Question

Find the coordinates of point which divides the line segment joining the points (a+b,a-b) and (a-b,a+b) in the ratio 3:2 interally

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Answer 1

• Coordinate of 0 =
  `\bigg( (5a-b)/(5),\:(5a+b)/(5)\bigg)`

Explanation:

• Let AOB be the line segment
• Point 'O' divides the line segment into m:n = 3:2

Given:

• Coordinate of a = (a+b, a-b)
• Coordinate of b = (a-b, a+b)
• Ratio = 3:2

ToFind:

• Coordinate of 0 (zero)

Solution:

ATQ,

• 
  `(m)/(n)=(3)/(2)`
• Coordinate of a = (x
  `_(1)`, y
  `_(1)`)
• Coordinate of b = (x
  `_(2)`, y
  `_(2)`)

As we know that,

`\implies\:\:x = (mx_(2) + nx_(1))/(m + n)`

`\implies\:\:x = ((a-b)3 + (a+b)2)/(3+2)`

`\implies\:\:x = (3a - 3b + 2a + 2b)/(5)`

`\implies\:\:\red{x = (5a - b)/(5)}`

Similarly,

`\implies\:\:y = (my_(2)+ny_(1))/(m+n)`

`\implies\:\:y = ((a+b)3+(a-b)2)/(3+2)`

`\implies\:\:y = (3a+3b+2a-2b)/(5)`

`\implies\:\:\red{y = (5a+b)/(5)}`

Hence,

Coordinate of 0 is
`\blue{\bigg( (5a-b)/(5),\:(5a+b)/(5)\bigg)}`