JEE Maths›Straight Lines›Visual Solution
Visual SolutionPYQ 2024 · Jan Shift 2Tricky
Question
The distance of the point (2,3)(2, 3) from the line 2x−3y+28=02x - 3y + 28 = 0, measured parallel to the line 3x−y+1=0\sqrt{3}x - y + 1 = 0, is equal to:
(A)424\sqrt{2}
(B)636\sqrt{3}
(C)3+423 + 4\sqrt{2}
(D)4+634 + 6\sqrt{3}
Solution Path
Parametric form along 60°60° direction from (2,3)(2,3). Substituting into 2x−3y+28=02x-3y+28=0 gives r=4+63r = 4+6\sqrt{3}.
01Question Setup
1/4
Find the distance from (2,3)(2,3) to 2x−3y+28=02x - 3y + 28 = 0, measured parallel to 3 x−y+1=0\sqrt{3}\,x - y + 1 = 0.
r=  ?r = \;?
02Parametric Form
2/4
Direction has slope 3\sqrt{3} (angle 60°60°). Parametric point: (2+r2,  3+r32)(2 + \frac{r}{2},\; 3 + \frac{r\sqrt{3}}{2}).
θ=60°\theta = 60°
03Solve for rKEY INSIGHT
3/4
Substitute parametric point into 2x−3y+28=02x - 3y + 28 = 0 and solve: r=4+63r = 4 + 6\sqrt{3}.
r=4+63r = 4 + 6\sqrt{3}
04Final Answer
4/4
Distance measured parallel to the given line is 4+634 + 6\sqrt{3}.
4+63\boxed{4 + 6\sqrt{3}}
Concepts from this question1 concepts unlocked
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Distance from Point to Line

EASY

Distance from (x₁, y₁) to line ax + by + c = 0 is |ax₁ + by₁ + c| / √(a² + b²).

d=∣ax1+by1+c∣a2+b2d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}

Used in area calculations, mirror image, and foot of perpendicular problems. High-frequency JEE formula.

Distance problemsArea of triangleFoot of perpendicular
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