Question

The distance between the two lines represented by the equation $9{x^2} - 24xy + {16^2} - 12x + 16y - 12 = 0$
A) 5
B) 8
C) $\dfrac{8}{5}$
D) $\dfrac{5}{8}$

Answer 1

Hint : We can first find the equations of the two parallel lines that are represented by the given equation and then find the distance between the two using distance formula for parallel lines given as:
$d = \left| {\dfrac{{{c_2} - {c_1}}}{{\sqrt {{a^2} + {b^2}} }}} \right|$ where,
d = distance between parallel lines
${c_1}$ = constant of the first equation of line obtained
${c_2}$ = constant of the second equation of line obtained
a and b = coefficients of x and y respectively ( will be same for both the lines )

Complete step-by-step answer :
The given equation of pair of parallel pair of lines is:
$9{x^2} - 24xy + {16^2} - 12x + 16y - 12 = 0$
For finding the equations of respective pair lines, we can write the given equation as:
${\left( {3x} \right)^2} - 2(3x)(4y) + {\left( 4 \right)^2} - 12x + 16y - 12 = 0$
Writing the first part of the equation in terms of square and the second by taking 4 as common.
$\Rightarrow {\left( {3x - 4y} \right)^2} - 4(3x - 4y) - 12 = 0$
Now, Let ( 3x – 4y ) be t ( to make calculations easier ), the equation becomes:
$\Rightarrow {t^2} - 4t - 12 = 0$ , factorizing 4, we get:
$\Rightarrow {t^2} - 6t + 2t - 12 = 0 \\\ \Rightarrow t(t - 6) + 2(t - 6) = 0 \\\ \Rightarrow (t + 2)(t - 6) = 0 \\\$
From here the values of t obtained are:
t + 2 = 0 => t = -2 ____ (1)
t – 6 = 0 => t = 6 ____ (2)
The value of t is 3x – 4y, from (1) and (2), we have:
$\Rightarrow$ 3x – 4y = - 2
$\Rightarrow$ 3x – 4y + 2 = 0
$\Rightarrow$ 3x – 4y = 6
$\Rightarrow$ 3x – 4y - 6 = 0
Thus, the equations of two parallel lines represented by the given equations are:
3x – 4y - 6 = 0
3x – 4y + 2 = 0

The distance between the two parallel lines is given by:
$d = \left| {\dfrac{{{c_2} - {c_1}}}{{\sqrt {{a^2} + {b^2}} }}} \right|$
Here,
a = 3
b = 4
${c_1} = - 6$
${c_2} = 2$
Substituting the values, we get:
$\Rightarrow d = \left| {\dfrac{{2 - ( - 6)}}{{\sqrt {{{(3)}^2} + {{(4)}^2}} }}} \right| \\\ \Rightarrow d = \left| {\dfrac{8}{{\sqrt {9 + 16} }}} \right| \\\ \Rightarrow d = \left| {\dfrac{8}{{\sqrt {25} }}} \right| \\\ \Rightarrow d = \left| {\dfrac{8}{5}} \right| \\\ \Rightarrow d = \dfrac{8}{5} \\\$
Therefore, distance between the two lines represented by the equation $9{x^2} - 24xy + {16^2} - 12x + 16y - 12 = 0$ is $\dfrac{8}{5}$
So, the correct answer is “Option C”.

Note : We can check if the two lines obtained are parallel or not is by the fact that -
Coefficients of x and y are the same for both the lines but the constant is different.
Parallel lines on X – Y plane can be represented as the following diagram and they can never intersect