
If two vertices of a triangle are \[\left( {6,4} \right),\left( {2,6} \right)\] and its centroid is \[\left( {4,6} \right)\], find the third vertex.
A. \[\left( {4,8} \right)\]
B. \[\left( {8,4} \right)\]
C. \[\left( {6,4} \right)\]
D. None of these
Answer
164.1k+ views
Hint:
In the given question, we need to find the third vertex using the given two vertices. We will use the formula of coordinates of the centroid for this by assuming the third vertex as \[\left( {x_3^{},y_3^{}} \right)\]. Also, by comparing coordinates and simplifying the equations, we will get the desired result.
Formula Used:
The coordinates of a centroid \[\left( {a,b} \right)\] are given by
\[\left( {a,b} \right) = \left[ {\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3}} \right),\left( {\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)} \right]\].
Here, the vertices of a triangle are \[\left( {x_1^{},y_1^{}} \right),\left( {x_2^{},y_2^{}} \right),\left( {x_3^{},y_3^{}} \right)\].
Complete step-by-step answer:
Let the three vertices of a triangle be \[\left( {x_1^{},y_1^{}} \right),\left( {x_2^{},y_2^{}} \right),\left( {x_3^{},y_3^{}} \right)\].
Given that, \[\left( {x_1^{},y_1^{}} \right) \equiv \left( {6,4} \right)\] and \[\left( {x_2^{},y_2^{}} \right) \equiv \left( {2,6} \right)\]
And coordinates of the centroid are \[\left( {a,b} \right) \equiv \left( {4,6} \right)\]
Now, we will use the formula of coordinates of centroid to find the third vertex of a triangle.
\[\left( {a,b} \right) = \left[ {\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3}} \right),\left( {\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)} \right]\]
Substituting the respective values, we get;
\[\left( {4,6} \right) = \left[ {\left( {\dfrac{{6 + 2 + {x_3}}}{3}} \right),\left( {\dfrac{{4 + 6 + {y_3}}}{3}} \right)} \right]\]
On comparing, we get;
\[\dfrac{{6 + 2 + {x_3}}}{3} = 4\] and \[\dfrac{{4 + 6 + {y_3}}}{3} = 6\]
Thus, by simplifying, we get;
\[6 + 2 + {x_3} = 12\] and \[4 + 6 + {y_3} = 18\]
By solving,
\[{x_3} = 4\] and \[{y_3} = 8\]
Hence, the third vertex of a triangle is \[\left( {x_3^{},y_3^{}} \right) \equiv \left( {4,8} \right)\]
If two vertices of a triangle are \[\left( {6,4} \right),\left( {2,6} \right)\] and its centroid is \[\left( {4,6} \right)\], then the third vertex of a triangle is \[\left( {4,8} \right)\].
Therefore, the correct option is (A).
Note:
Students may make mistakes while calculating the third vertex of a triangle. The centroid of a triangle is determined by intersecting its medians. Line segments of medians join the vertex to the opposite side's midpoint. Thus, all three medians intersect at the same point and the point of convergence is called the triangle's centroid.
In the given question, we need to find the third vertex using the given two vertices. We will use the formula of coordinates of the centroid for this by assuming the third vertex as \[\left( {x_3^{},y_3^{}} \right)\]. Also, by comparing coordinates and simplifying the equations, we will get the desired result.
Formula Used:
The coordinates of a centroid \[\left( {a,b} \right)\] are given by
\[\left( {a,b} \right) = \left[ {\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3}} \right),\left( {\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)} \right]\].
Here, the vertices of a triangle are \[\left( {x_1^{},y_1^{}} \right),\left( {x_2^{},y_2^{}} \right),\left( {x_3^{},y_3^{}} \right)\].
Complete step-by-step answer:
Let the three vertices of a triangle be \[\left( {x_1^{},y_1^{}} \right),\left( {x_2^{},y_2^{}} \right),\left( {x_3^{},y_3^{}} \right)\].
Given that, \[\left( {x_1^{},y_1^{}} \right) \equiv \left( {6,4} \right)\] and \[\left( {x_2^{},y_2^{}} \right) \equiv \left( {2,6} \right)\]
And coordinates of the centroid are \[\left( {a,b} \right) \equiv \left( {4,6} \right)\]
Now, we will use the formula of coordinates of centroid to find the third vertex of a triangle.
\[\left( {a,b} \right) = \left[ {\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3}} \right),\left( {\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)} \right]\]
Substituting the respective values, we get;
\[\left( {4,6} \right) = \left[ {\left( {\dfrac{{6 + 2 + {x_3}}}{3}} \right),\left( {\dfrac{{4 + 6 + {y_3}}}{3}} \right)} \right]\]
On comparing, we get;
\[\dfrac{{6 + 2 + {x_3}}}{3} = 4\] and \[\dfrac{{4 + 6 + {y_3}}}{3} = 6\]
Thus, by simplifying, we get;
\[6 + 2 + {x_3} = 12\] and \[4 + 6 + {y_3} = 18\]
By solving,
\[{x_3} = 4\] and \[{y_3} = 8\]
Hence, the third vertex of a triangle is \[\left( {x_3^{},y_3^{}} \right) \equiv \left( {4,8} \right)\]
If two vertices of a triangle are \[\left( {6,4} \right),\left( {2,6} \right)\] and its centroid is \[\left( {4,6} \right)\], then the third vertex of a triangle is \[\left( {4,8} \right)\].
Therefore, the correct option is (A).
Note:
Students may make mistakes while calculating the third vertex of a triangle. The centroid of a triangle is determined by intersecting its medians. Line segments of medians join the vertex to the opposite side's midpoint. Thus, all three medians intersect at the same point and the point of convergence is called the triangle's centroid.
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