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Chapter 14: Problem 1322

If a vertex of a triangle is \((1,1)\) and the mid-points of two sides through this vertex are \((-1,2)\) and \((3,2)\), then centroid of the triangle is (a) \([(1 / 3),(7 / 3)]\) (b) \([1,(7 / 3)]\) (c) \([-(1 / 3),(7 / 3)]\) (d) \([-1,(7 / 3)]\)

Short Answer

Expert verified
The centroid of the triangle is (1, 7/3), so the short answer is option (b) $[1,(\frac{7}{3})]$.

Step by step solution

01

Identify given information and notations

Given, a vertex of a triangle is A(1, 1) and the midpoints of two sides through this vertex are M1(-1, 2) and M2(3, 2). Let the other vertices of the triangle be B and C.
02

Find the coordinates of point B

Since M1 is the midpoint of AB, we can use the midpoint formula to find the coordinates of B: Midpoint formula: \(\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\) Substitute the coordinates of point A(1,1) and M1(-1,2), we have: \((-1, 2) = \left( \frac{1 + x_B}{2}, \frac{1 + y_B}{2} \right)\) Now, solve for x_B and y_B: \(x_B = 2 \times (-1) - 1 = -3\) \(y_B = 2 \times 2 - 1 = 3\) So, the coordinates of point B are (-3, 3).
03

Find the coordinates of point C

Similarly, since M2 is the midpoint of AC, we can use the midpoint formula to find the coordinates of C: Substitute the coordinates of point A(1,1) and M2(3,2), we have: \((3, 2) = \left( \frac{1 + x_C}{2}, \frac{1 + y_C}{2} \right)\) Now, solve for x_C and y_C: \(x_C = 2 \times 3 - 1 = 5\) \(y_C = 2 \times 2 - 1 = 3\) So, the coordinates of point C are (5, 3).
04

Calculate the centroid coordinates

The centroid of a triangle can be found by taking the average of its vertices' x-coordinates and y-coordinates: Centroid: \(\left( \frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3} \right)\) Substitute the coordinates of points A(1,1), B(-3,3), and C(5,3), we have: Centroid: \(\left( \frac{1 - 3 + 5}{3}, \frac{1 + 3 + 3}{3} \right)\) Simplifying the coordinates, we get: Centroid: \(\left( \frac{3}{3}, \frac{7}{3} \right) = (1, \frac{7}{3})\) The centroid of the triangle is (1, 7/3), so our answer is option (b).

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Most popular questions from this chapter

The equation of the bisector of acute angle between the lines \(3 x-4 y+7=0\) and \(-12 x-5 y+2=0\) is (a) \(11 \mathrm{x}-3 \mathrm{y}+9=0\) (b) \(3 x+11 y-13=0\) (c) \(3 x+11 y-3=0\) (d) \(11 x-3 y+2=0\)

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