# Show that 16x^2+24xy+py^2+24x+18y-5=0 represents a pair of parallel straight lines and find the distance between them.?

Dec 10, 2017

See below.

#### Explanation:

Let one of the lines be described as

${L}_{1} \to a x + b y + c = 0$

now, a parallel to ${L}_{1}$ can be denoted as

${L}_{2} \to \lambda a x + \lambda b y + d = 0$

Now equating

$16 {x}^{2} + 24 x y + p {y}^{2} + 24 x + 18 y - 5 = \left(a x + b y + c\right) \left(\lambda a x + \lambda b y + d\right)$

after grouping variables we have

{(c d =-5), (b d + b c lambda = 18), (b^2 lambda = p), (a d +a c lambda = 24), (2 a b lambda = 24), ( a^2 lambda = 16):}

Solving we have a set of solutions but we will focus only one
$a = \frac{4}{\sqrt{\lambda}} , b = \frac{3}{\sqrt{\lambda}} , c = \frac{3 + \sqrt{14}}{\sqrt{\lambda}} , d = \left(3 - \sqrt{14}\right) \lambda , p = 9$

so making $\lambda = 1$

$\left(\begin{matrix}a = 4 \\ b = 3 \\ c = 3 + \sqrt{14} \\ d = 3 - \sqrt{14} \\ p = 9\end{matrix}\right)$

The distance calculus between ${L}_{1}$ and ${L}_{2}$ is left as an exercise to the reader.

NOTE:

Considering ${p}_{1} \in {L}_{1}$ and ${p}_{2} \in {L}_{2}$, the distance between ${L}_{1}$ and ${L}_{2}$ can be computed as

$\left\mid \left\langle{p}_{2} - {p}_{1} , \hat{v}\right\rangle \right\mid = d$ where $\hat{v} = \frac{\left\{b , - a\right\}}{\sqrt{{a}^{2} + {b}^{2}}}$