Question

find the ratio in which Y axis divides the line segments joining the A(5,-6) and B(-1,-4) also find the coordinates of the point of division​

Answer 1

Question : -

Find the ratio in which y-axis divides the line segment joining the points A(5,-6) and B(-1,-4). Also find the co-ordinates of the point which divides the line ?

ANSWER

Given : -

A point on y-axis divides the line segment joining the points A(5,-6) and B(-1,-4)

Required to find : -

• Ratio in which the line got divided ?
• Co-ordinate of the point which divides the line segment AB ?

Formula used : -

Section formula

$\pink{ \sf{ \bf{p(x,y) = \left \lgroup \dfrac{m_1x_2+m_2x_1}{m_1+m_2} , \dfrac{ m_1 y_2 + m_2 y_1 }{ m_1 + m_2 }\right\rgroup}} }$

Solution : -

A point on y-axis divides the line segment joining the points A(5,-6) & B(-1,-4)

Since, it is mentioned that the point is on y-axis

The x co-ordinate of that point should be 0(zero).

This implies;

The points which divides the line segment AB be p(x,y)

Now,

Let's first find the ratio which in return can help us to find the y co-ordinate !

So,

According to problem;

$\sf{ A(x_1,y_1) = (5,-6) \qquad B(x_2,y_2) = (-1,-4) }$

Using the formula;

$\pink{ \sf{ \bf{p(x,y) = \left \lgroup \dfrac{m_1x_2+m_2x_1}{m_1+m_2} , \dfrac{ m_1 y_2 + m_2 y_1 }{ m_1 + m_2 }\right\rgroup}} }$

Substituting the values ;

$\sf p(0,y) = \left( \dfrac{ m_1( -1) + m_2( 5)}{m_1+m_2} , \dfrac{ m_1(- 4) + m_2 ( - 6)}{m_1+m_2} \right) \\ \\ \\ \sf p(0,y) = \left( \dfrac{-m_1+5m_2}{m_1+m_2} , \dfrac{-4m_1-6m_2}{m_1+m_2} \right) \\ \\ \red{ \mathscr{ By \ comparing \ the \ x \ co-ordinates \ on \ both \ sides }} \\ \\ \\ \sf 0 = \dfrac{-m_1+5m_2}{m_1+m_2} \\ \\ \rm{\blue{ By \: cross - multiplication} } \: \\ \\ \sf 0 = -m_1 + 5m_2 \\ \\ \sf m_1 = 5 m_2 \\ \\ \sf \dfrac{m_1}{m_2} = \dfrac{5}{1} \\ \\ \implies{\pink{\sf{\bf{ m_1:m_2 = 5:1 }}}}$

Now,

Substituting the value of ratio in the above formula we can find the y co-ordinate !

So,

$\sf p(0,y) = \left( \dfrac{ 5( -1) + 1( 5)}{5+1} , \dfrac{ 5(- 4) +1 ( - 6)}{5+1} \right) \\ \\ \\ \sf p(0,y) = \left( \dfrac{ - 5 + 5}{6} , \dfrac{ - 20 - 6}{6} \right) \\ \\ \\ \sf p(0,y) = \left( \dfrac{ 0}{6} , \dfrac{ - 26}{6} \right) \\ \\ \\ \sf p(0,y) = \left( 0, \dfrac{ - 26}{6} \right) \\ \\ \sf By \ comparing \ the \ co-ordinates \ on \ both \ sides \\ \\ \green{ \therefore { \implies { \sf {\bf { y = \dfrac{-26}{6} }}}}}$

Therefore,

• Ratio in which the point p(x, y) is 5:1
• The co-ordinate of the point which divides the line segment is p(0,[-26]/[6])

Answer 2

$\huge\bf{\underline{\underline{\gray{GIVEN:-}}}}$

• Y-axis divides the line segments joining the A(5,-6) and B(-1,-4) .

$\huge\bf{\underline{\underline{\gray{TO\:FIND:-}}}}$

1. The ratio in which the line get's divided .
2. Co-ordinate of the point which divides the line segment .

$\huge\bf{\underline{\underline{\gray{SOLUTION:-}}}}$

See the attachment diagram .

Here,

1. Let point P is on y-axis, which intersect by AB line .

• So the co-ordinates of the point P be (0 , y) [Let] .

Let,

• The ratio be 'k : 1' .

We know that,

${\color{blue}\bigstar}\:\bf{\red{\overbrace{\underbrace{\color{aqua}{P_x\:=\:\dfrac{(m\:x_2\:+\:n\:x_1)}{m\:+\:n}\:}}}}}$

Where,

• m : n = k : 1

• $\bf\red{(x_1\:,\:y_1)}$ = (5 , -6)

• $\bf\red{(x_2\:,\:y_2)}$ = (-1 , -4)

• $\bf\red{(x_1)}$ = 5

• $\bf\red{(x_1)}$ = -1

$\rm{\implies\:P_x\:=\:\dfrac{(k\times(-1)\:+\:1\times{5})}{k\:+\:1}\:}$

$\rm{\implies\:0\:=\:\dfrac{(-k\:+\:5)}{k\:+\:1}\:}$

$\rm{\implies\:\:-k\:+\:5\:=\:0\:}$

$\rm{\implies\:\:-k\:=\:-5\:}$

$\bf{\implies\:\:k\:=\:5\:}$

$\bf{\color{violet}{\implies\:k\::\:1\:=\:5\::\:1\:}}$

Again we know that,

$\purple\bigstar\:\bf{\blue{\overbrace{\underbrace{\green{P_y\:=\:\dfrac{(m\:y_2\:+\:n\:y_1)}{m\:+\:n}\:}}}}}$

Where,

• $\bf\red{(y_1)}$ = -6

• $\bf\red{(y_2)}$ = -4

$\rm{\implies\:P_y\:=\:\dfrac{\Big(5\times(-4)\:+\:1\times(-6)\Big)}{5\:+\:1}\:}$

$\rm{\implies\:P_y\:=\:\dfrac{\Big(-20\:+\:(-6)\Big)}{6}\:}$

$\rm{\implies\:P_y\:=\:\dfrac{\Big(-20\:-\:6\Big)}{6}\:}$

$\rm{\implies\:P_y\:=\:\dfrac{-26}{6}\:}$

$\bf{\implies\:P_y\:=\:\dfrac{-13}{3}\:}$

$\huge{\color{orange}\therefore}$ [1] The ratio in which the line get's divided is '5 : 1' .

$\\\huge{\color{orange}\therefore}$ [2] Co-ordinate of the point which divides the line segment is '$\bf{\Big(0\:,\:-\dfrac{13}{3}\Big)}$' .