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

Answer:

• Coordinate of 0 = $\bigg( \dfrac{5a-b}{5},\:\dfrac{5a+b}{5}\bigg)$

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,

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

As we know that,

$\implies\:\:x = \dfrac{mx_{2} + nx_{1}}{m + n}$

$\implies\:\:x = \dfrac{(a-b)3 + (a+b)2}{3+2}$

$\implies\:\:x = \dfrac{3a - 3b + 2a + 2b}{5}$

$\implies\:\:\red{x = \dfrac{5a - b}{5}}$

Similarly,

$\implies\:\:y = \dfrac{my_{2}+ny_{1}}{m+n}$

$\implies\:\:y = \dfrac{(a+b)3+(a-b)2}{3+2}$

$\implies\:\:y = \dfrac{3a+3b+2a-2b}{5}$

$\implies\:\:\red{y = \dfrac{5a+b}{5}}$

Hence,

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