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Answer:

Answer:

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 and Q be the point of trisection of the line segment joining the points A(−5,8) and B(10,−4).

So, AP=PQ=QB. That is, AP:PQ:QB=1:1:1

⇒AP:PB=1:2

Co-ordinates of the point P are (

1+2

1⋅10+2⋅−5

,

1+2

1⋅−4+2⋅8

)=(

3

10−10

,

3

12

)=(0,4)

Therefore point P lie on y−axis

Since AP:PQ:QB=1:1:1

⇒AQ:QB=2:1

Co-ordinates of the point Q are (

2+1

2⋅10+1⋅−5

,

2+1

2⋅−4+1⋅8

)=(

3

20−5

,

3

−8+8

)=(5,0)

Therefore point Q lie on x−axis

Hence, the line segment joining the given points A and B is trisected by the co-ordinate axes.

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