Question:
If a non-horizontal line \(L\) passes through the point \((4,-2)\) and the distance of \(L\) from the origin is \(2\) units, then the equation of the line \(L\) is
If a non-horizontal line \(L\) passes through the point \((4,-2)\) and the distance of \(L\) from the origin is \(2\) units, then the equation of the line \(L\) is
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Whenever a line is determined by both a point condition and a distance condition, first verify which options pass through the given point and then use the perpendicular distance formula.
Updated On: Jun 17, 2026
- \(4x+3y-10=0\)
- \(x+y-2\sqrt2=0\)
- \(3x+4y-4=0\)
- \(2x+3y-2=0\)
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The Correct Option is A
Solution and Explanation
Concept:
The perpendicular distance of a point \((x_1,y_1)\) from a line
\[
Ax+By+C=0
\]
is given by
\[
d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}.
\]
We use the condition that the line passes through a fixed point and is at a fixed distance from the origin.
Step 1: Check which options pass through \((4,-2)\).
Substitute \((4,-2)\) into each option. For option (A): \[ 4(4)+3(-2)-10 =16-6-10 =0. \] Hence option (A) passes through the point. For option (B): \[ 4-2-2\sqrt2\neq0. \] For option (C): \[ 12-8-4=0. \] Option (C) also passes through the point. For option (D): \[ 8-6-2=0. \] Option (D) also passes through the point.
Step 2: Use distance from origin condition.
Distance of the origin from line \[ Ax+By+C=0 \] is \[ \frac{|C|}{\sqrt{A^2+B^2}}. \] For option (A): \[ \frac{|-10|}{\sqrt{4^2+3^2}} = \frac{10}{5} =2. \] Hence option (A) satisfies the distance condition. For option (C): \[ \frac{|-4|}{\sqrt{3^2+4^2}} = \frac45 \neq2. \] For option (D): \[ \frac{|-2|}{\sqrt{2^2+3^2}} = \frac{2}{\sqrt{13}} \neq2. \] Thus only option (A) satisfies all conditions. Therefore the required equation is \[ \boxed{4x+3y-10=0}. \]
Step 1: Check which options pass through \((4,-2)\).
Substitute \((4,-2)\) into each option. For option (A): \[ 4(4)+3(-2)-10 =16-6-10 =0. \] Hence option (A) passes through the point. For option (B): \[ 4-2-2\sqrt2\neq0. \] For option (C): \[ 12-8-4=0. \] Option (C) also passes through the point. For option (D): \[ 8-6-2=0. \] Option (D) also passes through the point.
Step 2: Use distance from origin condition.
Distance of the origin from line \[ Ax+By+C=0 \] is \[ \frac{|C|}{\sqrt{A^2+B^2}}. \] For option (A): \[ \frac{|-10|}{\sqrt{4^2+3^2}} = \frac{10}{5} =2. \] Hence option (A) satisfies the distance condition. For option (C): \[ \frac{|-4|}{\sqrt{3^2+4^2}} = \frac45 \neq2. \] For option (D): \[ \frac{|-2|}{\sqrt{2^2+3^2}} = \frac{2}{\sqrt{13}} \neq2. \] Thus only option (A) satisfies all conditions. Therefore the required equation is \[ \boxed{4x+3y-10=0}. \]
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