need solution for this!!
Answers
Solution :
For a triangle OPQ lying on the co-ordinate plane , the vertices are -
O > (0,0)
P > (3,4)
Q > (6,0)
Now, there lies a point R inside the triangles such that the triangles OPR, PQR, OQR are of equal area .
We have to find the coordinates of R.
Let us assume that R initially has the coordinates (x,y)
Area of ∆ OPR = Area of ∆ PQR = Area of ∆ OQR
Hence point R is the centroid.
R_x coordinate : ( x_1 + x_2 + x_3)/3
> 9/3
> 3
R_y coordinate : ( y_1 + y_2 + y_3 )/3
> 4/3
Answer : The required co-ordinates are (3,4/3) .
This is the answer .
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Additional Information : Solving by finding the areas seperately will be quite long
Step-by-step explanation:
Given:-
O(0,0) P(3, 4) and Q(6,0) are the vertices of a triangle and R is the interior point in the triangle such that ar(∆OPR)=Aar(∆PQR)=ar(∆OQR)
To find:-
Find the coordinates of the point R
Solution:-
Given points are O(0,0) P(3, 4) and Q(6,0)
Let the Coordinates of the point R be (x,y)
We know that
Area of a triangle formed by the three points
(x1, y1),(x2, y2) and (x3, y3) is
(1/2) | x1(y2-y3)+x2(y3-y1)+x3(y1-y2) | sq.units
i) Area of ∆OPR :-
(x1, y1)=(0,0)=>x1 = 0 and y1 = 0
(x2, y2)=(3,4)=>x2 = 3 and y2 = 4
(x3, y3)=(x,y)=>x3 = x and y3 = y
Area = (1/2) |0(4-y)+3(y-0)+x(0-4) |
=>(1/2) | 0+3y-4x |
=>(1/2) | 3y-4x |
Area. = (3y-4x)/2 sq.units --------------(1)
ii)Area of ∆ PQR :-
(x1, y1)=(3,4)=>x1 = 3 and y1 = 4
(x2, y2)=(6,0)=>x2 = 6and y2 = 0
(x3, y3)=(x,y)=>x3 = x and y3 = y
Area = (1/2) | 3(0-y)+6(y-4)+x(4-0) |
=>(1/2) | -3y+6y-24+4x |
=>(1/2) | 3y+4x-24 |
Area = (4x+3y-24)/2 sq.units --------(2)
iii)Area of ∆OQR:-
(x1, y1)=(0,0)=>x1 = 0 and y1 = 0
(x2, y2)=(6,0)=>x2 = 6and y2 = 0
(x3, y3)=(x,y)=>x3 = x and y3 = y
Area = (1/2) | 0(0-y)+6(y-0)+x(0-0) |
=>(1/2) | 0+6y+0 |
=>6y/2
Area = 3y sq.units --------------(3)
ar(∆OPR)=Aar(∆PQR)=ar(∆OQR)
=> (3y-4x)/2 = (4x+3y-24)/2 = 3y
On taking (1)&(3)
=>(3y-4x)/2 = 3y
=>3y-4x = 6y
=>6y-3y+4x = 0
=>4x+3y = 0
=>4x = -3y-------(4)
On taking (2)&(3)
(4x+3y-24)/2 = 3y
= 4x+3y-24 = 6y
=>-3y+3y-24 = 6y (from (4))
=>6y = -24
=>y = -24/6
y = -4
from (4)
4x = -3(-4)
=>4x = 12
=>x = 12/4
=>x = 3
The coordinates of R = (3,-4)
Answer:-
The coordinates of the point R = (3,-4)
Check:-
If (x,y)=(3,-4) then
Area of ∆OPR =| (3y-4x)|/2 sq.units
=>|3(-4)-4(3) |/2
=>| -12-12 |/2
=> | -24 | /2
=>24/2
= 12 sq.units
Area of ∆PQR |(4x+3y-24)|/2
=>| 4(3)+3(-4)-24|/2
=>| 12-12-24 |/2
=>|-24|/2
=>24/2
=>12 sq.units
Area of ∆OQR = |3y|
=>| 3(-4)|
=>| -12|
=>12 sq.units
all are equal
so verified the given relations
Used formula:-
Area of a triangle formed by the three points
(x1, y1),(x2, y2) and (x3, y3) is
(1/2) | x1(y2-y3)+x2(y3-y1)+x3(y1-y2) | sq.units