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JEE Main 202622 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Straight Lines

JEE Main 2026 · 22 January · Evening Shift · Mathematics · Straight Lines

Among the statements (S 1) : If A(5,-1) and B(-2,3) are two...

Among the statements
(S1)(S 1) : If A(5,−1)A(5,-1) and B(−2,3)B(-2,3) are two vertices of a triangle, whose orthocentre is (0,0)(0,0), then its third vertex is (−4,−7)(-4,-7)
and
(S2) : If positive numbers 2a, b, c are three consecutive terms of an A.P., then the lines ax+by+c=0\mathrm{ax}+\mathrm{by}+\mathrm{c}=0 are concurrent at (2,−2)(2,-2),Among the statements (S 1) : If A(5,-1) and B(-2,3) are two vertices of a triangle, whose orthocentre is (0,0) , then

Options

  1. Aboth are incorrect
  2. Bonly (S2) is correct
  3. Conly (S1) is correct
  4. Dboth are correct

Correct answer

D. both are correct

Step-by-step solution

Statement 1
mAO⋅mBC=−1m_{AO} \cdot m_{BC} = -1
B(−2,3)B(-2, 3)
A(5,−1)C(h,k)A(5, -1) \qquad C(h, k)
Among the statements (S 1) : If A(5,-1) and B(-2,3) are two vertices of a triangle, whose orthocentre is (0,0) , then
⇒5h−k+13=0\Rightarrow 5h - k + 13 = 0 ....(1)
& mAB⋅mOC=−1m_{AB} \cdot m_{OC} = -1
⇒4k=7h\Rightarrow 4k = 7h ....(2)
⇒\Rightarrow third vertex is (−4,−7)(-4, -7)
∴\therefore Statement 1 is correct.
Statement 2
2a,b,c→2a, b, c \rightarrow A.P.
b=2a+c2b = \dfrac{2a + c}{2}
⇒2a−2b+c=0\Rightarrow 2a - 2b + c = 0
∴\therefore lines ax+by+c=0ax + by + c = 0 are concurrent then
x2=y−2=11\dfrac{x}{2} = \dfrac{y}{-2} = \dfrac{1}{1}
x=2x = 2 and y=−2y = -2
∴\therefore Point of concurrency is (2,−2)(2, -2)
∴\therefore Statement 2 is correct.
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