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

ANSWER

Let the required ratio be m:n

here,x

1

=5,x

2

=−1,y

1

=−6,y

2

=−4

Then, According to question, we have ;

m+n

mx

2

+mx

1

=0

⇒m(−1)+n(5)=0

⇒5n=m

⇒m=5n

⇒

n

m

=

1

5

⇒m:n=5:1

Coordinate of the point of division

=(0,

m+n

my

2

+my

1

)

=(0,

6

5(−4)+1(−6)

)

=(0,

6

−20−6

)

=(0,

3

−13

)

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Answer 2

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

$\pink{ \sf{ \bf{p(x,y) = \lgroup \dfrac{m_1x_2+m_2x_1}{m_1+m_2} , \dfrac{ m_1 y_2 + m_2 y_1 }{ m_1 + m_2 }\rgroup}} }p(x,y)=⟮ </p><p>m </p><p>1</p><p> </p><p> +m </p><p>2</p><p> </p><p> </p><p>m </p><p>1</p><p> </p><p> x </p><p>2</p><p> </p><p> +m </p><p>2</p><p> </p><p> x </p><p>1</p><p> </p><p> </p><p> </p><p> , </p><p>m </p><p>1</p><p> </p><p> +m </p><p>2</p><p> </p><p> </p><p>m </p><p>1</p><p> </p><p> y </p><p>2</p><p> </p><p> +m </p><p>2</p><p> </p><p> y </p><p>1</p><p> </p><p> </p><p> </p><p> ⟯$

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;

$\sf{ A(x_1,y_1) = (5,-6) \qquad B(x_2,y_2) = (-1,-4) }A(x </p><p>1</p><p> </p><p> ,y </p><p>1</p><p> </p><p> )=(5,−6)B(x </p><p>2</p><p> </p><p> ,y </p><p>2</p><p> </p><p> )=(−1,−4)$

Using the formula;

$\pink{ \sf{ \bf{p(x,y) = \lgroup \dfrac{m_1x_2+m_2x_1}{m_1+m_2} , \dfrac{ m_1 y_2 + m_2 y_1 }{ m_1 + m_2 }\rgroup}} }p(x,y)=⟮ </p><p>m </p><p>1</p><p> </p><p> +m </p><p>2</p><p> </p><p> </p><p>m </p><p>1</p><p> </p><p> x </p><p>2</p><p> </p><p> +m </p><p>2</p><p> </p><p> x </p><p>1</p><p> </p><p> </p><p> </p><p> , </p><p>m </p><p>1</p><p> </p><p> +m </p><p>2</p><p> </p><p> </p><p>m </p><p>1</p><p> </p><p> y </p><p>2</p><p> </p><p> +m </p><p>2</p><p> </p><p> y </p><p>1</p><p> </p><p> </p><p> </p><p> ⟯$

Substituting the values ;

$\begin{gathered}\sf p(0,y) = ( \dfrac{ m_1( -1) + m_2( 5)}{m_1+m_2} , \dfrac{ m_1(- 4) + m_2 ( - 6)}{m_1+m_2} ) \\ \\ \\ \sf p(0,y) = ( \dfrac{-m_1+5m_2}{m_1+m_2} , \dfrac{-4m_1-6m_2}{m_1+m_2} ) \\ \\ \red{ \mathscr{ By \ comparing \ the \ x \ co-ordinates \ on \ both \ sides }} \\ \\ \\ \sf 0 = \dfrac{-m_1+5m_2}{m_1+m_2} \\ \\ \rm{\blue{ By \: cross - multiplication} } \: \\ \\ \sf 0 = -m_1 + 5m_2 \\ \\ \sf m_1 = 5 m_2 \\ \\ \sf \dfrac{m_1}{m_2} = \dfrac{5}{1} \\ \\ \implies{\pink{\sf{\bf{ m_1:m_2 = 5:1 }}}}\end{gathered}</p><p>p(0,y)=( </p><p>m </p><p>1</p><p> </p><p> +m </p><p>2</p><p> </p><p> </p><p>m </p><p>1</p><p> </p><p> (−1)+m </p><p>2</p><p> </p><p> (5)</p><p> </p><p> , </p><p>m </p><p>1</p><p> </p><p> +m </p><p>2</p><p> </p><p> </p><p>m </p><p>1</p><p> </p><p> (−4)+m </p><p>2</p><p> </p><p> (−6)</p><p> </p><p> )</p><p>p(0,y)=( </p><p>m </p><p>1</p><p> </p><p> +m </p><p>2</p><p> </p><p> </p><p>−m </p><p>1</p><p> </p><p> +5m </p><p>2</p><p> </p><p> </p><p> </p><p> , </p><p>m </p><p>1</p><p> </p><p> +m </p><p>2</p><p> </p><p> </p><p>−4m </p><p>1</p><p> </p><p> −6m </p><p>2</p><p> </p><p> </p><p> </p><p> )$

Now,

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

So,

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

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])