Let the three vertices of the triangle be, and.
The three midpoints are given. Let these points be, and.
Let us now equate these points using the earlier mentioned formula,
Equating the individual components we get,
Using the midpoint of another side we have,
Equating the individual components we get,
Using the midpoint of the last side we have,
Equating the individual components we get,
Adding up all the three equations which have variable ‘x’ alone we have,
Substituting in the above equation we have,
Therefore,
And
Adding up all the three equations which have variable ‘y’ alone we have,
Substituting in the above equation we have,
Therefore,
And
Therefore the co-ordinates of the three vertices of the triangle are.
The point P is the point of trisection of the line segment AB. So, P divides AB in the ratio 1: 2
Now we will use section formula to find the co-ordinates of unknown point A as,
Equate the individual terms on both the sides. We get,
Similarly, the point Q is the point of trisection of the line segment AB. So, Q divides AB in the ratio 2: 1
Now we will use section formula to find the co-ordinates of unknown point A as,
Equate the individual terms on both the sides. We get,
The point P is the point of trisection of the line segment AB. So, P divides AB in the ratio 1: 2
Now we will use section formula to find the co-ordinates of unknown point A as,
Therefore, co-ordinates of point P is(3,−2)
It is given that point P lies on the line whose equation is
So point A will satisfy this equation.
So,
Therefore mid-point D of side BC can be written as,
Now equate the individual terms to get,
So co-ordinates of D is (ii)We have to find the co-ordinates of a point P which divides AD in the ratio 2: 1 internally.Now according to the section formula if any point P divides a line segment joining and in the ratio m: n internally than,
P divides AD in the ratio 2: 1. So,
(iii)We need to find the mid-point of sides AB and AC. Let the mid-points be F and E for the sides AB and AC respectively.
Therefore mid-point F of side AB can be written as,
So co-ordinates of F is
Similarly mid-point E of side AC can be written as,
So co-ordinates of E is
Q divides BE in the ratio 2: 1. So,
Similarly, R divides CF in the ratio 2: 1. So,
(iv)We observe that that the point P, Q and R coincides with the centroid. This also shows that centroid divides the median in the ratio 2: 1.
Since ABCD is a parallelogram, the diagonals bisect each other. Therefore the mid-point of the diagonals of the parallelogram will coincide.
In general to find the mid-point of two pointsand we use section formula as,
The mid-point of the diagonals of the parallelogram will coincide.
So,
Therefore,
Now equate the individual terms to get the unknown value. So,
Similarly,
Therefore,
So, P divides the line segment joining the points A(3, −5) and B(−4, 8) in the ratio k : 1. Using the section formula, we get
Coordinates of P =
Since P lies on the line x + y = 0, so
Hence, the value of k is 1/2 .
∴ Coordinates of P = ( -6 , 2)
Suppose P divides the line segment joining the points C(−9, −4) and D(−4, y) in the ratio k : 1.
Using section formula, we get
Coordinates of P =
( -6 , 2)
Now,
So, P divides the line segment CD in the ratio 3 : 2.
Putting k
⇒ 3y - 8 =10
⇒ 3y=18
⇒y=6
Hence, the value of y is 6.
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