Question

Find the co-ordinates of the points of trisection of the line segment the A(6,-2) and B(-8,10).
Please write the steps

Answer 1

Step-by-step explanation:

Using the section formula, if a point (x,y) divides the line joining the points (x

1

,y

1

) and (x

2

,y

2

) in the ratio m:n, then

(x,y)=(

m+n

mx

2

+nx

1

,

m+n

my

2

+ny

1

)

Let P(x

1

,y

1

) and Q (x

2

,y

2

) divide the line AB into 3 equal parts.

AP=PQ=QB

PB

AP

=

PQ+QB

AP

=

AP+AP

AP

=

2AP

AP

=

2

1

So, P divides AB in the ratio 1:2.

Therefore,

x

1

=

2+1

1×(−4)+2×2

=0

y

1

=

2+1

1×(−8)+2×7

=2

(x

1

y

1

)=(0,2)

Also,

QB

AQ

=

QB

AP+PB

=

QB

QB+QB

=

1

2

x

2

=

2+1

2×(−4)+1×2

=−2

y

2

=

2+1

2×(−8)+1×7

=−3

(x

2

,y

2

)=(−2,−3)

Thus, the coordinates of the point of intersection are:

P(x

1

,y

1

)=(0,2)

Q(x

2

,y

2

)=(−2,−3)

Answer 2

We know that trisection divides the line segment in the ratio 1:2 or 2:1 internally. Let the line segment be PQ. Let A (x, y) divides the line segment PQ in the ratio 1:2. Now by using the section formula let us find the A (x, y) coordinates, where the ratio m: n=1:2 and point are P (-3, 4) and Q (4, 5).