In which ratio does the X-axis divide the lin | vedclass

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(D) Let the $X$-axis divide the line segment $AB$ in the ratio $k:1$ at point $P(x, 0)$.Using the section formula, the coordinates of point $P$ are gi

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$2:5$

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(D) Let the $X$-axis divide the line segment $AB$ in the ratio $k:1$ at point $P(x, 0)$.Using the section formula, the coordinates of point $P$ are given by:$P = \left( \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \right)$Substituting the values $A(1, 2)$ and $B(4, -5)$:$P = \left( \frac{4k + 1}{k + 1}, \frac{-5k + 2}{k + 1} \right)$Since $P$ lies on the $X$-axis, its $y$-coordinate must be $0$:$\frac{-5k + 2}{k + 1} = 0$$-5k + 2 = 0$$5k = 2$$k = \frac{2}{5}$Thus, the ratio is $2:5$.

In which ratio does the $X$-axis divide the line segment joining $A(1, 2)$ and $B(4, -5)$ starting from $A$?

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