JEE 2022 Mathematics PYQ — The distance of the origin from the centroid of the triangle whos… | Mathem Solvex | Mathem Solvex
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JEE 2022 — Mathematics PYQ
JEE | Mathematics | 2022
The distance of the origin from the centroid of the triangle whose two sides have the equations x−2y+1=0 and 2x−y−1=0 and whose orthocenter is (37,37) is:
Choose the correct answer:
A.
2
B.
2
C.
22
(Correct Answer)
D.
4
Correct Answer:
22
Explanation
Solution
1. Find the first vertex (A)
The first vertex is the intersection of the two given side equations:
L1:x−2y+1=0
L2:2x−y−1=0
Multiplying L1 by 2: 2x−4y+2=0.
Subtracting L2 from this: (2x−4y+2)−(2x−y−1)=0⟹−3y+3=0⟹y=1.
Substituting y=1 into L1: x−2(1)+1=0⟹x=1.
So, vertex A=(1,1).
2. Find vertices B and C
Let vertex B lie on L1 and vertex C lie on L2.
Since B is on x−2y+1=0, let B=(2b−1,b).
Since C is on 2x−y−1=0, let C=(c,2c−1).
The orthocenter H(37,37) is the intersection of altitudes. The altitude from B must be perpendicular to side AC (which is L2).
Slope of L2 (2x−y−1=0) is m2=2.
Therefore, the slope of altitude BH must be mBH=−21.
(2b−1)−37b−37=−21
6b−103b−7=−21⟹6b−14=−6b+10⟹12b=24⟹b=2
Thus, B=(2(2)−1,2)=(3,2).
Similarly, the altitude from C must be perpendicular to side AB (which is L1).
Slope of L1 (x−2y+1=0) is m1=21.
Therefore, the slope of altitude CH must be mCH=−2.
c−37(2c−1)−37=−2
3c−76c−10=−2⟹6c−10=−6c+14⟹12c=24⟹c=2
Thus, C=(2,2(2)−1)=(2,3).
3. Find the Centroid (G)
The coordinates of the centroid G(x,y) are:
G=(3xA+xB+xC,3yA+yB+yC)
G=(31+3+2,31+2+3)=(2,2)
4. Distance from Origin (0,0)
Distance OG=(2−0)2+(2−0)2
OG=4+4=8=22
Correct Option: (C) 22
Explanation
Solution
1. Find the first vertex (A)
The first vertex is the intersection of the two given side equations:
L1:x−2y+1=0
L2:2x−y−1=0
Multiplying L1 by 2: 2x−4y+2=0.
Subtracting L2 from this: (2x−4y+2)−(2x−y−1)=0⟹−3y+3=0⟹y=1.
Substituting y=1 into L1: x−2(1)+1=0⟹x=1.
So, vertex A=(1,1).
2. Find vertices B and C
Let vertex B lie on L1 and vertex C lie on L2.
Since B is on x−2y+1=0, let B=(2b−1,b).
Since C is on 2x−y−1=0, let C=(c,2c−1).
The orthocenter H(37,37) is the intersection of altitudes. The altitude from B must be perpendicular to side AC (which is L2).
Slope of L2 (2x−y−1=0) is m2=2.
Therefore, the slope of altitude BH must be mBH=−21.
(2b−1)−37b−37=−21
6b−103b−7=−21⟹6b−14=−6b+10⟹12b=24⟹b=2
Thus, B=(2(2)−1,2)=(3,2).
Similarly, the altitude from C must be perpendicular to side AB (which is L1).
Slope of L1 (x−2y+1=0) is m1=21.
Therefore, the slope of altitude CH must be mCH=−2.
