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JEE Main 201912 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
12 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Straight Lines

JEE Main 2019 · 12 April · Evening Shift · Mathematics · Straight Lines

A triangle has a vertex at ( 1, 2 ) and the mid points of the two...

A triangle has a vertex at (1,2)( 1, 2 ) and the mid points of the two sides through it are (−1,1)( - 1, 1 ) and (2,3)( 2, 3 ) . Then the centroid of this triangle is:A triangle has a vertex at ( 1, 2 ) and the mid points of the two sides through it are ( - 1, 1 ) and ( 2, 3 ) . Then

Options

  1. A(13,1)\left(\dfrac{1}{3}, 1\right)
  2. B(1,73)\left(1, \dfrac{7}{3}\right)
  3. C(13,2)\left(\dfrac{1}{3}, 2\right)
  4. D(13,53)\left(\dfrac{1}{3}, \dfrac{5}{3}\right)

Correct answer

C. (13,2)\left(\dfrac{1}{3}, 2\right)

Step-by-step solution

A triangle has a vertex at ( 1, 2 ) and the mid points of the two sides through it are ( - 1, 1 ) and ( 2, 3 ) . Then
Let co-ordinates of B=(x1,y1)B = \left({x}_{1}, {y}_{1}\right)
∴x1+12=−1 and y1+22=1∴ \dfrac{{x}_{1} + 1}{2} = - 1 \text{ and } \dfrac{{y}_{1} + 2}{2} = 1
⇒x1=−3,y1=0⇒ {x}_{1} = - 3, {y}_{1} = 0
∴B=(−3,0)∴ B = ( - 3, 0 )
Let co-ordinates of C=(x2,y2)C = \left({x}_{2}, {y}_{2}\right)
∴x2+12=2 and y2+22=3∴ \dfrac{{x}_{2} + 1}{2} = 2 \text{ and } \dfrac{{y}_{2} + 2}{2} = 3
∴C=(3,4)∴ C = \left(3, 4\right)
Let centroid G=(h,k)G = \left(h, k\right)
h=1+x1+x23=1−3+33=13h = \dfrac{1 + {x}_{1} + {x}_{2}}{3} = \dfrac{1 - 3 + 3}{3} = \dfrac{1}{3}
k=2+y1+y23=2+0+43=2k = \dfrac{2 + {y}_{1} + {y}_{2}}{3} = \dfrac{2 + 0 + 4}{3} = 2
∴∴ Centroid, G = \left(\dfrac{1}{3}, 2\right)
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