Question

Class 10th coordinate geometry Find the ratio in which y-axis divides the line segment joining the points A(5, -6) and B(-1, -4). Also find the coordinates of the point of division​

Answer 1

Let, the ratio be k:1

A(5, -6) _____________________B(-1, -4)

Given that :

The ratio in which y-axis divides the line segment joining the given points. On y-axis the value of x will be 0 then,

Let the coordinates of the point of division be (0, y)

As we know that :

Coordinates of the point

$\frac{( mx_{2} + n.x_{1})}{m + n} and \: \frac{(my_{2} + n.y_{1})}{m + n}$

$Where,x_{1} = 5 \\ x_{2} = - 1 \\ y_{1} = - 6 \\ y_{2} = - 4$

and m = k and n = 1

According to the question :

$\frac{( kx_{2} + 1.x_{1})}{k + 1} = 0$

$\frac{k.( - 1) + 1.5}{k + 1} = 0 \\ \\ = > \frac{ - k + 5}{k + 1} = 0 \\ \\ = > - k + 5 = 0 \\ \\ = > k = 5$

Hence, the ratio will be 5:1

Now, On putting the value of k in the formula :

$y = \frac{k.( - 4) + 1.( - 6)}{k + 1} \\ \\ = > y = \frac{5.( - 4) + 1.( - 6)}{5 + 1} \\ \\ = > y = \frac{ - 20 - 6}{6} \\ \\ = > y = \frac{ - 26}{6} = \frac{ -1 3}{3}$

So, the coordinates of the point of division will be (0,-13/3)

_____________________________

Hope it helps ☺

Answer 2

$\huge\bf{\underline{\underline{\gray{GIVEN:-}}}} </p><p>$

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

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

The ratio in which the line get's divided .

Co-ordinate of the point which divides the line segment .

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

See the attachment diagram .

Here,

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,

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

Where,

m : n = k : 1

$\bf\red{(x_1\:,\:y_1)}(x = (5 , -6) \\ </p><p>\bf\red{(x_2\:,\:y_2)}(x = (-1 , -4) \\ </p><p>\bf\red{(x_1)}(x = 5) \\ </p><p>\bf\red{(x_1)}(x = -1)$

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

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

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

$\begin{gathered}\rm{\implies\:\:-k\:=\:-5\:} \\ \end{gathered}$

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

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

Again we know that,

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

Where,

$\bf\red{(y_1)}= -6$

$\bf\red{(y_2)}= -4$

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

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

$\begin{gathered}\rm{\implies\:P_y\:=\:\dfrac{\Big(-20\:-\:6\Big)}{6}\:} \\ \end{gathered} </p><p>$

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

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

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

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