Question

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

Answer 1

Given:

Points of line (-5,6) and (-1,-4)

Y axis divides the line segment.

To find : Ratio and point of Intersection.

Let k:1 be the ratio. Since the line segment is divided by Y axis , the X Coordinate at that point will be zero.

Using section formula, we get

$0 = \frac{-k-5}{k+1}$

⇒ -k -5 = 0

or, k = -5

Ratio is -5:1 or 5:1 externally.

The coordinate of Intersection

$y = \frac{20+6}{-4}$

or, y = 26/-4 = -13/2

or, y = -6.5

∴ Coordinates are (0,-6.5)

Answer 2

$\displaystyle\large\underline{\sf\red{Given}}$

✭ Point A = (5, -6)

✭ Point B = (-1, - 4)

$\displaystyle\large\underline{\sf\blue{To \ Find}}$

◈ Ratio in which y axis divides the line segment and The point of intersection

$\displaystyle\large\underline{\sf\gray{Solution}}$

It is given that the line segment is divided by y axis so,

›› The point of intersection = (0, y)

›› Also let the line segment be divided in the ratio k : 1

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$\underline{\bigstar\:\textsf{According to the given Question :}}$

As per the section formula,

$\displaystyle\underline{\boxed{\sf (x,y)=\bigg\lgroup\dfrac{m_1x_2+m_2x_1}{m_1+m_2} +\dfrac{m_1y_2+m_2y_1}{m_1+m_2}\bigg\rgroup}}$

Where,

• x = 0
• y = y
• x₁ = 5
• x₂ = -1
• y₁ = -6
• y₂ = -4
• m₁ = k
• m₂ = 1

Substituting the given values,

➝ $\displaystyle\sf (0,y)=\bigg\lgroup\dfrac{-1k+5}{k+1} +\dfrac{-4k-6}{k+1}\bigg\rgroup$

➝ $\displaystyle\sf \dfrac{-1k+5}{k+1}=0$

➝ $\displaystyle\sf -1k +5 = 0$

➝ $\displaystyle\sf -k = - 5$

➝ $\displaystyle\sf\purple{k = 5}$

$\displaystyle\sf \therefore$ The line segment is divided in the ratio 5 : 1

Now on equate the y coordinate,

➳ $\displaystyle\sf \dfrac{-4k - 6}{k+1}=y$

➳ $\displaystyle\sf -4 × 5 - 6= (5 + 1 )y$

➳ $\displaystyle\sf -20 - 6 = 6y$

➳ $\displaystyle\sf \dfrac{-26}{6} = y$

➳ $\displaystyle\sf\pink{y = \dfrac{-13}{3}}$

$\displaystyle\therefore\:\underline{\sf The \ point \ of \ intersection \ is \ (0 ,\dfrac{-13}{3})}$

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