Question

Find the coordinates of the point C which divides the points A (-3, -1), B (-1, 0) in a ratio of 2:3 externally, with the help of section formula.

Answer 1

★ Given:-        

• $\dashrightarrow\tt{\bigg(\:x_{1}\:‚\:y_{1}\:\bigg)\:=\:\bigg(\:-3\:‚\:-1\:\bigg)}$

• $\dashrightarrow\tt{\bigg(\:x_{2}\:‚\:y_{2}\:\bigg)\:=\:\bigg(\:-1\:‚\:0\:\bigg)}$

• $\dashrightarrow\tt{m\::\:n\:=\:2\::\:3}$

★ To find:-    

• $\dashrightarrow\tt{ The \:coordinates \:of\: the \:point\: C\:\bigg(\:x\:‚\:y\:\bigg)}$

★ Solution:-      

Here, we have given that the points A (-3, -1), B (-1, 0) in a ratio of 2:3 externally, and we have to find out the coordinates of the point C.

So, to find out the coordinates of point C we will use the section formula.

                  ______________

$\qquad\tt{:\implies\:C\:\bigg(\:x\:,\:y\:\bigg)\:=\:\left(\:\dfrac{mx_{2}\:-\:nx_{1}}{m\:-\:n}\:,\:\dfrac{my_{2}\:-\:ny_{1}}{m\:-\:n}\:\right)}$

$\qquad\tt{:\implies\:C\:\bigg(\:x\:,\:y\:\bigg)\:=\:\left\{\:\dfrac{(\:\:2\:(-1)\:\:-\:\:3\:(-3)}{2\:-\:3}\:,\:\dfrac{2\:(0)\:-\:3\:(-1)}{2\:-\:3}\:\right\}}$

$\qquad\tt{:\implies\:C\:\bigg(\:x\:,\:y\:\bigg)\:=\:\left\{\:\dfrac{(-2\:+\:9)}{-1}\:,\:\dfrac{0\:-\:3}{-1}\:\right\}}$

$\qquad\tt{:\implies\:C\:\bigg(\:x\:,\:y\:\bigg)\:=\:\left\{\:\dfrac{(-2\:+\:9)}{-1}\:,\:\dfrac{3}{-1}\:\right\}}$

$\qquad\tt{:\implies\:C\:\bigg(\:x\:,\:y\:\bigg)\:=\:\left\{\:\dfrac{7}{-1}\:,\:\dfrac{3}{-1}\:\right\}}$

$\qquad\tt{:\implies\:C\:\bigg(\:x\:,\:y\:\bigg)\:=\:\left\{\:\dfrac{-7}{1}\:,\:\dfrac{-3}{1}\:\right\}}$

$\qquad\tt{:\implies\:C\:\bigg(\:x\:,\:y\:\bigg)\:=\:\bigg(\:-7\:,\:-3\:\bigg)}$

. ° .  the coordinates of the point C is ( -7 , -3 ).

                         ______________

      ★ Important Section formula ★

Section formula for internal division is:

$\qquad\tt{:\implies\:P\:\bigg(\:x\:,\:y\:\bigg)\:=\:\left(\:\dfrac{mx_{2}\:+\:nx_{1}}{m\:+\:n}\:,\:\dfrac{my_{2}\:+\:ny_{1}}{m\:+\:n}\:\right)}$

Section formula for external division is:

$\qquad\tt{:\implies\:P\:\bigg(\:x\:,\:y\:\bigg)\:=\:\left(\:\dfrac{mx_{2}\:-\:nx_{1}}{m\:-\:n}\:,\:\dfrac{my_{2}\:-\:ny_{1}}{m\:-\:n}\:\right)}$

Midpoint formula is:

$\qquad\tt{:\implies\:M\:\bigg(\:x\:,\:y\:\bigg)\:=\:\left(\:\dfrac{x_{2}\:+\:x_{1}}{2}\:,\:\dfrac{y_{2}\:-\:y_{1}}{2}\:\right)}$

Answer 2

Answer:

Refer to the attachement given above...