Question

If the point C(-1,2) divides internally the line segment joining A (2,5) and B in
the ratio 3: 4,find the co-ordinates of B.​

Answer 1

☯ Let the Co-ordinate of point B be (x, y) and AC: BC = 3:4.

$\star$ DIAGRAM:

⠀⠀$\setlength{\unitlength}{14mm}\begin{picture}(7,5)(0,0)\thicklines\put(0,0){\line(1,0){5}}\put(5.1, - 0.3){\sf B}\put( - 0.2, - 0.3){\sf A}\put(5.2, 0){\sf (x, y)}\put( - 0.7, 0){\sf (2,5)}\put(2.3, 0.2){\sf C}\put(2.2, - 0.3){\sf (-1,2)}\put(5, 0){\circle*{0.1}}\put(2.4, 0){\circle*{0.1}}\put(0, 0){\circle*{0.1}}\put(1,0.2){\sf 3}\put(3.5, 0.2){\sf 4}\end{picture}$⠀⠀

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━

⠀⠀⠀

Given that,

• The point C(-1, 2) divide internally the line segment joining points A(2, 5) and B in the ratio 3:4.

$\underline{\bigstar\:\boldsymbol{Using\:section\:formula\::}}$

$\dag\:\boxed{\sf{\pink{\Big(x, y \Big) = \Bigg(\dfrac{mx_2 + nx_1}{m + n} \dfrac{my_2 + ny_1}{m + n}\Bigg)}}}$

$:\implies\sf \dfrac{3x + 4(2)}{3 + 4} = -1 \\\\\\:\implies\sf \dfrac{3x + 8}{7} \\\\\\\\:\implies\sf 3x + 8 = -7 \\\\\\:\implies\sf x = \cancel\dfrac{-15}{\:3} \\\\\\:\implies{\underline{\boxed{\frak{\purple{x = -5}}}}}\:\bigstar$

Similarly,

⠀⠀⠀

$:\implies\sf \dfrac{3y + 4(5)}{3 + 4} = 2\\\\\\:\implies\sf \dfrac{3y + 20}{7} = 2 \\\\\\:\implies\sf 3y + 20 = 14\\\\\\:\implies\sf 3y = -6\\\\\\:\implies\sf y = \cancel\dfrac{-6}{\:3}\\\\\\:\implies{\underline{\boxed{\frak{\purple{y = -2}}}}}\:\bigstar$

⠀⠀

$\therefore\:{\underline{\sf{Hence, \ the \ Co-ordinate \ of \ point \ B \ is\: \bf{\Big(-5, -2 \Big)}.}}}$

Answer 2

$\underline\green{\bold{Given \: Question :- }}$

• If the point C(-1,2) divides internally the line segment joining A (2,5) and B in the ratio 3: 4,find the co-ordinates of B.

___________________________________________

$\huge \orange{AηsωeR} ✍$

${ \boxed {\bf{Given}}}$

• The point C(-1,2) divides internally the line segment joining A (2,5) and B in the ratio 3: 4,

${ \boxed {\bf{To Find}}}$

• The co-ordinates of B.

${ \boxed {\bf{Formula \: used :- }}}$

☆ Section Formula

Let us consider a line segment joining the points

$\sf \: A(x_1,y_1) \: and \: B(x_2,y_2)$

Let C(x, y) be the point which divides the line segment joining A and B, in the ration m : n, then coordinates of C is given by

$\bf \:( x, y) = (\dfrac{nx_1+mx_2}{m + n} , \dfrac{ny_1+my_2}{m + n} )$

${ \boxed {\bf{Solution}}}$

$\bf \:Let \: coordinates \: of \: B \: be \: (a, b).$

$\begin{gathered}\bf\red{According \: to \: statement}\end{gathered}$

☆ The point C(-1,2) divides internally the line segment joining A (2,5) and B in the ratio 3: 4.

Using Section Formula,

$\bf \:( x, y) = (\dfrac{nx_1+mx_2}{m + n} , \dfrac{ny_1+my_2}{m + n} )$

☆ On substituting the values of

$\bf \:  x_1 =2 ,y_1 =5 ,x_2=a ,y_2=b \\ \bf \:  x= - 1,y =2 ,m= 3 ,n=4$

☆ we get

$\bf \:( - 1, 2) = (\dfrac{4 \times 2 + 3 \times a}{3 + 4} , \dfrac{4 \times 5 + 3 \times b}{3 + 4} )$

$\bf \:( - 1, 2) = (\dfrac{8 + 3a}{7} , \dfrac{20 + 3 b}{7} )$

$\bf \:\dfrac{8 + 3a}{7} = - 1 \: and \: \dfrac{20 + 3 b}{7} = 2$

$\bf \:  ⟼ 8 + 3a = - 7 \: and \: 20 + 3b = 14$

$\bf \:  ⟼ 3a = - 15 \: and \: 3b = - 6$

$\bf\implies \:a = - 5 \: and \: b \: = - 2$

$\large{\boxed{\boxed{\bf{Hence, \: coordinates \: of \: B \: is \: (-5, -2)}}}}$