Question

Find the equation of one of the lines parallel to \[4x - 3y = 5\] and at a unit distance from the point (-1,-4).
A. \[3x + 4y - 3 = 0\]
B. \[3x + 4y + 3 = 0\]
C. \[4x - 3y + 3 = 0\]
D. \[4x - 3y - 3 = 0\]

Answer 1

Hint:We know the equation of a line that is parallel to \[ax + by + c = 0\] is \[ax + by + k = 0\] where k is a constant. First, we will assume the equation of line that is parallel to \[4x - 3y = 5\] using the above concept. Then applying the formula of the distance of a line \[ax + by + c = 0\] from a point \[\left( {{x_1},{y_1}} \right)\]is \[\left| {\dfrac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|\]. Equate the distance with 1 and calculate the value of \[k\].

Formula Used: The equation of a line that is parallel to \[ax + by + c = 0\] is \[ax + by + k = 0\] where is \[k\] a constant.
the formula of the distance of a line \[ax + by + c = 0\] from a point \[\left( {{x_1},{y_1}} \right)\]is \[\left| {\dfrac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|\].

Complete step by step solution:
Given equation of line is \[4x - 3y = 5\].
The equation of a line that is parallel to \[4x - 3y = 5\] is \[4x - 3y + k = 0\].
Calculate the distance of the line \[4x - 3y + k = 0\] from the point (-1,-4).
Applying the formula \[\left| {\dfrac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|\].
The distance of the line \[4x - 3y + k = 0\] from the point (-1,-4) is \[\left| {\dfrac{{4 \cdot \left( { - 1} \right) - 3 \cdot \left( { - 4} \right) + k}}{{\sqrt {{4^2} + {{\left( { - 3} \right)}^2}} }}} \right|\]
                                                                                                                     \[ = \left| {\dfrac{{ - 4 + 12 + k}}{{\sqrt {16 + 9} }}} \right|\]
                                                                                                                   \[ = \left| {\dfrac{{8 + k}}{5}} \right|\]
Now equating the distance with 1.
\[\left| {\dfrac{{8 + k}}{5}} \right| = 1\]
Applying the absolute function \[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}{ - x}&{x < 0}\\x&{x \ge 0}\end{array}} \right.\]
Either or,
\[ \Rightarrow \dfrac{{8 + k}}{5} = 1\] or, \[ \Rightarrow \dfrac{{8 + k}}{5} = - 1\]
Solving the above equations
\[ \Rightarrow 8 + k = 5\] or, \[ \Rightarrow 8 + k = - 5\]
Subtract 8 from both sides
\[ \Rightarrow k = 5 - 8\] or, \[ \Rightarrow k = - 5 - 8\]
\[ \Rightarrow k = - 3\] or, \[ \Rightarrow k = - 13\]
Putting \[k = - 3\] in the equation \[4x - 3y + k = 0\].
\[4x - 3y - 3 = 0\]
Putting \[k = - 13\] in the equation \[4x - 3y + k = 0\].
\[4x - 3y - 13 = 0\]
The equations of lines are \[4x - 3y - 3 = 0\] and \[4x - 3y - 13 = 0\].

Hence option D is the correct option.

Note: Students often confused the equation of line parallel to a line. Sometimes they take the equation of line in the form \[bx - ay + k = 0\] which is an incorrect equation. \[bx - ay + k = 0\] is the equation of line perpendicular to \[ax + by + c = 0\]. The equation of line parallel to \[ax + by + c = 0\] is \[ax + by + k = 0\].