Question

Find the ratio in which y axis divides the loline segment joining the point A (5,-6),B (1,-4)

Answer 1

Given

Line segment joining two points (5,-6) and (1,-4) is divided by y-axis.

To Find:

• Ratio in which y-axis divides the line segment joining the points (5,-6) and (1,- 4)

Sectional Formula

$\setlength{\unitlength}{0.9 cm}}\begin{picture}(12,4)\thicklines\put(6,6){\line(1,0){5.5}}\put(5.6,5.9){ P }\put(11.7,5.9){ Q }\put(5.4,5.5){ (x_1\,,\,y_1) }\put(11.4,5.5){ (x_2\,,\,y_2) }\put(8,6){\circle*{0.2}}\put(7.8,6.3){ R }\put(7.4,5.5){ (x\,,\,y) }\put(6.6,6.3){ m }\put(9.3,6.3){ n }\put(11.7,5.9){ Q }\end{picture}$

Let P(x₁,y₁) and Q(x₂,y₂) be two points. Let the point R(x,y) divide the line segment joining the points P and Q internally in the ratio m:n, then

$\sf{(x,y)=\bigg(\dfrac{mx_{2}+nx_{1}}{m+n},\dfrac{my_{2}+ny_{1}}{m+n}\bigg)}$

Solution:

Let the y-axis divides the given line segment in λ:1 and point of intersection of y-axis and given line segment be (0,b)

• (0,b) because on y-axis, x-coordinate is always 0 and y-coordinate can be any constant(say b).  

Diagram:

$\setlength{\unitlength}{0.9 cm}}\begin{picture}(12,4)\thicklines\put(6,6){\line(1,0){5.5}}\put(5.6,5.9){ P }\put(11.7,5.9){ Q }\put(5.4,5.5){ (5\,,\,-6) }\put(11.4,5.5){ (1\,,\,-4) }\put(8,6){\circle*{0.2}}\put(7.8,6.3){ R }\put(7.4,5.5){ (0\,,\,b) }\put(6.6,6.3){ \lambda }\put(9.3,6.3){ 1 }\put(11.7,5.9){ Q }\end{picture}$

Now, by using sectional formula, we get

$\longrightarrow\sf{(0,b)=\bigg(\dfrac{\lambda(1)+5}{\lambda+1},\dfrac{\lambda(-4) -6}{\lambda +1}\bigg)}$

$\longrightarrow\sf{(0,b)=\bigg(\dfrac{\lambda+5}{\lambda+1},\dfrac{-4\lambda-6}{\lambda +1}\bigg)}$

On comparing x-coordinate of both the sides we get,

$\sf{\dfrac{\lambda+5}{\:\:\:\lambda+1}=0}$-----------(1)

From (1), we get

✏ λ=-5

So, y-axis divides the given line segment in 5:1 externally

Hence, the required ratio is 5:1

Answer 2

Given:-

y-axis divides the line segment joining two points (5,-6) and (1,-4).

To find:-

The ratio in which y-axis divides the line segment joining the points (5,-6) and (1,- 4).

Solution:-

• Let y-axis divide the segment in the ratio of α:1 and let the point of intersection in y-axis be (0,a). (∴it is on y-axis, hence its x-coordinate is 0)

• Let AB be the line segment that joins the points (5,-6) & (1,-4) where the coordinates of A are A(5,-6) and the coordinates of B are B(1,-4).

• The coordinates of the point of intersection are P(0,a).

$\setlength{\unitlength}{0.9 cm}}\begin{picture}(12,4)\thicklines\put(6,6){\line(1,0){5.5}}\put(5.6,5.9){ A }\put(11.7,5.9){ B }\put(5.4,5.5){ (5\,,\,-6) }\put(11.4,5.5){ (1\,,\,-4) }\put(8,6){\circle*{0.2}}\put(7.8,6.3){ P }\put(7.4,5.5){ (0\,,\,a) }\put(6.6,6.3){ \alpha }\put(9.3,6.3){ 1 }\put(11.7,5.9){ B }\end{picture}$

We took the ratio as α:1, so by section formula,

m = k

n = 1

x₁ = 5 & x₂ = 1

y₁ = -6 & y₂ = -4

x = 0 & y = a

Using section formula,

$\sf{(x,y)=\bigg(\dfrac{mx_{2}+nx_{1}}{m+n},\dfrac{my_{2}+ny_{1}}{m+n}\bigg)}$

$\longrightarrow\sf{(0,a)=\bigg(\dfrac{\alpha(1)+5}{\alpha +1},\dfrac{\alpha (-4) -6}{\alpha +1}\bigg)}\\\longrightarrow\sf{(0,a)=\bigg(\dfrac{\alpha +5}{\alpha +1},\dfrac{-4\alpha -6}{\alpha +1}\bigg)}$

On comparing y-coordinate of both the sides we get,

$\bold{\frac{\alpha +5}{\alpha +1}=0 }$

α + 5 = 0

α = -5

So, the ratio is:

α:1

= -5:1

= 5:1