JEE Maths›Straight Lines›Visual Solution
Visual SolutionPYQ 2024 · Apr Shift 2Standard
Question
Let A(−1,1)A(-1, 1) and B(2,3)B(2, 3) be two points and PP be a variable point above the line ABAB such that the area of △PAB\triangle PAB is 10. If the locus of PP is ax+by=15ax + by = 15, then 5a+2b5a + 2b is:
(A)66
(B)−65-\frac{6}{5}
(C)44
(D)−125-\frac{12}{5}
Solution Path
Area formula gives ∣−2h+3k−5∣=20|-2h+3k-5| = 20. Locus: −2x+3y=25-2x+3y = 25, rewrite as ax+by=15ax+by = 15. Then 5a+2b=−12/55a+2b = -12/5.
01Question Setup
1/4
Locus of point PP such that the area of triangle PABPAB equals 10, where A(−1,1)A(-1,1) and B(2,3)B(2,3). Express as ax+by=15ax + by = 15 and find 5a+2b5a + 2b.
ax+by=15,  5a+2b= ?ax + by = 15,\; 5a + 2b = \,?
02Area Formula
2/4
Use the determinant formula for area with P(h,k)P(h,k), A(−1,1)A(-1,1), B(2,3)B(2,3). Set the absolute value equal to 20 (since area = 10 and formula gives twice the area).
∣−2h+3k−5∣=20|-2h + 3k - 5| = 20
03Locus LineKEY INSIGHT
3/4
Taking the case −2x+3y−5=20-2x + 3y - 5 = 20 (P above AB): −2x+3y=25-2x + 3y = 25. Rewrite as ax+by=15ax + by = 15 by dividing by 25/15=5/325/15 = 5/3.
a=−65,  b=95a = -\frac{6}{5},\; b = \frac{9}{5}
04Final Answer
4/4
5a+2b=5⋅(−6/5)+2⋅(9/5)=−6+18/5=−12/55a + 2b = 5 \cdot (-6/5) + 2 \cdot (9/5) = -6 + 18/5 = -12/5.
5a+2b=−125\boxed{5a + 2b = -\frac{12}{5}}
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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