Question

Find the ratio in which the line segment joining A(1,-5) and B (-4, 5) is divided by the x-axis. Also find the co-ordinate of the point of division.

Answer 1

Let the ratio in which the line segment joining A (1, - 5) and B ( - 4, 5) is divided by x-axis be k: 1
Therefore, the coordinates of the point of division is (-4k+1/k+1,5k-5/k+1)
We know that y-coordinate of any point on x-axis is 0
Therefore,
(5k-5/k+1)= 0
k = 1
Therefore, x-axis divides it in the ratio 1:1.
Division point = ()
= (-4(1)+1/1+1,5(1)-5/1+1)
= (-3/2, 0)

Answer 2

$\bf\huge\underline{Question}$

Find the ratio in which the line segment joining A(1,-5) and B (-4, 5) is divided by the x-axis. Also find the co-ordinate of the point of division.

$\bf\huge\underline{Solution}$

The given points are A(1, -5) and B(-4, 5).

Let the required ratio = k : 1 and the required point be P( x, y).

∴ Its y-coordinate is 0.

x = $\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}$ and 0 = $\dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}$

=> x = $\dfrac{k(-4) + 1(1)}{k + 1}$ and 0=$\dfrac{k(5) + 1(-5)}{k + 1}$

=> x = $\dfrac{-4k + 1}{k + 1}$ and 0 = $\dfrac{5k - 5}{k + 1}$

=> x(k + 1) = -4k + 1 and 5k - 5 = 0 => k=1

Now, x(k + 1) = -4k + 1

=> x(1 + 1) = -4 + 1⠀⠀⠀⠀ ⠀⠀⠀ ⠀[∵ k = 1]

=> 2x = -3

=> x = -$\dfrac{3}{2}$

∴ The required ratio is k : 1 = 1 : 1

Coordinates of P are (x, 0) = ($\dfrac{-3}{2}$, 0)