Question

The equation of one of the lines parallel to $4x-3y=5$ and at a unit distance from the point $(-1,-4)$ is

• A. $3x+4y–3=0$
• B. $3x+4y+3=0$
• C. $4x–3y+3=0$
• D. $4x–3y–3=0$

Answer 1

The correct option is D

$4x–3y–3=0$

Explanation For The Correct Option:

Finding the equation of line

The given equation of line,

$$\begin{gathered} 4x–3y=5 \\ \Rightarrow y=\left(\frac{4}{3}\right)x–\left(\frac{5}{3}\right)...\left(i\right) \end{gathered}$$

Therefore the slope is $m=\frac{4}{3}$

Consider the line which is parallel to the given line is $y=m_{1}x+c$

Since the slopes are equal, $m_1=m=\frac{4}{3}$

$$\begin{gathered} \therefore y=\left(\frac{4}{3}\right)x+c \\ \Rightarrow 3y–4x–3c=0 \\ \Rightarrow 3y–4x+k=0....\left(ii\right)[Putting-3c=k] \end{gathered}$$

Given that the point $(-1,-4)$ is at a unit distance from this line.

We know that the distance between a point and a line is given by

$d=\left|\frac{(a{x}_1+b{y}_1+c)}{\sqrt{(a^2+b^2)}}\right|$

Substituting the values,

$$\begin{gathered} \Rightarrow d=\left|\frac{(4–12+k)}{\sqrt{(3^2+4^2)}}\right| \\ \Rightarrow 1=\left|\frac{(-8+k)}{5}\right| \\ \Rightarrow \pm 5=-8+k \\ \Rightarrow k=3,ork=13 \end{gathered}$$

Putting the value of $k=3$ into $\left(ii\right)$ the required equation is

$\Rightarrow 4x–3y–3=0$

Hence, the correct answer is Option (D)