Question

The line segment joining the point A(3,2)and B(5,1)is devided at a point p in the ratio 1:2initaly find the co- ordinate of 'p'​

Answer 1

Answer:-

Given:

The line segment joining the points A(3 , 2) , B(5 , 1) is divided at a point P internally in the ratio 1 : 2.

Using section formula;

If The line segment joining the points (x₁ , y₁) , (x₂ , y₂) is divided by the point P internally in the ratio m : n , then co - ordinates of point P are :

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

Let,

• m = 1

• n = 2

• x₁ = 3

• y₁ = 2

• x₂ = 5

• y₂ = 1

Hence,

$\implies \sf \: (x \: \: ,\: \: y) = \bigg( \dfrac{(1)(5) + (2)(3) }{1 + 2} \: \: , \: \: \dfrac{(1)(1) + (2)(2)}{1 + 2} \bigg) \\ \\ \implies \sf \: (x \: \: , \: \: y) = \bigg( \dfrac{5 + 6 }{3} \: \: ,\: \: \dfrac{1 + 4}{3} \bigg) \\ \\ \implies \boxed{ \sf \: (x \: \: , \: \: y) = \bigg( \dfrac{11}{3} \: \: , \: \: \dfrac{5}{3} \bigg)}$

∴ The co - ordinates of the point are (11/3 , 5/3)

Answer 2

$\tt \huge \underline{solution}$

$\tt{Given \: that \: point \: divides \: the \: } \\ \tt{ line \: segment \: joining \: the \: points } \\ \tt{A \: (3,2) \: and \: B \: (5,1) \: in \: the \: ratio \: } \\ \tt{ 1:2} </p><p> \\ \tt {The \: coordinate \: of \: P \: is} , \\ \tt{By \: section \: formula}</p><p></p><p>$

$\\ \tt{( \frac{mx2 + nx1}{m + n} \: \: \: ,\frac{my2 + ny1}{m + n}) }$

$\tt{here \: m : n = 1 : 2 \: (x1 \: y1) = (3 \: \: \: 2)}$

$\\ \tt{and \: (x2 \: y2) = (5 \: \: 1)}$

$\\ \tt{( \frac{1 \times 5 \ + 2\times 3}{1 + 2} \: \: \frac{1 \times 1 + 2 \times 2}{1 + 2} ) }$

$\tt{ \:the \: coordinate \: are\: }$

$\\ \tt{ (\frac{11}{3} \: \: \: \: \: \frac{5}{3}) }$

$\\ \tt{3 (\frac{11}{3}) - 18 (\frac{5}{3} ) + k =0 }$

$\tt{k -19 = 0 \implies \: k = 19 }$