The ratio in which the straight line 3x + 4y | vedclass

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(C) Let the line $3x + 4y = 6$ divide the line segment joining the points $P(2, -1)$ and $Q(1, 1)$ in the ratio $k:1$. Using the section formula,the c

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$4:1$

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(C) Let the line $3x + 4y = 6$ divide the line segment joining the points $P(2, -1)$ and $Q(1, 1)$ in the ratio $k:1$. Using the section formula,the coordinates of the point of intersection are $\left(\frac{1(2) + k(1)}{k+1}, \frac{1(-1) + k(1)}{k+1}\right) = \left(\frac{2+k}{k+1}, \frac{k-1}{k+1}\right)$. Since this point lies on the line $3x + 4y = 6$,we substitute these coordinates into the equation: $3\left(\frac{2+k}{k+1}\right) + 4\left(\frac{k-1}{k+1}\right) = 6$. Multiplying by $(k+1)$,we get: $3(2+k) + 4(k-1) = 6(k+1)$. $6 + 3k + 4k - 4 = 6k + 6$. $7k + 2 = 6k + 6$. $k = 4$. Thus,the ratio is $4:1$.

The ratio in which the straight line $3x + 4y = 6$ divides the join of the points $(2, -1)$ and $(1, 1)$ is

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