Find the ratio in which the line joining poin | vedclass

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(B) The points are $A(-1, -1)$ and $B(2, 1)$. The equation of the line $AB$ is given by $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$.Substituting

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$7: 5$ externally

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(B) The points are $A(-1, -1)$ and $B(2, 1)$. The equation of the line $AB$ is given by $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$.Substituting the values,we get $y + 1 = \frac{1 - (-1)}{2 - (-1)}(x - (-1)) \Rightarrow y + 1 = \frac{2}{3}(x + 1)$.This simplifies to $3y + 3 = 2x + 2$,or $2x - 3y - 1 = 0$.Let the line $AB$ divide the line segment joining $C(3, 4)$ and $D(1, 2)$ in the ratio $\lambda: 1$ at point $P$. The coordinates of $P$ are given by the section formula: $P = \left(\frac{1(3) + \lambda(1)}{1 + \lambda}, \frac{1(4) + \lambda(2)}{1 + \lambda}\right) = \left(\frac{3 + \lambda}{1 + \lambda}, \frac{4 + 2\lambda}{1 + \lambda}\right)$.Since point $P$ lies on the line $2x - 3y - 1 = 0$,we substitute the coordinates of $P$ into the equation:$2\left(\frac{3 + \lambda}{1 + \lambda}\right) - 3\left(\frac{4 + 2\lambda}{1 + \lambda}\right) - 1 = 0$.Multiplying by $(1 + \lambda)$,we get $2(3 + \lambda) - 3(4 + 2\lambda) - (1 + \lambda) = 0$.$6 + 2\lambda - 12 - 6\lambda - 1 - \lambda = 0$.$-5\lambda - 7 = 0 \Rightarrow \lambda = -7/5$.The negative sign indicates that the line divides the segment externally in the ratio $7: 5$.

Find the ratio in which the line joining points $A(-1, -1)$ and $B(2, 1)$ divides the line segment joining points $C(3, 4)$ and $D(1, 2)$.

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