Straight Lines and Pair of Straight Lines

The $x$-intercept is $-\frac{r}{p}=-\frac{8}{2}=-4$. and the $y$-intercept is $\frac{-r}{q}=\frac{-8}{4}=-2$

$ \begin{aligned} \text { Thus, the area of the triangle } & =\frac{1}{2} \times x_{\text {intercept }} \times y_{\text {intercept }} \\ & =\frac{1}{2} \times 4 \times 2=4 \end{aligned} $

$ \begin{aligned} &\text { The equations of the four lines are }\\ &\begin{aligned} & a x+b y+c=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & a x-b y+c=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & a x+b y-c=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ & a x-b y-c=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv) \end{aligned} \end{aligned} $

Clearly, lines (i) and (iii) are parallel. Also, lines (ii) and (iv) are also parallel.

So, the given lines form a parallelogram whose vertices are $A\left(\frac{c}{a}, 0\right), B\left(0, \frac{c}{b}\right), C\left(\frac{-c}{a}, 0\right)$ and $D\left(0, \frac{-c}{b}\right)$.

Clearly, diagonal $A C$ and $B D$ of this 11 gm are at right angle to each other. So, the given lines form a rhombus $A B C D$.

$ \begin{aligned} \therefore \text { Area of rhombus } A B C D & =\frac{1}{2}(A C \times B D) \\ & =\frac{1}{2}\left(\frac{2 c}{a} \times \frac{2 c}{b}\right)=\frac{2 c^2}{a b} \text { sq. units } \end{aligned} $

The equation $2 h x y+2 g x+2 f y+c=0$ will represents a pair of straight lines if

$ \begin{aligned} & 0+2 f g h-0 \times f^2-0 \times g^2-c h^2=0 \\ & \Rightarrow \quad 2 f g=c h \\ & \Rightarrow \quad c=\frac{2 f g}{h} \\ & \quad 2 h x y+2 g x+2 f y+\frac{2 f g}{h}=0 \\ & \Rightarrow h^2 x y+g h x+f h y+f g=0 \\ & \Rightarrow h y(h x+f)+g(h x+f)=0 \end{aligned} $

$ \Rightarrow(h x+f)(h y+g)=0 $

Thus, the equations of lines represented by the given equation are $h x+f=0$ and $h y+g=0$.

These two lines and the coordinate axes form a rectangle $O A B C$.

∴ Area of rectangle $O A B C=O A \times O C$

$ =\left|\frac{-f}{h}\right| \times\left|\frac{-g}{h}\right|=\frac{|f g|}{h^2} $

Given $ a, c, b $ are in GP

$ \Rightarrow c^{2}=a b $

According to question,

$ \therefore $ Area of $ \triangle O A B=\left|\frac{1}{2} \times O A \times O B\right| $

$ \begin{array}{r} =\left|\frac{1}{2} \times\left(\frac{-c}{b}\right) \times\left(\frac{-c}{a}\right)\right| \\ =\frac{1}{2} \times \frac{c^{2}}{a b}=\frac{1}{2} \text { [by Eq. (i)] } \end{array} $

Equations of the bisectors of the angles between the given straight lines are given by

${{3x + 4y + 7} \over {\sqrt {9 + 16} }} = \pm {{12x + 5y - 8} \over {\sqrt {144 + 25} }}$

$ \Rightarrow 13(3x + 4y + 7) = \pm 5(12x + 5y - 8)$

$ \Rightarrow 39x + 52y + 91 = \pm (60x + 25y - 40)$

Taking positive signs,

$39x + 52y + 91 = 60x + 25y - 40$

$ \Rightarrow - 21x + 27y + 131 = 0$

$ \Rightarrow 21x - 27y - 131 = 0$

Taking negative signs,

$(39x + 52y + 91) = - (60x + 25y - 40)$

$ \Rightarrow 99x + 77y + 51 = 0$

A line passing through (1, 2) and parallel to the line 3x $-$ y = 10 is

3(x $-$ 1) $-$ (y $-$ 2) = 0

$\Rightarrow$ 3x $-$ 3 $-$ y + 2 = 0 $\Rightarrow$ 3x $-$ y $-$ 1 = 0

The intersection point of line

x + y + 5 = 0 ..... (i)

and 3x $-$ y $-$ 1 = 0 ..... (ii)

are ($-$1, $-$4)

$\therefore$ Distance from the line x + y + 5 to the point (1, 2) measured parallel to the line 3x $-$ y = 10 is

$\sqrt {{{( - 1 - 1)}^2} + {{( - 4 - 2)}^2}} $

$\sqrt {4 + 36} = \sqrt {40} = 2\sqrt {10} $