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The mid-points of the sides of a triangle are ( 5 , 7 , 1 1 ) , ( 0 , 8 , 5 ) , and ( 2,3,–1 ) . Find its vertices.

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    Answered by

    Payal Gupta | Contributor-Level 10

    2 months ago

    This is a short answer type question as classified in NCERT Exemplar

    Let  the  coordinates  of  the  vertices  of  ΔABC  beA(x1,y1,z1),  B(x2,y2,z2)  and  C(x3,y3,z3)  respectively. Since D(5,7,11)  mid point of  BC∴        5=x2+x32       ⇒x2+x3=10                                   …(i)           7=y2+y32       ⇒y2+y3=14                                …(ii)         11=z2+z32       ⇒z2+z3=22                                  …(iii)E(0,8,5)  is  the  mid point of  AB∴        0=x1+x22       ⇒x1+x2=0                                    …(iv)           8=y1+y22       ⇒y1+y2=16                                  …(v)           5=z1+z22       ⇒z1+z2=10                                     …(vi)

    Similarly,  F(2,3,−1)  is  the  mid point of  AC∴         2=x1+x32       ⇒x1+x3=4                                    …(vii)            3=y1+y32       ⇒y1+y3=6                                   …(viii)         −1=z1+z32       ⇒z1+z3=−2                                    …(ix)Adding  eq.(i),(iv)  and  (vii)  we  get,     2x1+2x2+2x3=10+0+4⇒        x1+x2+x3=7                                                                …(x)Subtracting  (i)  from  (x)  we  get,              x1=7−10=−3Subtracting  (iv)  from  (x)  we  get,              x3=7−0=7Subtracting  (vii)  from  (x)  we  get,              x2=7−4=3Adding  eq.(ii),(v)  and  (viii)  we  get,     2(y1+y2+y3)=14+16+6⇒       y1+y2+y3=18                                                              …(xi)Subtracting  (ii)  from  (xi)  we  get,              y1=18−14=4Subtracting  (v)  from  (xi)  we  get,              y3=18−16=2Subtracting  (viii)  from  (xi)  we  get,              y2=18−6=12Similarly,  Adding  eq.(iii),(vi)  and  (ix)  we  get,     2(z1+z2+z3)=22+10−2⇒       z1+z2+z3=15                                                              …(xii)Subtracting  (iii)  from  (xii)  we  get,              z1=15−22=−7Subtracting  (vi)  from  (xii)  we  get,              z3=15−10=5Subtractin<

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)
Adding (i) and (ii)
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
Then which of the following is true?
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alok kumar singh

f (x)= {sinx, 0≤x<π/2; 1, π/2≤x≤π 2+cosx, x>π}
f' (x)= {cosx, 0π}
f' (π/2? ) = 0
f' (π/2? ) = 0
f' (π? ) = 0
f' (π? ) = 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

∴ 3 l + m − 2 n = 0 ∴ 2 l − 5 m − n = 0 ∴ l − 1 1 = m − 1 = n − 1 7             

Equation of required plane

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

          

The distance of the point (1, 1, 9) form the point of intersection of the line x − 3 1 = y − 4 2 = z − 5 2 and the plane  x + y + z = 17 is:
A
alok kumar singh

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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