
Find the centroid of a triangle whose vertices are \[(2,1),(5,2)\], and \[(3,4)\].
A.\[\left( {\dfrac{{10}}{3},\dfrac{7}{3}} \right)\]
B. \[\left( {\dfrac{5}{3},\dfrac{7}{3}} \right)\]
C. \[\left( {\dfrac{{10}}{3},\dfrac{5}{3}} \right)\]
D. \[\left( {\dfrac{5}{3},\dfrac{5}{3}} \right)\]
Answer
162.3k+ views
Hints We will use the centroid formula to find the centroid of the triangle. First, we will add all x-coordinate of the points and divide them by 3 to get the x-coordinate of the centroid. Then we will add all y-coordinate of the points and divide it by 3 to get the y-coordinate of the centroid.
Formula used
The formula of the centroid of the triangle is,
\[\left( {\dfrac{{{a_1} + {a_2} + {a_3}}}{3},\dfrac{{{b_1} + {b_2} + {b_3}}}{3}} \right)\] , where three vertices of the triangle are \[\left( {{a_1},{b_1}} \right),\left( {{a_2},{b_2}} \right),\left( {{a_3},{b_3}} \right)\].
Complete step by step solution
The x-coordinate of (2,1) is 2.
The x-coordinate of (5,2) is 5.
The x-coordinate of (3,4) is 3.
Now we add all x-coordinate
The sum of x-coordinate is (2 + 5 + 3) = 10
The x-coodinate of the centroid is \[\dfrac{10}{3}\]
The y-coordinate of (2,1) is 1.
The y-coordinate of (5,2) is 2.
The y-coordinate of (3,4) is 4.
Now we add all y-coordinate
The sum of y-coordinate is (1 + 2 + 4) = 7
The y-coordinate of the centroid is \[\dfrac{7}{3}\]
The coordinate of the centroid is \[(\dfrac{10}{3}, \dfrac{7}{3})\].
The correct option is “A”.
Additional informationhe median is the line that joins the middle point of every vertex with the opposite vertex of the triangle. Three medians of the triangle divide the triangle into six equal parts. The centroid is the center point of the triangle. The point at which the three medians of the triangle intersect is the centroid of the triangle. Basically, the point of intersection of the medians of the triangle is the centroid. The centroid of a triangle always lies inside the triangle irrespective of the shape of the triangle. If three vertices of a triangle are given we can find the positions of the median and centroid easily.
Note Students often do mistakes to find the centroid. They thought the midpoint of the line by joining any two vertices is the centroid of the triangle. It is not the correct way to find the centroid. We have to use the formula \[\left( {\dfrac{{{a_1} + {a_2} + {a_3}}}{3},\dfrac{{{b_1} + {b_2} + {b_3}}}{3}} \right)\] , where three vertices of the triangle are \[\left( {{a_1},{b_1}} \right),\left( {{a_2},{b_2}} \right),\left( {{a_3},{b_3}} \right)\].
Formula used
The formula of the centroid of the triangle is,
\[\left( {\dfrac{{{a_1} + {a_2} + {a_3}}}{3},\dfrac{{{b_1} + {b_2} + {b_3}}}{3}} \right)\] , where three vertices of the triangle are \[\left( {{a_1},{b_1}} \right),\left( {{a_2},{b_2}} \right),\left( {{a_3},{b_3}} \right)\].
Complete step by step solution
The x-coordinate of (2,1) is 2.
The x-coordinate of (5,2) is 5.
The x-coordinate of (3,4) is 3.
Now we add all x-coordinate
The sum of x-coordinate is (2 + 5 + 3) = 10
The x-coodinate of the centroid is \[\dfrac{10}{3}\]
The y-coordinate of (2,1) is 1.
The y-coordinate of (5,2) is 2.
The y-coordinate of (3,4) is 4.
Now we add all y-coordinate
The sum of y-coordinate is (1 + 2 + 4) = 7
The y-coordinate of the centroid is \[\dfrac{7}{3}\]
The coordinate of the centroid is \[(\dfrac{10}{3}, \dfrac{7}{3})\].
The correct option is “A”.
Additional informationhe median is the line that joins the middle point of every vertex with the opposite vertex of the triangle. Three medians of the triangle divide the triangle into six equal parts. The centroid is the center point of the triangle. The point at which the three medians of the triangle intersect is the centroid of the triangle. Basically, the point of intersection of the medians of the triangle is the centroid. The centroid of a triangle always lies inside the triangle irrespective of the shape of the triangle. If three vertices of a triangle are given we can find the positions of the median and centroid easily.
Note Students often do mistakes to find the centroid. They thought the midpoint of the line by joining any two vertices is the centroid of the triangle. It is not the correct way to find the centroid. We have to use the formula \[\left( {\dfrac{{{a_1} + {a_2} + {a_3}}}{3},\dfrac{{{b_1} + {b_2} + {b_3}}}{3}} \right)\] , where three vertices of the triangle are \[\left( {{a_1},{b_1}} \right),\left( {{a_2},{b_2}} \right),\left( {{a_3},{b_3}} \right)\].
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