Question

Suppose \(\triangle ABC\) is an isosceles triangle with \(\angle C=90^\circ\), \(A=(2,3)\) and \(B=(4,5)\). Then the centroid of the triangle is

• A. \(\left(\dfrac{13}{8},\dfrac{8}{3}\right)\)
• B. \(\left(\dfrac{11}{3},\dfrac{10}{3}\right)\)
• C. \(\left(\dfrac{10}{3},\dfrac{13}{3}\right)\)
• D. \(\left(\dfrac{10}{3},\dfrac{11}{3}\right)\)

Hint:

The centroid of a triangle is found by averaging the \(x\)-coordinates and \(y\)-coordinates of its three vertices.

Answer 1

The Correct Option is D

Step 1: Understand the triangle.
Since \(\triangle ABC\) is isosceles and \[ \angle C=90^\circ, \] the sides \(AC\) and \(BC\) are equal.
Thus, \(AB\) is the hypotenuse.

Step 2: Find the midpoint of \(AB\).
Given, \[ A=(2,3),\qquad B=(4,5) \] The midpoint of \(AB\) is \[ M=\left(\frac{2+4}{2},\frac{3+5}{2}\right) \] \[ M=(3,4) \]

Step 3: Find the possible point \(C\).
The vector from \(M\) to \(B\) is \[ (4-3,5-4)=(1,1) \] A perpendicular vector of the same length is \[ (1,-1) \] So, one possible position of \(C\) is \[ C=M+(1,-1) \] \[ C=(3+1,4-1)=(4,3) \]

Step 4: Find the centroid.
The centroid of a triangle with vertices \[ (x_1,y_1),\quad (x_2,y_2),\quad (x_3,y_3) \] is \[ \left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right) \] Using \[ A=(2,3),\quad B=(4,5),\quad C=(4,3), \] we get \[ G=\left(\frac{2+4+4}{3},\frac{3+5+3}{3}\right) \] \[ G=\left(\frac{10}{3},\frac{11}{3}\right) \]

Step 5: Final conclusion.
Hence, the centroid is \[ \boxed{\left(\frac{10}{3},\frac{11}{3}\right)} \] which corresponds to option (4).