Question

need solution for this!!​

Answer 1

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

Answer 2

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