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Q.PQR is a triangle with vertices P(2a,2,6)P(2a, 2, 6), Q(−4,3b,−10)Q(-4, 3b, -10) and R(8,14,2c)R(8, 14, 2c). If the centroid lies at the origin, calculate the values of a,b,ca, b, c. OR Show that the points (−2,3,5)(-2, 3, 5), (1,2,3)(1, 2, 3) and (7,0,−1)(7, 0, -1) are collinear.

Meghalaya MboseMBOSE Meghalaya 11th Board 2019Subjective· 4mImportance★★★★★
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Setting the centroid formula equal to the origin gives a=−2a=-2, b=−163b=-\dfrac{16}{3}, c=2c=2.

The centroid of a triangle with vertices (x1,y1,z1)(x_1,y_1,z_1), (x2,y2,z2)(x_2,y_2,z_2), (x3,y3,z3)(x_3,y_3,z_3) is:

(x1+x2+x33, y1+y2+y33, z1+z2+z33)\left(\dfrac{x_1+x_2+x_3}{3},\ \dfrac{y_1+y_2+y_3}{3},\ \dfrac{z_1+z_2+z_3}{3}\right)

Here P(2a,2,6)P(2a,2,6), Q(−4,3b,−10)Q(-4,3b,-10), R(8,14,2c)R(8,14,2c), and the centroid is given to be the origin (0,0,0)(0,0,0).

x-coordinate:

2a+(−4)+83=0⇒2a+4=0⇒a=−2\dfrac{2a+(-4)+8}{3}=0 \Rightarrow 2a+4=0 \Rightarrow a=-2

y-coordinate:

2+3b+143=0⇒16+3b=0⇒b=−163\dfrac{2+3b+14}{3}=0 \Rightarrow 16+3b=0 \Rightarrow b=-\dfrac{16}{3}

z-coordinate:

6+(−10)+2c3=0⇒−4+2c=0⇒c=2\dfrac{6+(-10)+2c}{3}=0 \Rightarrow -4+2c=0 \Rightarrow c=2

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