Question

Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of ΔABC. (i) The median from A meets BC at D. Find the coordinates of point D. (ii) Find the coordinates of the point P on AD such that AP: PD = 2:1 (iii) Find the coordinates of point Q and R on medians BE and CF respectively such that BQ: QE = 2:1 and CR: RF = 2:1. (iv) What do you observe? (v) If A(x1, y1), B(x2, y2), and C(x3, y3) are the vertices of ΔABC, find the coordinates of the centroid of the triangle.

Answer 1

Solution:

(1) According to the question since AD is median on Side BC,so it's bisect the side

or we can say the point D is mid-point of side BC

Apply mid-point formula D(x,y)

$x = \frac{6 + 1}{2} = \frac{7}{2} \\ \\ y = \frac{5 + 4}{2} = \frac{9}{2}$
D(7/2,9/2)

(ii) AP:PD = 2:1

To find the coordinates of point P. apply section formula

$x = \frac{2 \times \frac{7}{2} + 1 \times4 }{2 + 1} \\ \\ = \frac{11}{3} \\ \\ y = \frac{2 \times \frac{9}{2} + 1 \times2 }{3} \\ \\ = \frac{11}{3} \\ \\ P( \frac{11}{3}, \frac{11}{3} )$

(iii)As it is not clear in question,but accordingly BE is median on Side AC,and the point Q will intersect in the ratio 2:1

Point E is mid-point of Side AC

E(5/2,3)

Coordinates of point Q
$x = \frac{2 \times \frac{5}{2} +6 \times 1 }{3} \\ \\ x = \frac{11}{3} \\ \\ y = \frac{2 \times 3 + 1 \times5 }{3} \\ \\ y = \frac{11}{3} \\ \\ Q( \frac{11}{3}, \frac{11}{3} )$

By the same way we can find the coordinates of point R(11/3,11/3)

iv) We observed that all the points P,Q,R are concide ,and where the medians intersect is known as centroid of triangle.

v) Coordinates of centroid of triangle

$x = \frac{x1 + x2 + x3}{3} \\ \\ y = \frac{y1 + y2 + y3}{3}$
Here in ∆ABC

$x = \frac{6 + 4 + 1}{3} = \frac{11}{3} \\ \\ y = \frac{2 + 5 + 4}{3} \\ \\ = \frac{11}{3}$
coordinates of centroid are (11/3,11/3)

Hope it helps you.

Answer 2

Answer:

Solution:

(1) According to the question since AD is median on Side BC,so it's bisect the side

or we can say the point D is mid-point of side BC

Apply mid-point formula D(x,y)

D(7/2,9/2)

(ii) AP:PD = 2:1

To find the coordinates of point P. apply section formula

(iii)As it is not clear in question,but accordingly BE is median on Side AC,and the point Q will intersect in the ratio 2:1

Point E is mid-point of Side AC

E(5/2,3)

Coordinates of point Q

By the same way we can find the coordinates of point R(11/3,11/3)

iv) We observed that all the points P,Q,R are concide ,and where the medians intersect is known as centroid of triangle.

v) Coordinates of centroid of triangle

Here in ∆ABC

coordinates of centroid are (11/3,11/3)

Hope it helps you.