Answer

Verified

446.7k+ views

**Hint:**A triangle ABC is placed so that the mid-points of the sides are on the x,y,z axes. Lengths of the intercepts made by the plane containing the triangle on these axes are respectively\[\alpha ,\beta ,\gamma \]. Coordinates of the centroid of the triangle ABC are

\[( - \alpha /3,\beta /3,\gamma /3)\]

\[(\alpha /3, - \beta /3,\gamma /3)\]

\[(\alpha /3,\beta /3, - \gamma /3)\]

\[(\alpha /3,\beta /3,\gamma /3)\]

**Complete step by step solution:**

1) \[\,ABC\] is placed so that the mid-points of the sides are on the x,y,z axes

i.e.:D,E,F are the mid-points of CA , BC , BA respectively and D,E,F lie on the Z,Y,X axes.

2) Lengths of the intercepts made by the plane containing the triangle on these axes are \[\alpha ,\beta ,\gamma \]respectively which means the coordinates of D,E,F are \[F(\alpha ,0,0),E(0,\beta ,0)and\,D(0,0,\gamma )\].

Here, \[A = ({x_1},{y_1},{z_1})\]

\[B = ({x_2},{y_2},{z_2})\]

\[C = ({x_3},{y_3},{z_3})\]

And \[F = (\alpha ,0,0)\]

\[E = (0,\beta ,0)\]

\[D = (0,0,\gamma )\]

Step 1: To make a relation between the coordinates of A,B,C with \[\alpha ,\beta ,\gamma \]

i)= > \[F = (\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2},\dfrac{{{z_1} + {z_2}}}{2})\] [ F is the mid point of AB]

= > \[F = (\alpha ,0,0)\]

i.e.: \[\dfrac{{{x_1} + {x_2}}}{2} = \alpha ,\,\,\dfrac{{{y_1} + {y_2}}}{2} = 0,\,\,\dfrac{{{z_1} + {z_2}}}{2} = 0\]

ii) = > \[E = (\dfrac{{{x_2} + {x_3}}}{2},\dfrac{{{y_2} + {y_3}}}{2},\dfrac{{{z_2} + {z_3}}}{2})\] [ E is the mid point of BC]

= > \[F = (0,\beta ,0)\]

i.e.: \[\dfrac{{{x_2} + {x_3}}}{2} = 0,\,\,\dfrac{{{y_2} + {y_3}}}{2} = \beta ,\,\,\dfrac{{{z_2} + {z_3}}}{2} = 0\]

=> \[D = (\dfrac{{{x_3} + {x_1}}}{2},\dfrac{{{y_3} + {y_1}}}{2},\dfrac{{{z_3} + {z_1}}}{2})\]

= > \[D = (0,0,\gamma )\]

i.e.: \[\dfrac{{{x_3} + {x_1}}}{2} = 0,\,\,\dfrac{{{y_3} + {y_1}}}{2} = 0,\,\,\dfrac{{{z_3} + {z_1}}}{2} = \gamma \]

From (i) , We get,

\[{x_1} + {x_2} = 2\alpha \]

\[{y_1} + {y_2} = 0\]

\[{z_1} + {z_2} = 0\]

From (ii), we get,

\[{x_2} + {x_3} = 0\]

\[{y_2} + {y_3} = 2\beta \]

\[{z_2} + {z_3} = 0\]

From (iii), we get,

\[{x_3} + {x_1} = 0\]

\[{y_3} + {y_1} = 0\]

\[{z_3} + {z_1} = 2\gamma \]

From all the above equations, we can conclude that,

\[{x_1} + {x_2} + {x_3} = \alpha \]

\[{y_1} + {y_2} + {y_3} = \beta \]

\[{z_1} + {z_2} + {z_3} = \gamma \]

Hence, we get the coordinates of the points A,B,C in terms of \[\alpha ,\beta ,\gamma \]

Step 2: We will now find out the coordinate of centroid of the triangle coordinate of the centroid of the triangle is given by

\[(\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3},\dfrac{{{z_1} + {z_2} + {z_3}}}{3})\]

= \[(\dfrac{\alpha }{3},\dfrac{\beta }{3},\dfrac{\gamma }{3})\]

Hence, the coordinate of the centroid of the triangle ABC is \[(\dfrac{\alpha }{3},\dfrac{\beta }{3},\dfrac{\gamma }{3})\].

So, \[D(\dfrac{\alpha }{3},\dfrac{\beta }{3},\dfrac{\gamma }{3})\] is the correct answer.

**Note:**Diagram should be drawn properly. As because, a correct diagram will only lead you to a correct answer. Also, equations should be written correctly.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE