Question

The points A(x₁, y₁)_, B(x₂, y₂) and C(x₃, y₃) are the vertices of ∆ABC.  Find the coordinates of points Q and R on medians BE and CF, respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1

Answer 1

According to the question,

The vertices of ΔABC = A, B and C

Coordinates of A, B and C = A(x₁, y₁), B(x₂, y₂), C(x₃, y₃)

Let the coordinates of a point Q be (p, q)

Given,

The point Q(p, q),

Divide the line joining `"B"(x_2, y_2)` and `"E"((x_1 + x_3)/2, (y_1 + y_3)/2)` in the ratio 2 : 1,

Then,

Coordinates of Q = `[(2 xx ((x_1 + x_3)/2) + 1 xx x_2)/(2 + 1), (2 xx ((y_1 + y_3)/2) + 1 xx y_2)/(2 + 1)]`

= `((x_1 + x_2 + x_3)/3, (y_1 + y_2 + y_3)/3)`

Since, BE is the median of side CA,

So BE divides AC into two equal parts.

∴ Mid-point of AC = Coordinate of E;

E = `((x_1 + x_3)/2, (y_1 + y_3)/2)`

So, the required coordinate of point Q;

Q = `((x_1 + x_2 + x_3)/3, (y_1 + y_2 + y_3)/3)`

Now,

Let the coordinates of a point E be (⍺, β)

Given,

Point `"R"(alpha, beta)` divide the line joining `"C"(x_3, y_3)` and `"F"((x_1 + x_2)/2, (y_1 + y_2)/2)` in the ratio 2 : 1,

Then the coordinates of R;

= `[(2 xx ((x_1 + x_2)/2) + 1 xx x_3)/(2 + 1), (2 xx ((y_1 + y_2)/2) + 1 xx y_3)/(2 + 1)]`

= `((x_1 + x_2 + x_3)/3, (y_1 + y_2 + y_3)/3)`

Since, CF is the median of side AB.

So, CF divides AB into two equal parts.

∴ Mid-point of AB = Coordinates of F;

F = `((x_1 + x_2)/2, (y_1 + y_2)/2)`

So, the required coordinate of point R;

= `((x_1 + x_2 + x_3)/3, (y_1 + y_2 + y_3)/3)`