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If the midpoints of the sides of a triangle AB,BC,CA are D(1,2,3),E(3,0,1),F(1,1,4) , then the centroid of the triangle ABC is ____.

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    Payal Gupta | Contributor-Level 10

    2 months ago

    This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

    L e t     t h e     v e r t i c e s     o f     t h e     Δ A B C     b e A ( x 1 , y 1 , z 1 ) ,     B ( x 2 , y 2 , z 2 )     a n d     C ( x 3 , y 3 , z 3 ) D  is  the  mid-point  of  AB ∴                         1 = x 1 + x 2 2         ⇒ x 1 + x 2 = 2                                                       … ( i )                             2 = y 1 + y 2 2         ⇒ y 1 + y 2 = 4                                                     … ( i i ) a n d     − 3 = z 1 + z 2 2         ⇒ z 1 + z 2 = − 6                                                     … ( i i i ) E  is  the  mid-point  of  BC ∴                         3 = x 2 + x 3 2         ⇒ x 2 + x 3 = 6                                                       … ( i v )                             0 = y 2 + y 3 2         ⇒ y 2 + y 3 = 0                                                       … ( v ) a n d             1 = z 2 + z 3 2         ⇒ z 2 + z 3 = 2                                                             … ( v i ) F  is  the  mid-point  of  AC ∴             − 1 = x 1 + x 3 2         ⇒ x 1 + x 3 = − 2                                                       … ( v i i )                           1 = y 1 + y 3 2         ⇒ y 1 + y 3 = 2                                                           … ( v i i i ) a n d     − 4 = z 1 + z 3 2         ⇒ z 1 + z 3 = − 8                                                         … ( i x ) A d d i n g     e q . ( i ) , ( i v )     a n d     ( v i i )     w e     g e t                       2 ( x 1 + x 2 + x 3 ) = 2 + 6 − 2 ∴                                 x 1 + x 2 + x 3 = 3                                                                                                     … ( x ) A d d i n g     e q . ( i i ) , ( v )     a n d     ( v i i i )     w e     g e t                       2 ( y 1 + y 2 + y 3 ) = 4 + 0 + 2 ∴                                 y 1 + y 2 + y 3 = 3                                                                                                     … ( x i ) A d d i n g     e q . ( i i i ) , ( v i )     a n d     ( i x )     w e     g e t                       2 ( z 1 + z 2 + z 3 ) = − 6 + 2 − 8 ∴                                 z 1 + z 2 + z 3 = − 6                                                                                                   … ( x i i ) S u b t r a c t i n g     ( i )     f r o m     e q . ( x )     w e     g e t                                       x 3 = 3 − 2 = 1             ⇒ x 3 = 1 S u b t r a c t i n g     ( i v )     f r o m     e q . ( x )     w e     g e t                                       x 1 = 3 − 6 = − 3             ⇒ x 1 = − 3 S u b t r a c t i n g     ( v i i )     f r o m     e q . ( x )     w e     g e t                                       x 2 = 3 − ( − 2 ) = 5             ⇒ x 2 = 5

Similar Questions for you

The distance of the point P(3,4,4) from the point of intersection of the line joining the points Q(3,-4,-5) and R(2,-3,1) and the plane 2x + y + z = 7, is equal to………..
A
alok kumar singh

Direction ratio of line (1, -1, -6)
Equation of line (x−3)/1 = (y+4)/-1 = (z+5)/-6 =k
x=k+3, y=−k−4, z=−6k−5
Solving with plane k=−2
⇒x=1, y=−2, z=7
⇒Distance=√ (3−1)²+6²+3²=√49=7

For real numbers α and β ≠ 0, if the point of intersection of the straight lines (x-α)/1 = (y-1)/2 = (z-1)/3 and (x-4)/β = (y-6)/3 = (z-7)/3 lies on the plane x+2y-z = 8, then α - β is equal to:
A
alok kumar singh

Any point on line (1)
x=α+k
y=1+2k
z=1+3k
Any point on line (2)
x=4+Kβ
y=6+3K
Z=7+3K?
⇒1+2k=6+3K, as the intersect
∴1+3k=7+3K?
⇒K=1, K? =−1
x=α+1; x=4−β
⇒y=3; y=3
z=4; z=4
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x+2y−z=8
⇒α+1+6−4=8 . (i)
and 4−β+6−4=8 . (ii)
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α+5−β+12−8=16
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Let f : [0,∞) → be a function defined by f(x) = { max{sint: 0≤ t ≤x}, 0≤x≤π, 2 + cos x, x > π
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A
alok kumar singh

f (x)= {sinx, 0≤x<π/2; 1, π/2≤x≤π 2+cosx, x>π}
f' (x)= {cosx, 0π}
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f' (π/2? ) = 0
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The equation of the plane passing through the point (1, 2, -3) and perpendicular to the planes 3x + y – 2z = 5 and
A
alok kumar singh

Let direction ratio of the normal to the required plane are l, m, n

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Equation of required plane

11 (x – 1) + 1 (y – 2) + 17 (z + 3) = 0

          

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A
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Any point on line  x − 3 1 = y − 4 2 = z − 5 2 = r is P (r + 3, 2r + 4, 2r + 5) lies on x + y + z = 17,

5r + 12 = 17

r = 1

P ( 4 , 6 , 7 ) A ( 1 , 1 , 9 ) d i s t     A P = 9 + 2 5 + 4 = 3 8  

 

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