The straight line 3x + y = 9 divides the line | vedclass

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(B) Let the line $3x + y - 9 = 0$ divide the line segment joining $A(1, 3)$ and $B(2, 7)$ in the ratio $k: 1$. Using the section formula, the coordina

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$3: 4$ internally

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(B) Let the line $3x + y - 9 = 0$ divide the line segment joining $A(1, 3)$ and $B(2, 7)$ in the ratio $k: 1$. Using the section formula, the coordinates of the dividing point are $(\frac{2k+1}{k+1}, \frac{7k+3}{k+1})$. Since this point lies on the line $3x + y = 9$, we have: $3(\frac{2k+1}{k+1}) + (\frac{7k+3}{k+1}) = 9$ $6k + 3 + 7k + 3 = 9(k + 1)$ $13k + 6 = 9k + 9$ $4k = 3$ $k = \frac{3}{4}$. Since $k > 0$, the division is internal in the ratio $3: 4$.

The straight line $3x + y = 9$ divides the line segment joining the points $(1, 3)$ and $(2, 7)$ in the ratio

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