Question

The distance between the parallel lines $8x + 6y + 5 = 0$ and $4x + 3y - 25 = 0$ is
A. $\dfrac{7}{2}$
B. $\dfrac{9}{2}$
C. $\dfrac{{11}}{2}$
D. $\dfrac{5}{4}$

Answer 1

Hint: In this problem, first we will write the given equations in the form $y = mx + {c_1}$ and $y = mx + {c_2}$. Then, we will find the distance between two given parallel lines by using the formula $d = \left| {\dfrac{{{c_1} - {c_2}}}{{\sqrt {1 + {m^2}} }}} \right|$ where $m$ is the slope of line.

Complete step-by-step solution:
In this problem, the equation of two parallel lines are given. We need to find the distance between these lines. The following equations are given:
$
  8x + 6y + 5 = 0 \cdots \cdots \left( 1 \right) \\
  4x + 3y - 25 = 0 \cdots \cdots \left( 2 \right) \\
 $
Divide by $2$ on both sides of equation $\left( 1 \right)$ and write the equation in the form $y = mx + {c_1}$. Therefore, we get
$
  4x + 3y + \dfrac{5}{2} = 0 \\
   \Rightarrow 3y = - 4x - \dfrac{5}{2} \\
   \Rightarrow y = - \dfrac{4}{3}x - \dfrac{5}{6} \cdots \cdots \left( 3 \right) \\
 $
Now we will write the equation $\left( 2 \right)$in the form $y = mx + {c_2}$. Therefore, we get
$
  3y = - 4x + 25 \\
   \Rightarrow y = - \dfrac{4}{3}x + \dfrac{{25}}{3} \cdots \cdots \left( 4 \right) \\
 $
Let us compare the equation $\left( 3 \right)$ with $y = mx + {c_1}$. Therefore, we get $m = - \dfrac{4}{3}\;,\;{c_1} = - \dfrac{5}{6}$.
Let us compare the equation $\left( 4 \right)$ with$y = mx + {c_2}$. Therefore, we get $m = - \dfrac{4}{3}\;,\;{c_2} = \dfrac{{25}}{3}$.
Now we will use the formula $d = \left| {\dfrac{{{c_1} - {c_2}}}{{\sqrt {1 + {m^2}} }}} \right|$ to find the distance between two parallel lines where $m$ is the slope of line. Therefore, we get
$
  d = \left| {\dfrac{{ - \dfrac{5}{6} - \dfrac{{25}}{3}}}{{\sqrt {1 + {{\left( { - \dfrac{4}{3}} \right)}^2}} }}} \right| \\
   \Rightarrow d = \dfrac{{\left| {\dfrac{{ - 5 - 50}}{6}} \right|}}{{\sqrt {1 + \dfrac{{16}}{9}} }} \\
   \Rightarrow d = \dfrac{{\left| { - \dfrac{{55}}{6}} \right|}}{{\sqrt {\dfrac{{25}}{9}} }} \\
   \Rightarrow d = \dfrac{{\dfrac{{55}}{6}}}{{\dfrac{5}{3}}} \\
   \Rightarrow d = \dfrac{{55 \times 3}}{{6 \times 5}} \\
   \Rightarrow d = \dfrac{{11}}{2} \\
 $
Therefore, we can say that the distance between the parallel lines $8x + 6y + 5 = 0$ and $4x + 3y - 25 = 0$ is $\dfrac{{11}}{2}$. Therefore, option C is true.

Note: General equation of line is given by $ax + by + c = 0$. In this problem, if we consider the given lines in the form $Ax + By + {c_1} = 0$ and $Ax + By + {c_2} = 0$ then we can find distance between these two lines by using the formula $d = \dfrac{{\left| {{c_1} - {c_2}} \right|}}{{\sqrt {{A^2} + {B^2}} }}$. The equation $y = mx + c$ is called the slope-intercept form. Note that slopes of two parallel lines are equal. Two parallel lines will never meet. That is, there is no intersection point if lines are parallel. If two lines intersect at a right angle then we can say that the lines are perpendicular.