Find the lengths of the medians of a triangle whose vertices are A(-1, 1), B(5, -3) and C(3, 5) .
13. Find the lengths of the medians of a triangle whose vertices are A(-1, 1), B(5, -3) and C(3, 5) .
Let the medians meet the lines BC, AC and AB at points be
\[P\left( x_1 , y_1 \right)\]
\[Q\left( x_2 , y_2 \right)\] and
\[R\left( x_3 , y_3 \right)\]respectively.
P is thus the mid point of line BC
\[P\left( x_1 , y_1 \right) = \left( \frac{5 + 3}{2}, \frac{5 - 3}{2} \right) = \left( 4, 1 \right)\]
\[AP = \sqrt{\left( -4 - 1 \right)^2 + \left( 1 - 1 \right)^2} = \sqrt{25} = 5\]
Q is the mid point of line AC.
\[Q\left( x_2 , y_2 \right) = \left( \frac{- 1 + 3}{2}, \frac{1 + 5}{2} \right) = \left( 1, 3 \right)\]
\[BQ = \sqrt{\left( 5 - 1 \right)^2 + \left( - 3 - 3 \right)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}\]
R is the mid point of AB.
\[R\left( x_3 , y_3 \right) = \left( \frac{- 1 + 5}{2}, \frac{1 - 3}{2} \right) = \left( 2, - 1 \right)\]
\[RC = \sqrt{\left( 3 - 2 \right)^2 + \left( -1 - 5 \right)^2} = \sqrt{1 + 36} = \sqrt{37}\]
Explanation:-
The solution calculates the coordinates of the midpoints of the sides of a triangle, and then uses these midpoints to find the lengths of the medians of the triangle.
Let’s consider a triangle with vertices A, B, and C, and let the medians meet the lines BC, AC, and AB at points P, Q, and R respectively.
The first step is to calculate the coordinates of point P, which is the midpoint of line BC. The coordinates of B and C are (5,-3) and (-1,1) respectively. Therefore, the coordinates of P can be found by taking the average of the x-coordinates and the y-coordinates of B and C separately:
x-coordinate of P = (5 + (-1))/2 = 2
y-coordinate of P = (-3 + 1)/2 = -1
So, P has coordinates (2,-1).
Next, the length of median AP is calculated using the distance formula, which gives the distance between two points. The coordinates of A are (-4,1). Therefore, the length of AP can be found using the distance formula:
AP = sqrt((-4 – 2)^2 + (1 – (-1))^2) = sqrt(36 + 4) = sqrt(40) = 2sqrt(10)
Moving on to point Q, which is the midpoint of line AC. The coordinates of A and C are (-4,1) and (-1,1) respectively. Therefore, the coordinates of Q can be found by taking the average of the x-coordinates and the y-coordinates of A and C separately:
x-coordinate of Q = (-4 + (-1))/2 = -2.5
y-coordinate of Q = (1 + 1)/2 = 1
So, Q has coordinates (-2.5,1).
Next, the length of median BQ is calculated using the distance formula. The coordinates of B are (5,-3). Therefore, the length of BQ can be found using the distance formula:
BQ = sqrt((5 – (-2.5))^2 + (-3 – 1)^2) = sqrt(89) = sqrt(99 + 22) = 3sqrt(17)
Finally, point R is the midpoint of line AB. The coordinates of A and B are (-4,1) and (5,-3) respectively. Therefore, the coordinates of R can be found by taking the average of the x-coordinates and the y-coordinates of A and B separately:
x-coordinate of R = (-4 + 5)/2 = 0.5
y-coordinate of R = (1 + (-3))/2 = -1
So, R has coordinates (0.5,-1).
Lastly, the length of median CR is calculated using the distance formula. The coordinates of C are (-1,1). Therefore, the length of CR can be found using the distance formula:
CR = sqrt((0.5 – (-1))^2 + (-1 – 1)^2) = sqrt(2.5^2 + 2^2) = sqrt(6.25 + 4) = sqrt(10.25) = sqrt(41)/2
In summary, the coordinates of the midpoints of the sides of the triangle are P(2,-1), Q(-2.5,1), and R(0.5,-1). The lengths of the medians of the triangle are AP = 2sqrt(10), BQ = 3sqrt(17), and CR = sqrt(41)/2.
Chapter 5. Co-ordinate Geometry – Problem set 5 (Page 122)