Question

us bindu ke nirdeshank gyat kijiye jo binduo (-1,7) aur (4,-3) ko milane wale rekhakhand ko 2:3 ke anupat me vobhajit karta hai.​

Answer 1

Question:-

Find the co - ordinates of the point which divides the line segment joining the points ( - 1 , 7) & (4 , - 3) in the ratio 2 : 3.

Answer:-

Let the co - ordinates of the point be (x , y) .

Given:

(x , y) divides the line segment joining the points ( - 1 , 7) & (4 , - 3) in the ratio 2 : 3.

Using section formula ,

i.e., The co - ordinates of the point which divides the line segment joining the points $\sf (x_1 , y_1)$ and $\sf (x_2 , y_2)$ in the ratio m : n are given by :

$\sf (x \: , \: y) \: = \: \bigg( \dfrac{mx _{2} + nx _{1}}{m + n} \: \: , \: \: \dfrac{my _{2} + ny _{1} }{m + n} \bigg)$

Let,

• m = 2

• n = 3

• x1 = - 1

• y1 = 7

• x2 = 4

• y2 = - 3

Hence,

$\sf \: (x \:, \: y) = \bigg( \dfrac{(2)(4) + (3)( - 1)}{2 + 3} \: \:, \: \: \dfrac{(2)(-3) + (3)( 7)}{2 + 3} \bigg)$

$\sf \implies \: (x \: , \: y) = \bigg( \dfrac{8 - 3}{5} \: \: ,\: \: \dfrac{-6 +21 }{5} \bigg)$

$\sf \implies \: (x \: ,\: y) = \bigg( \dfrac{5}{5} \: ,\: \dfrac{15}{5} \bigg)$

→ (x , y) = (1 , 3)

Therefore, the co - ordinates of the point are (1 , 3).

Answer 2

To Find :

• we have to find the cordinates of points which divides the line segment in Ratio of 2:3.

Solution :

Let p(x ,y) be the required point.

Using section formula :

$\boxed { \red{\underline{\bf\{{\frac{m_1x_2+m_2x_1}{m_1+ m_2},\frac{m_1y_2+m_2y_1}{m_1+ m_2}\}}}}}$

• m1 = 2
• m2 = 3
• x1 = -1
• x2 = 4
• y1 = 7
• y2 = -3

$\large\sf{ x = \{ \frac{2 \times 4 + 3 \times ( - 1)}{2 + 3} \} ,y = \{ \frac{2 \times ( - 3) + 3 \times 7}{2 + 3} \}}$

$\large(\sf{ x = \frac{8 - 3}{5} }$,$\sf{ y = \frac{-6 + 21}{5} ) }$

$\large(\sf{ x = \frac{5}{5} }$,$\sf{ y = \frac{15}{5} ) }$

$\large(\sf{ x = 1 }$,$\sf{ y = 3})$

$\large\dag \large { \red{\underline{\bf{Hence }}}}$

${ \green{\textrm{(1,3) are the required point. }}}$

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