
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
A. \[( - \alpha /3,\beta /3,\gamma /3)\]
B. \[(\alpha /3, - \beta /3,\gamma /3)\]
C. \[(\alpha /3,\beta /3, - \gamma /3)\]
D. \[(\alpha /3,\beta /3,\gamma /3)\]
Answer
486.3k+ 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.
\[( - \alpha /3,\beta /3,\gamma /3)\]
\[(\alpha /3, - \beta /3,\gamma /3)\]
\[(\alpha /3,\beta /3, - \gamma /3)\]
\[(\alpha /3,\beta /3,\gamma /3)\]
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
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

What is a transformer Explain the principle construction class 12 physics CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

What are the major means of transport Explain each class 12 social science CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?
