Question

The distance between $4x + 3y = 11$ and $8x + 6y = 15$ is
A.$\dfrac{7}{2}$
B.$4$
C.$\dfrac{7}{{10}}$
D.None of these

Answer 1

Hint : Parallel lines are those lines that never meet each other. When the distance between a pair of lines is the same throughout, it can be called parallel lines Distance between two parallel lines is the perpendicular distance from any point to one of the lines.

Complete step-by-step answer :
The method for calculating the distance between two parallel lines is as follows:
Ensure whether the equations of the given parallel lines are in slope-intercept form i.e. $y = mx + c$
The intercepts i.e. ${c_1}$ and ${c_2}$ and the slope value which is common for both the lines has to be determined.
After obtaining the above values, substitute them in the slope-intercept equation to find y.
Finally, put all the above values in the distance formula to find the distance between two parallel lines.

We are given the equations of two lines as $4x + 3y = 11$ and $8x + 6y = 15$
General equation of a line is $ax + by + c = 0$
Hence we get ,
$4x + 3y = 11$ can be rewritten as $4x + 3y - 11 = 0$
And $8x + 6y = 15$ can be rewritten as $8x + 6y - 15 = 0$
Distance of a line from origin is $p = \dfrac{c}{{\sqrt {{a^2} + {b^2}} }}$
For the first line we have ${a_1} = 4,{b_1} = 3,{c_1} = - 11$
Therefore we get ${p_1} = \dfrac{{{c_1}}}{{\sqrt {{a_1}^2 + {b_1}^2} }} = \left| {\dfrac{{ - 11}}{{\sqrt {{4^2} + {3^2}} }}} \right| = \dfrac{{11}}{5}$
Similarly for the second line we have ${a_2} = 8,{b_2} = 6,{c_2} = - 15$
Therefore we get ${p_2} = \dfrac{{{c_2}}}{{\sqrt {{a_2}^2 + {b_2}^2} }} = \left| {\dfrac{{ - 15}}{{\sqrt {{8^2} + {6^2}} }}} \right| = \dfrac{{15}}{{10}}$
Therefore $d = \left| {{p_1} - {p_2}} \right| = \left| {\dfrac{{11}}{5} - \dfrac{{15}}{{10}}} \right| = \dfrac{7}{{10}}units$
Therefore the distance between the given lines $= \dfrac{7}{{10}}units$
Therefore option ( $3$ ) is the correct answer.
So, the correct answer is “Option 3”.

Note : Alternate method is as follows:
We are given the equations of two lines as $4x + 3y = 11$ and $8x + 6y = 15$
Slope of line $4x + 3y = 11$ is $- \dfrac{4}{3}$ .
Slope of line $8x + 6y = 15$ is $- \dfrac{4}{3}$ .
Since the slopes are the same. Therefore lines are parallel to each other.
General equation of a line is $ax + by + c = 0$
Hence we get ,
$4x + 3y = 11$ can be rewritten as $4x + 3y - 11 = 0$
And $8x + 6y = 15$ can be rewritten as $4x + 3y - \dfrac{{15}}{2} = 0$
Distance between two parallel lines $d = \left| {\dfrac{{{c_1} - {c_2}}}{{\sqrt {{a^2} + {b^2}} }}} \right|$
Here we have $a = 4,b = 3,{c_1} = - 11,{c_2} = - \dfrac{{15}}{2}$
Therefore $d = \left| {\dfrac{{{c_1} - {c_2}}}{{\sqrt {{a^2} + {b^2}} }}} \right| = \left| {\dfrac{{ - 11 + \dfrac{{15}}{2}}}{{\sqrt {{4^2} + {3^2}} }}} \right| = \dfrac{7}{{10}}units$
Therefore option ( $3$ ) is the correct answer.