contestada

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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Respuesta :

AestheticDevil

Answer:

  • Coordinate of 0 = [tex]\bigg( \dfrac{5a-b}{5},\:\dfrac{5a+b}{5}\bigg)[/tex]

Step-by-step 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,

  • [tex]\dfrac{m}{n}=\dfrac{3}{2}[/tex]
  • Coordinate of a = (x[tex]_{1}[/tex], y[tex]_{1}[/tex])
  • Coordinate of b = (x[tex]_{2}[/tex], y[tex]_{2}[/tex])

As we know that,

[tex]\implies\:\:x = \dfrac{mx_{2} + nx_{1}}{m + n}[/tex]

[tex]\implies\:\:x = \dfrac{(a-b)3 + (a+b)2}{3+2}[/tex]

[tex]\implies\:\:x = \dfrac{3a - 3b + 2a + 2b}{5}[/tex]

[tex]\implies\:\:\red{x = \dfrac{5a - b}{5}}[/tex]

Similarly,

[tex]\implies\:\:y = \dfrac{my_{2}+ny_{1}}{m+n}[/tex]

[tex]\implies\:\:y = \dfrac{(a+b)3+(a-b)2}{3+2}[/tex]

[tex]\implies\:\:y = \dfrac{3a+3b+2a-2b}{5}[/tex]

[tex]\implies\:\:\red{y = \dfrac{5a+b}{5}}[/tex]

Hence,

Coordinate of 0 is [tex]\blue{\bigg( \dfrac{5a-b}{5},\:\dfrac{5a+b}{5}\bigg)}[/tex]

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