**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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