Straight Lines

If the pair of straight lines $x^2 - 2 kxy - y^2 = 0$ bisect the angle between the pair of straight lines $x^2 - 2 \ell x y - y^2$, show that the later pair also bisects the angle between the former.

Comparing the equation $\;\;$ $x^2 - 2 \ell x y - y^2$ $\;\;\; \cdots \; (1)$

with the standard equation $\;\;$ $ax^2 = 2hxy + by^2 = 0$ $\;\;\; \cdots \; (2)$ $\;$

gives $\;\;$ $a = 1$, $\;\;\;$ $h = - \ell$, $\;\;\;$ $b = -1$

Equation of angle bisector of pair of lines given by equation $(2)$ is

$\dfrac{x^2 - y^2}{ab} = \dfrac{xy}{h}$ $\;\;\; \cdots \; (3)$

Substituting the values of $a$, $b$ and $h$ in equation $(3)$, the equation of angle bisector of the pair of lines given by equation $(1)$ is

$\dfrac{x^2 - y^2}{1 \times \left(-1\right)} = \dfrac{xy}{- \ell}$

i.e. $\ell x^2 - \ell y^2 = xy$ $\implies$ $\ell x^2 - xy - \ell y^2 = 0$ $\;\;\; \cdots \; (4)$

Given: The pair of straight lines $x^2 - 2kxy - y^2 = 0$ $\;\;\; \cdots \; (5)$

is the angle bisector the pair of lines given by equation $(1)$.

Since equations $(4)$ and $(5)$ represent the same angle bisector, comparing the like terms in both equations gives

$\dfrac{\ell}{1} = \dfrac{1}{2k} = \dfrac{\ell}{1}$ $\implies$ $k = \dfrac{1}{2 \ell}$ $\;\;\; \cdots \; (6)$

Comparing equation $(5)$ with the standard equation $\;\;$ $Ax^2 + 2 Hxy + By^2 = 0$

gives $\;\;$ $A = 1$, $\;\;\;$ $H = -k$, $\;\;\;$ $B = -1$

Equation of angle bisector of pair of lines given by equation $(5)$ is

$\dfrac{x^2 - y^2}{1 \times \left(-1\right)} = \dfrac{xy}{-k}$

i.e. $kx^2 - ky^2 = xy$ $\implies$ $k x^2 - xy - ky^2 = 0$ $\;\;\; \cdots \; (7)$

Substituting the value of $k$ from equation $(6)$ in equation $(7)$ gives

$\dfrac{x^2}{2 \ell} - xy - \dfrac{y^2}{2 \ell} = 0$

i.e. $x^2 - 2 \ell xy - y^2 = 0$ $\;\;\;$ which are pair of lines given by equation $(1)$.

$\therefore$ $\;$ The pair of equations given by $(1)$ bisect the angle between the pair of lines given by equation $(5)$.

Hence proved.

Straight Lines

Prove that one of the straight lines given by $ax^2 + 2hxy + by^2 = 0$ will bisect the angle between the coordinate axes if $\left(a + b\right)^2 = 4h^2$

Let $m_1$ and $m_2$ be the slopes of the lines given by the equation $\;\;$ $ax^2 + 2hxy + by^2 = 0$ $\;\;\; \cdots \; (1)$

The two lines given by equation $(1)$ are

$y - m_1 x = 0$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $y - m_2 x = 0$ $\;\;\; \cdots \; (2b)$

Now, $\;$ $m_1 + m_2 = \dfrac{-2h}{b}$ $\;\;\; \cdots \; (3a)$ $\;$ and $\;\;$ $m_1 m_2 = \dfrac{a}{b}$ $\;\;\; \cdots \; (3b)$

Let the line given by equation $(2a)$ bisect the coordinate axes.

Then the angle between the line and the coordinate axis (say, X axis) is $\dfrac{\pi}{4}$

$\therefore$ $\;$ Slope of line given by equation $(2a)$ is $m_1 = \tan \left(\dfrac{\pi}{4}\right) = 1$

Substituting $m_1 = 1$ in equation $(3a)$ gives

$1 + m_2 = \dfrac{-2h}{b}$ $\implies$ $m_2 = \dfrac{-2h}{b} - 1$ $\;\;\; \cdots \; (4a)$

Substituting $m_1 = 1$ in equation $(3b)$ gives

$m_2 = \dfrac{a}{b}$ $\;\;\; \cdots \; (4b)$

$\therefore$ $\;$ We have from equations $(4a)$ and $(4b)$,

$\dfrac{-2h}{b} - 1 = \dfrac{a}{b}$

i.e. $- 2h - b = a$ $\implies$ $a + b = - 2h$ $\implies$ $\left(a + b\right)^2 = 4 h^2$

Hence, one of the straight lines given by $ax^2 + 2hxy + by^2 = 0$ will bisect the angle between the coordinate axes if $\left(a + b\right)^2 = 4h^2$.

Straight Lines

Show that the equation $4x^2 + 4 xy + y^2 - 6x - 3y - 4 = 0$ represents a pair of parallel lines. Find the distance between them.

Comparing the given equation $\;$ $4x^2 + 4 xy + y^2 - 6x - 3y - 4 = 0$ $\;\;\; \cdots \; (1)$

with the standard equation $\;$ $ax^2 + 2 hxy + by^2 + 2 gx + 2 fy + c = 0$ $\;\;\; \cdots \; (2)$ $\;$ gives

$\begin{aligned} a = 4, & \hspace{1em} 2 f = -3 \implies f = \dfrac{-3}{2} \\\\ b = 1, & \hspace{1em} 2 g = -6 \implies g = -3 \\\\ c = -4, & \hspace{1em} 2 h = 4 \implies h = 2 \end{aligned}$

Two straight lines represented by the standard equation $(2)$ are parallel if $\;\;$ $af^2 = bg^2$

Now, $a f^2 = 4 \times \left(\dfrac{-3}{2}\right)^2 = 9$ $\;\;\; \cdots \; (3a)$

and $bg^2 = 1 \times \left(-3\right)^2 = 9$ $\;\;\; \cdots \; (3b)$

$\therefore$ $\;$ We have from equations $(3a)$ and $(3b)$ $\;$ $af^2 = bg^2$

$\implies$ Equation $(1)$ represents a pair of parallel lines.

Distance between the pair of parallel lines $= d = 2 \times \sqrt{\dfrac{g^2 - ac}{a \left(a + b\right)}}$

i.e. $d = 2 \times \sqrt{\dfrac{\left(-3\right)^2 - 4 \times \left(-4\right)}{4 \left(4 + 1\right)}} = 2 \times \sqrt{\dfrac{9 + 16}{4 \times 5}} = \sqrt{5}$

$\therefore$ $\;$ Distance between the pair of parallel lines represented by equation $(1)$ is $d = \sqrt{5}$ units.

Straight Lines

A $\triangle OPQ$ is formed by the pair of straight lines $x^2 - 4 xy + y^2 = 0$ and the line $PQ$. The equation of $PQ$ is $x + y - 2 = 0$. Find the equation of the median of $\triangle OPQ$ drawn from the origin.

$x^2 - 4xy + y^2 = 0$ $\;\;\; \cdots \; (1)$

represents a pair of straight lines passing through the origin $O \left(0,0\right)$.

Equation $(1)$ can be written as

$x^2 - 4xy + 4y^2 - 3y^2 = 0$

i.e. $\left(x - 2y\right)^2 - \left(\sqrt{3}y\right)^2 = 0$

i.e. $\left(x - 2y + \sqrt{3}y\right) \left(x - 2y - \sqrt{3}y\right) = 0$

$\therefore$ $\;$ Equation $(1)$ represents the pair of lines

$x - y \left(2 - \sqrt{3}\right) = 0$ and $x - y \left(2 + \sqrt{3}\right) = 0$

i.e. $x = y \left(2 - \sqrt{3}\right)$ $\;\;\; \cdots \; (2a)$ and

$x = y \left(2 + \sqrt{3}\right)$ $\;\;\; \cdots \; (2b)$

Equation of line $PQ$ is $\;$ $x + y - 2 = 0$ $\;\;\; \cdots \; (3)$

Solving equations $(2a)$ and $(3)$ simultaneously we have,

$y \left(2 - \sqrt{3}\right) + y = 2$ $\implies$ $y = \dfrac{2}{3 - \sqrt{3}}$ $\;\;\; \cdots \; (4a)$

Substituting the value of $y$ from equation $(4a)$ in equation $(2a)$ we get,

$x = \left(\dfrac{2}{3 - \sqrt{3}}\right) \times \left(2 - \sqrt{3}\right)$

i.e. $x = \dfrac{2 \times \left(2 - \sqrt{3}\right) \times \left(3 + \sqrt{3}\right)}{\left(3 - \sqrt{3}\right) \times \left(3 + \sqrt{3}\right)}$

i.e. $x = \dfrac{2 \times \left(6 + 2 \sqrt{3} - 3 \sqrt{3} - 3\right)}{9 - 3}$ $\implies$ $x = \dfrac{3 - \sqrt{3}}{3}$

$\therefore$ $\;$ The point of intersection of equations $(2a)$ and $(3)$ is $A \left(\dfrac{3 - \sqrt{3}}{3}, \dfrac{2}{3 - \sqrt{3}}\right)$

Solving equations $(2b)$ and $(3)$ simultaneously we have,

$y \left(2 + \sqrt{3}\right) + y = 2$ $\implies$ $y = \dfrac{2}{3 + \sqrt{3}}$ $\;\;\; \cdots \; (4b)$

Substituting the value of $y$ from equation $(4b)$ in equation $(2b)$ we get,

$x = \left(\dfrac{2}{3 + \sqrt{3}}\right) \times \left(2 + \sqrt{3}\right)$

i.e. $x = \dfrac{2 \times \left(2 + \sqrt{3}\right) \times \left(3 - \sqrt{3}\right)}{\left(3 + \sqrt{3}\right) \times \left(3 - \sqrt{3}\right)}$

i.e. $x = \dfrac{2 \times \left(6 - 2 \sqrt{3} + 3 \sqrt{3} - 3\right)}{9 - 3}$ $\implies$ $x = \dfrac{3 + \sqrt{3}}{3}$

$\therefore$ $\;$ The point of intersection of equations $(2b)$ and $(3)$ is $B \left(\dfrac{3 + \sqrt{3}}{3}, \dfrac{2}{3 + \sqrt{3}}\right)$

Midpoint of $AB$ is $\;$ $M = \left(\dfrac{\dfrac{3 - \sqrt{3} + 3 + \sqrt{3}}{3}}{2}, \dfrac{\dfrac{2}{3 - \sqrt{3}} + \dfrac{2}{3 + \sqrt{3}}}{2}\right)$

i.e. $M = \left(1, \dfrac{6 + 2 \sqrt{3} + 6 - 2 \sqrt{3}}{2 \times \left(3 - \sqrt{3}\right) \left(3 + \sqrt{3}\right)}\right)$

i.e. $M = \left(1, \dfrac{12}{2 \times 6}\right)$ $\implies$ $M = \left(1,1\right)$

$\therefore$ $\;$ Equation of median $MO$ is

$y - 0 = \left(\dfrac{1 - 0}{1 - 0}\right) \left(x - 0\right)$

$\therefore$ $\;$ The required equation of the median of $\triangle OPQ$ drawn from the origin is $\;$ $y = x$

Straight Lines

Find $\;$ $p$ $\;$ and $\;$ $q$ $\;$ if the equation $\;$ $6x^2 + 5xy - py^2 + 7x + qy - 5 = 0$ $\;$ represents a pair of perpendicular lines.

Comparing the equation $\hspace{1em}$ $6x^2 + 5xy - py^2 + 7x + qy - 5 = 0$ $\;\;\; \cdots \; (1)$

with the standard equation $\hspace{1em}$ $ax^2 + 2 hxy + by^2 + 2gx + 2 fy + c = 0$ $\;$ gives

$\begin{aligned} a = 6, & \hspace{1em} 2 f = q \implies f = \dfrac{q}{2} \\\\ b = - p, & \hspace{1em} 2 g = 7 \implies g = \dfrac{7}{2} \\\\ c = -5, & \hspace{1em} 2 h = 5 \implies h = \dfrac{5}{2} \end{aligned}$

Given: The two lines represented by equation $(1)$ are perpendicular to each other.

Now, a pair of lines are perpendicular when $\;\;\;$ $a + b = 0$

Substituting the values of $a$ and $b$ we have,

$6 - p = 0$ $\implies$ $p = 6$

Condition for equation $(1)$ to represent a pair of lines is: $\hspace{1em}$ $abc + 2 fgh - af^2 - bg^2 - ch^2 = 0$

Substituting the values of $a$, $b$, $c$, $f$, $g$ and $h$ we have,

$6 \times \left(-6\right) \times \left(-5\right) + 2 \times \left(\dfrac{q}{2}\right) \times \left(\dfrac{7}{2}\right) \times \left(\dfrac{5}{2}\right)$
$\hspace{3em}$ $- 6 \times \left(\dfrac{q}{2}\right)^2 - \left(-6\right) \times \left(\dfrac{7}{2}\right)^2 - \left(-5\right) \times \left(\dfrac{5}{2}\right)^2 = 0$

i.e. $180 + \dfrac{35q}{4} - \dfrac{6q^2}{4} + \dfrac{294}{4} + \dfrac{125}{4} = 0$

i.e. $6 q^2 - 35 q - 1139 = 0$

i.e. $q = \dfrac{35 \pm \sqrt{35^2 + 4 \times 6 \times 1139}}{2 \times 6}$

i.e. $q = \dfrac{35 \pm \sqrt{28561}}{12} = \dfrac{35 \pm 169}{12}$

$\implies$ $q = 17$ $\;$ or $\;$ $q = \dfrac{-67}{6}$

Straight Lines

If the slope of one of the straight lines $ax^2 + 2hxy + by^2 = 0$ is three times the other, then show that $3h^2 = 4ab$.

Let $m_1$ and $m_2$ be the slopes of the two lines.

Given: $\hspace{1em}$ $m_1 = 3m_2$ $\;\;\; \cdots \; (1)$

For a pair of lines,

$m_1 + m_2 = \dfrac{-2h}{b}$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $m_1 m_2 = \dfrac{a}{b}$ $\;\;\; \cdots \; (2b)$

$\begin{aligned} \text{Now, } \left(m_1 - m_2\right)^2 & = \left(m_1 + m_2\right)^2 - 4 m_1 m_2 \\\\ & = \dfrac{4h^2}{b^2} - \dfrac{4a}{b} \hspace{1em} \left[\text{from equations }(2a) \text{ and } (2b)\right] \\\\ & = \dfrac{4 \left(h^2 - ab\right)}{b^2} \end{aligned}$

i.e. $m_1 - m_2 = \pm \dfrac{2 \sqrt{h^2 - ab}}{b}$

Consider the case $\hspace{1em}$ $m_1 - m_2 = \dfrac{2 \sqrt{h^2 - ab}}{b}$ $\;\;\; \cdots \; (2c)$

Adding equations $(2a)$ and $(2c)$ gives

$2 m_1 = \dfrac{-2h + 2 \sqrt{h^2 - ab}}{b}$ $\implies$ $m_1 = \dfrac{-h + \sqrt{h^2 - ab}}{b}$ $\;\;\; \cdots \; (3a)$

Substituting the value of $m_1$ from equation $(3a)$ in equation $(2a)$ gives

$m_2 = \dfrac{-2h}{b} + \dfrac{h - \sqrt{h^2 -ab}}{b}$ $\implies$ $m_2 = \dfrac{-h - \sqrt{h^2 - ab}}{b}$ $\;\;\; \cdots \; (3b)$

$\therefore$ $\;$ In view of equations $(1)$, $(3a)$ and $(3b)$ we have

$\dfrac{-h + \sqrt{h^2 - ab}}{b} = - 3 \times \left(\dfrac{h + \sqrt{h^2 - ab}}{b}\right)$

i.e. $- h + \sqrt{h^2 - ab} = - 3h - 3 \sqrt{h^2 - ab}$

i.e. $4 \sqrt{h^2 - ab} = - 2h$ $\implies$ $2\sqrt{h^2 - ab} = - h$ $\;\;\; \cdots \; (4)$

Squaring both sides of equation $(4)$ we have,

$4 \left(h^2 - ab\right) = h^2$

i.e. $4 h^2 - 4ab = h^2$ $\implies$ $3h^2 = 4ab$ $\;\;$ Hence proved.

Note:
The same result can be obtained by taking $\;$ $m_1 - m_2 = \dfrac{-2 \sqrt{h^2 - ab}}{b}$ $\;$ in equation $(2c)$.
The values of $m_1$ and $m_2$ will be interchanged.

Straight Lines

Find the equation of the pair of straight lines passing through the point $\left(1,3\right)$ and perpendicular to the lines $2x - 3y + 1 = 0$ and $5x + y - 3 = 0$

The given lines are

$2x - 3y + 1 = 0$ $\;$ i.e. $y = \dfrac{2}{3}x + \dfrac{1}{3}$ $\;\;\; \cdots \; (1)$

and $5x + y - 3 = 0$ $\;$ i.e. $y = - 5 x + 3$ $\;\;\; \cdots \; (2)$

$\therefore$ $\;$ Slope of line given by equation $(1)$ is $= m_1 = \dfrac{2}{3}$

and slope of line given by equation $(2)$ is $= m_2 = -5$

Let the slopes of the required pair of lines be $m_3$ and $m_4$.

Since the required lines are perpendicular to the given lines, their slopes are

$m_3 = \dfrac{-1}{m_1} = \dfrac{-3}{2}$ $\;$ and $\;$ $m_4 = \dfrac{-1}{m_2} = \dfrac{1}{5}$

Now, the required lines pass through the point $\left(1,3\right)$.

$\therefore$ $\;$ The equations of the individual lines are

$\left(y - 3\right) = \dfrac{-3}{2} \left(x - 1\right)$ $\;$ and $\;$ $\left(y - 3\right) = \dfrac{1}{5} \left(x - 1\right)$

i.e. $2y - 6 = - 3x + 3$ $\;$ and $\;$ $5y - 15 = x - 1$

i.e. $3x + 2 y - 9 = 0$ $\;$ and $\;$ $x - 5y + 14 = 0$

$\therefore$ $\;$ The combined equation of the required lines is

$\left(3x + 2y - 9\right) \left(x - 5y + 14\right) = 0$

i.e. $3x^2 - 15xy + 42x + 2 xy - 10 y^2 + 28 y - 9x + 45 y - 126 = 0$

i.e. $3x^2 - 13xy - 10y^2 + 33x + 73y - 126 = 0$ $\;\;\; \cdots \; (3)$

Equation $(3)$ gives the required equation of the pair of straight lines.