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 (x_{1}, y_{1}), B (x_{2}, y_{2}) and C (x_{3}, y_{3}) are the vertices of triangle ABC, find the coordinates of the centroid of the triangle.

Asked by Pragya Singh | 1 year ago | 138

**(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 (x_{1}, y_{1}), B (x_{2}, y_{2}) and C (x_{3}, y_{3}) 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 agoIn the determine whether the given quadratic equations have real roots and if so, And the roots 3x^{2} – 2x + 2 = 0

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

Answer the following questions:-

**(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 \( P(\sqrt{2} , \sqrt{2}), Q(- \sqrt{2}, – \sqrt{2)} and\; R (-\sqrt{6} , \sqrt{6} )\).

Find a point on the x-axis which is equidistant from the points (7, 6) and (-3, 4).

Find the coordinates of the circumcentre of the triangle whose vertices are (3, 0), (-1, -6) and (4, -1). Also, find its circumradius.