Question

In what ratio does x-axis divide the line segment joining points (3, 2) and (4, -1)

Answer 1

Let us name the point (3, 2) as A, and the point (4, -1) as B.

∴ A(3, 2) and B(4, -1).

For the x-axis to intersect AB, the y-coordinate has to be 0.

Therefore the point that divides AB that lies on the x-axis will have the coordinate (x, 0). Let this point be C(x, 0).

We can find the ratio in which the point C(x, 0) divides the line AB using the Section formula.

$\sf C(x, y) = \Bigg[ \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} , \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \Bigg]$

Here, m₁ : m₂ is the ratio in which C(x, 0) divides AB.

Also, the x-coordinate is equal to (m₁x₂ + m₂x₁)/(m₁ + m₂)

And, the y-coordinate is equal to (m₁y₂ + m₂y₁)/(m₁ + m₂).

$\sf \Longrightarrow C(x, 0) = \Bigg[ \dfrac{m_1(4) + m_2(3)}{m_1 + m_2} , \dfrac{m_1(-1) + m_2(2)}{m_1 + m_2} \Bigg]$

$\sf \Longrightarrow C(x, 0) = \Bigg[ \dfrac{4m_1 + 3m_2}{m_1 + m_2} , \dfrac{-m_1 + 2m_2}{m_1 + m_2} \Bigg]$

We know that the y-coordinate is equal to (m₁y₂ + m₂y₁)/(m₁ + m₂).

$\sf \Longrightarrow 0 = \dfrac{-m_1 + 2m_2}{m_1 + m_2}$

$\sf \Longrightarrow 0 = -m_1 + 2m_2$

$\sf \Longrightarrow m_1 = 2m_2$

$\sf \Longrightarrow \dfrac{m_1}{m_2} = \dfrac{2}{1}$

Therefore, the x-axis divides the line-segment joining the points (3, 2) and (4, -1) in the ratio 2 : 1.

Answer 2

Answer:

Let us name the point (3, 2) as A, and the point (4, -1) as B.

∴ A(3, 2) and B(4, -1).

For the x-axis to intersect AB, the y-coordinate has to be 0.

Therefore the point that divides AB that lies on the x-axis will have the coordinate (x, 0). Let this point be C(x, 0).

We can find the ratio in which the point C(x, 0) divides the line AB using the Section formula.

$\sf C(x, y) = \Bigg[ \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} , \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \Bigg]$

Here, m₁ : m₂ is the ratio in which C(x, 0) divides AB.

Also, the x-coordinate is equal to (m₁x₂ + m₂x₁)/(m₁ + m₂)

And, the y-coordinate is equal to (m₁y₂ + m₂y₁)/(m₁ + m₂).

$\sf \Longrightarrow C(x, 0) = \Bigg[ \dfrac{m_1(4) + m_2(3)}{m_1 + m_2} , \dfrac{m_1(-1) + m_2(2)}{m_1 + m_2} \Bigg]$

$\sf \Longrightarrow C(x, 0) = \Bigg[ \dfrac{4m_1 + 3m_2}{m_1 + m_2} , \dfrac{-m_1 + 2m_2}{m_1 + m_2} \Bigg]$

We know that the y-coordinate is equal to (m₁y₂ + m₂y₁)/(m₁ + m₂).

$\sf \Longrightarrow 0 = \dfrac{-m_1 + 2m_2}{m_1 + m_2}$

$\sf \Longrightarrow 0 = -m_1 + 2m_2$

$\sf \Longrightarrow m_1 = 2m_2$

$\sf \Longrightarrow \dfrac{m_1}{m_2} = \dfrac{2}{1}$

Therefore, the x-axis divides the line-segment joining the points (3, 2) and (4, -1) in the ratio 2 : 1.