}({{a}^{2}}+{{b}^{2}})}$

The line $a{{x}^{2}}+b{{y}^{2}}+2hxy=0$ & $\ell \times +my+n$

= 0 'form issosectes triangle if

$\frac{{{\ell }^{2}}-{{m}^{2}}}{\ell m}=\frac{a-b}{h}$

Area of the triangle = $\left| \frac{{{n}^{2}}\sqrt{{{h}^{2}}-ab}}{a{{m}^{2}}-2h\ell m+b{{\ell }^{2}}} \right|$

• Equation of the bisector of the angle between the pair of straight line $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ can be written as

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

• The equation of the lines joining the origin to the point of intersection of a given line & a given curve.

Let the straight line is

$y=mx+c$

$\Rightarrow y-mx=c\Rightarrow \frac{y-mx}{c}=1$

& The equation of the curve is

$a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2\text{f}y+K=0$ Let the line cuts the curve at point P & Q then the joint equation of OP & OQ be written as

$a{{x}^{2}}+2hxy+b{{y}^{2}}+(2gx+2\text{f}y)+K{{\left( \frac{y-mx}{c} \right)}^{2}}=0$

From this method we can make any pair of equation to be homogenous with the help of the equation of the line.

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