Question

Find the ratio in which the y-axis divides the line segment joining the points
(5, -6) and (-1. -4). Also find the coordinates of the point of intersection.

Answer 1

Answer:

Ratio is...1:5 and

co-ordinate is..(0,-25/6).

Let us consider the ratio as k:1.

As the line passes through the y-axis

so the coordinates at y-axis will be (0,y).....(i)

now by using formula.... x=(mx1 + nx2)/(m+n)

thereby using equation (i)

0=(5k+(-1))/1+k

0=5k-1

1=5k

1/5=k...

ratio is K:1

i.e. 1:5

now putting this in y ordinate

y=((5×1)+(-6)×5)/(1+5)

y=(5-30)/6

coordinates is (0,-25/6)

y= -25/6

Answer 2

To find:

• Ratio in which y-axis divides the line segment joining points (5, -6) and (-1, -4).
• And coordinates of point of intersection

Formula required:

• Section formula

$\purple{\bigstar}\;\;\boxed{\sf{(x,y)=\bigg(\dfrac{mx_2+nx_1}{m+n}\;,\;\dfrac{my_2+ny_1}{m+n}\bigg)}}$

[ where (x,y) are the coordinates of point which divide line segment joining points (x₁, y₁) and (x₂, y₂) in the ratio m : n ]

Solution:

• Let, us assume that $\sf{\dfrac{m}{n}=k}$

• then, m : n = k : 1 will be the ratio in which given line segment is divided

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

[ x coordinate would be zero because point of division lies in y-axis ]

Now,

Using Section Formula

$\implies\sf{(0,y)=\bigg(\dfrac{(k)(-1)+(1)(5)}{k+1}\;,\;\dfrac{(k)(-4)+(1)(-6)}{k+1}\bigg)}$

so,

$\implies\sf{0=\dfrac{(k)(-1)+(1)(5)}{k+1}\;,\;y=\dfrac{(k)(-4)+(1)(-6)}{k+1}}$

Taking

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

$\implies\sf{0=-k+5}$

$\implies\boxed{\boxed{\purple{\sf{k=5}}}}$

Now, taking

$\implies\sf{y=\dfrac{(k)(-4)+(1)(-6)}{k+1}}$

putting value of k

$\implies\sf{y=\dfrac{(5)(-4)+(1)(-6)}{5+1}}$

$\implies\sf{y=\dfrac{-20-6}{6}}$

$\implies\boxed{\boxed{\purple{\sf{y=\dfrac{-13}{3}}}}}$

Hence,

• Ratio in which line segment is divided is, k : 1 = 5 : 1.
• and, Coordinates of point of intersection are $\bf{\bigg(0,\dfrac{-13}{3}\bigg)}$.