Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of ∆ ABC.

(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?

[Note: The point which is common to all the three medians is called the centroid

and this point divides each median in the ratio 2 : 1.]

(v) If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of triangle ABC, find the coordinates of the centroid of the triangle.

Asked by Pragya Singh | 1 year ago |  138

Solution :-

(i) Coordinates of D can be calculated as follows:

Coordinates of D = $$\dfrac{ (6+1)}{2},\dfrac{(5+4)}{2}$$ = ($$\dfrac{7}{2}$$, $$\dfrac{9}{2}$$)

So, D is ($$\dfrac{7}{2}$$, $$\dfrac{9}{2}$$)

(ii) Coordinates of P can be calculated as follows:

Coordinates of P =

$$2\dfrac{7}{2}+\dfrac{1(4)}{(2+1)},2\dfrac{9}{2}+\dfrac{ 1(2)}{(2 + 1)}= \dfrac{11}{3}, \dfrac{11}{3}$$

= ($$\dfrac{11}{3}$$, $$\dfrac{11}{3}$$)

So, P is ($$\dfrac{11}{3}$$, $$\dfrac{11}{3}$$)

(iii) Coordinates of E can be calculated as follows:

Coordinates of E = $$\dfrac{(4+1)}{2}$$$$\dfrac{(2+4)}{2}$$

= ($$\dfrac{5}{2}$$, $$\dfrac{6}{2}$$) = ($$\dfrac{5}{2}$$ , 3)

So, E is ($$\dfrac{5}{2}$$ , 3)

Point Q and P would be coincident because medians of a triangle intersect each

other at a common point called centroid. Coordinate of Q can be given as follows:

Coordinates of Q = $$2\dfrac{5}{2}+\dfrac{1(6)}{(2 + 1)}$$

2(3) +$$\dfrac{ 1(5)}{(2 + 1) }$$+  = ($$\dfrac{11}{3}$$,$$\dfrac{11}{3}$$)

F is the mid- point of the side AB

Coordinates of F = $$\dfrac{(4+6)}{2},\dfrac{ (2+5)}{2}=5,\dfrac{7}{2}$$

Point R divides the side CF in ratio 2:1

Coordinates of R = $$\dfrac{2(5) + 1(1)}{(2 + 1)}$$

$$2\dfrac{7}{2}+\dfrac{1(4)}{(2 + 1)}$$ = ($$\dfrac{11}{3}$$, $$\dfrac{11}{3}$$)

(iv) Coordinates of P, Q and R are same which shows that medians intersect

each other at a common point, i.e. centroid of the triangle.

(v) If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of triangle ABC,

the coordinates of centroid can be given as follows:

x = $$\dfrac{(x_1 + x_2 + x_3)}{3}$$ and y =$$\dfrac{ (y_1 + y_2 + y_3)}{3}$$

Answered by Abhisek | 1 year ago

Related Questions

In the determine whether the given quadratic equations have real roots and if so, And the roots 3x2 – 2x + 2 = 0

In the determine whether the given quadratic equations have real roots and if so, And the roots 3x2 – 2x + 2 = 0

Find the point on x-axis which is equidistant from the points (-2, 5) and (2, -3).

Find the point on x-axis which is equidistant from the points (-2, 5) and (2, -3).

Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square.

(i) Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square.

(ii) Prove that the points A (2, 3), B (-2, 2), C (-1, -2) and D (3, -1) are the vertices of a square ABCD.

(iii) Name the type of triangle PQR formed by the point $$P(\sqrt{2} , \sqrt{2}), Q(- \sqrt{2}, – \sqrt{2)} and\; R (-\sqrt{6} , \sqrt{6} )$$