The line segment joining the points A(3, 2) a | vedclass

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(A) Given that the line segment joining the points $A(3, 2)$ and $B(5, 1)$ is divided at the point $P$ in the ratio $1: 2$.Using the section formula f

Vedclass · Accepted Answer

$19$

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(A) Given that the line segment joining the points $A(3, 2)$ and $B(5, 1)$ is divided at the point $P$ in the ratio $1: 2$.Using the section formula for internal division,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 $m_1 = 1, m_2 = 2, x_1 = 3, y_1 = 2, x_2 = 5, y_2 = 1$:$P = \left( \frac{1(5) + 2(3)}{1 + 2}, \frac{1(1) + 2(2)}{1 + 2} \right) = \left( \frac{5 + 6}{3}, \frac{1 + 4}{3} \right) = \left( \frac{11}{3}, \frac{5}{3} \right)$Since the point $P\left( \frac{11}{3}, \frac{5}{3} \right)$ lies on the line $3x - 18y + k = 0$,it must satisfy the equation:$3\left( \frac{11}{3} \right) - 18\left( \frac{5}{3} \right) + k = 0$$11 - 6(5) + k = 0$$11 - 30 + k = 0$$-19 + k = 0$$k = 19$Thus,the required value of $k$ is $19$.

The line segment joining the points $A(3, 2)$ and $B(5, 1)$ is divided at the point $P$ in the ratio $1: 2$,and it lies on the line $3x - 18y + k = 0$. Find the value of $k$.

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