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Question
Show that the line segment joining the points (–5, 8) and (10, −4) is trisected by the co-ordinate axes.
Solution
Let the points A (–5, 8) and B (10, −4).
Let P and Q be the two points on the axis which trisect the line joining the points A and B.
∵ AP = PQ = QB
∴ AP : PB = 1 : 2 and AQ : QB = 2 : 1
Now, co-ordinates of P will be,
`x = (1 xx 10 + 2 xx (-5))/(1 + 2)`
= `(10 - 10)/3`
= 0
y = `(1 xx (-4) + 2 xx 8)/(1 + 2)`
= `(-4 + 16)/3`
= `12/3`
= 4
∴ Co-ordinates of P are (0, 4)
Co-ordinates of Q will be,
`x = (2 xx 10 + 1 xx (-5))/(2 + 1)`
= `(20 - 5)/3`
= `15/3`
= 5
`y = (2 xx (-4) + 1 xx 8)/(2 + 1)`
= `(-8 + 8)/3`
= `0/3`
= 0
∴ Co-ordinates of Q are (5, 0)
Hence Proved.
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