JEE Maths›Straight Lines›Visual Solution
Visual SolutionPYQ 2024 · Apr Shift 1Tricky
Question
A ray of light coming from the point P(1,2)P(1, 2) gets reflected from the point QQ on the x-axis and then passes through the point R(4,3)R(4, 3). If the point S(h,k)S(h, k) is such that PQRSPQRS is a parallelogram, then hk2hk^2 is equal to:
(A)7070
(B)8080
(C)6060
(D)9090
Solution Path
Reflect PP in x-axis to find Q=(115,0)Q=(\tfrac{11}{5},0). Parallelogram gives S=(145,5)S=(\tfrac{14}{5},5). Answer: hk2=70hk^2 = 70.
01Question Setup
1/4
Ray from P(1,2)P(1,2) reflects at QQ on x-axis, passes through R(4,3)R(4,3). PQRSPQRS is a parallelogram. Find hk2hk^2.
hk2=  ?hk^2 = \;?
02Image Method
2/4
Reflect P(1,2)P(1,2) in x-axis: P′=(1,−2)P'=(1,-2). Line P′RP'R has slope 53\frac{5}{3}. At y=0y=0: Q=(115,0)Q = (\frac{11}{5}, 0).
Q=(115,0)Q = (\tfrac{11}{5}, 0)
03ParallelogramKEY INSIGHT
3/4
S=P+R−Q=(145,5)S = P + R - Q = (\frac{14}{5}, 5). So hk2=145×25=70hk^2 = \frac{14}{5} \times 25 = 70.
hk2=70hk^2 = 70
04Final Answer
4/4
hk2=70hk^2 = 70.
70\boxed{70}
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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