Question

P and Q are the points on the line segment joining the points A(3, -1) and B(-6, 5) such that AP=PQ=QB. Find the Co-ordinate of P and Q.​

Answer 1

Given :-

→ P and Q are the points on the line segment joining the points A(3, -1) and B(-6, 5) such that AP=PQ=QB.

To find :-

→ The co-ordinates of P and Q.

Solution :-

Given that

The coordinates of the point A = (3,-1)

The coordinates of the point B = (-6,5)

Let the coordinates of the point P be (x,y)

Let the coordinates of the point Q be (r,s)

Given that AP = PQ = QB

A________P________Q________B

We have,

AP : PQ = 1:1

PQ : QB = 1:1

AP : PB = 1:2

AQ : QB = 2:1

P divides AB in the ratio 1:2

We know that

Section formula: The coordinates of the point which divides the linesegment joining the points (x1, y1) and (x2, y2) in the ratio m1:m2 internally is ({m1x2+m2x1}/(m1+m2)),{m1y2+m2y1}/(m1+m2))

Let (x1, y1) = A(3,-1) => x1 = 3 and y1 = -1

Let (x2, y2) = B(-6,5) => x2 = -6 and y2 = 5

Let m1:m2 = AP:PB = 1:2 => m1 = 1 and

m2 = 2

The coordinates of the point P(x,y)

= ( {(1×-6)+(2×3)}/(1+2) , {1×5)+(2×-1)}/(1+2) )

=> (x,y) = ( (-6+6)/3 , (5-2)/3 )

=> (x,y) = (0/3,3/3)

=> (x,y)= (0,1)

The coordinates of P = (0,1)

And

Q divides AB in the ratio 2:1

Let (x1, y1) = A(3,-1) => x1 = 3 and y1 = -1

Let (x2, y2) = B(-6,5) => x2 = -6 and y2 = 5

Let m1:m2 = AQ:QB = 2:1 => m1 = 2 and m2 = 1

The coordinates of the point Q(r,t)

= ( {(2×-6)+(1×3)}/(2+1) , {2×5)+(1×-1)}/(2+1) )

=> (r,t) = ( (-12+3)/3 , (10-1)/3 )

=> (r,t) = (-9/3,9/3)

=> (r,t) = (-3,3)

The coordinates of Q = (-3,3)

Answer :-

•The coordinates of P = (0,1)

•The coordinates of Q = (-3,3)

Used formulae:-

Section Formula:-

• The coordinates of the point which divides the linesegment joining the points (x1, y1) and (x2, y2) in the ratio m1:m2 internally is

({m1x2+m2x1}/(m1+m2)),{m1y2+m2y1}/(m1+m2))

Note :-

P and Q are the Trisectional points of the given linesegment AB.

• The points which divides the linesegment into three equal parts or in the ratio 1:2 or 2:1 are called Trisectional Points or Points of Trisection.

Answer 2

Step-by-step explanation:

Given :-

→ P and Q are the points on the line segment joining the points A(3, -1) and B(-6, 5) such that AP=PQ=QB.

To find :-

→ The co-ordinates of P and Q.

Solution :-

Given that

The coordinates of the point A = (3,-1)

The coordinates of the point B = (-6,5)

Let the coordinates of the point P be (x,y)

Let the coordinates of the point Q be (r,s)

Given that AP = PQ = QB

A________P________Q________B

We have,

AP : PQ = 1:1

PQ : QB = 1:1

AP : PB = 1:2

AQ : QB = 2:1

P divides AB in the ratio 1:2

We know that

Section formula: The coordinates of the point which divides the linesegment joining the points (x1, y1) and (x2, y2) in the ratio m1:m2 internally is ({m1x2+m2x1}/(m1+m2)),{m1y2+m2y1}/(m1+m2))

Let (x1, y1) = A(3,-1) => x1 = 3 and y1 = -1

Let (x2, y2) = B(-6,5) => x2 = -6 and y2 = 5

Let m1:m2 = AP:PB = 1:2 => m1 = 1 and

m2 = 2

The coordinates of the point P(x,y)

= ( {(1×-6)+(2×3)}/(1+2) , {1×5)+(2×-1)}/(1+2) )

=> (x,y) = ( (-6+6)/3 , (5-2)/3 )

=> (x,y) = (0/3,3/3)

=> (x,y)= (0,1)

The coordinates of P = (0,1)

And

Q divides AB in the ratio 2:1

Let (x1, y1) = A(3,-1) => x1 = 3 and y1 = -1

Let (x2, y2) = B(-6,5) => x2 = -6 and y2 = 5

Let m1:m2 = AQ:QB = 2:1 => m1 = 2 and m2 = 1

The coordinates of the point Q(r,t)

= ( {(2×-6)+(1×3)}/(2+1) , {2×5)+(1×-1)}/(2+1) )

=> (r,t) = ( (-12+3)/3 , (10-1)/3 )

=> (r,t) = (-9/3,9/3)

=> (r,t) = (-3,3)

The coordinates of Q = (-3,3)

Answer :-

•The coordinates of P = (0,1)

•The coordinates of Q = (-3,3)

Used formulae:-

Section Formula:-

• The coordinates of the point which divides the linesegment joining the points (x1, y1) and (x2, y2) in the ratio m1:m2 internally is

({m1x2+m2x1}/(m1+m2)),{m1y2+m2y1}/(m1+m2))

Note :-

P and Q are the Trisectional points of the given linesegment AB.

The points which divides the linesegment into three equal parts or in the ratio 1:2 or 2:1 are called Trisectional Points or Points of Trisection.

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