Question:

The point $(2, 1)$ is translated parallel to the line $L : x-y = 4$ by $2\sqrt{3}$ units. If the new point $Q$ lies in the third quadrant, then the equation of the line passing through $Q$ and perpendicular to $L$ is :

Updated On: Apr 28, 2025
  • $x +y = 2 - \sqrt{6}$
  • $x +y = 3 - 3\sqrt{6}$
  • $x +y = 3 - 2\sqrt{6}$
  • $2x + 2y = 1 - \sqrt{6}$
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The Correct Option is C

Solution and Explanation

$x-y=4$ To find equation of R slope of L = 0 is 1 $\Rightarrow$ slope of QR = - 1 Let QR is y = mx + c $y=-x+c$ $x+y-c=0$ distance of QR from (2,1) is $2\sqrt{3}$ $2\sqrt{3}=\frac{\left|2+1-c\right|}{\sqrt{2}}$ $2\sqrt{6}=\left|3-c\right|$ $c-3=\pm2\sqrt{6}\,c=3\pm2\sqrt{6}$ Line can $be x+y=3\pm2\sqrt{6}$ $x+y=3-2\sqrt{6}$
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Concepts Used:

Equation of a Line in Space

In a plane, the equation of a line is given by the popular equation y = m x + C. Let's look at how the equation of a line is written in vector form and Cartesian form.

Vector Equation

Consider a line that passes through a given point, say ‘A’, and the line is parallel to a given vector '\(\vec{b}\)‘. Here, the line ’l' is given to pass through ‘A’, whose position vector is given by '\(\vec{a}\)‘.  Now, consider another arbitrary point ’P' on the given line, where the position vector of 'P' is given by '\(\vec{r}\)'.

\(\vec{AP}\)=𝜆\(\vec{b}\)

Also, we can write vector AP in the following manner:

\(\vec{AP}\)=\(\vec{OP}\)–\(\vec{OA}\)

𝜆\(\vec{b}\) =\(\vec{r}\)–\(\vec{a}\)

\(\vec{a}\)=\(\vec{a}\)+𝜆\(\vec{b}\)

\(\vec{b}\)=𝑏1\(\hat{i}\)+𝑏2\(\hat{j}\) +𝑏3\(\hat{k}\)