In what ratio does the line 2x + 3y + 7 = 0 d | vedclass

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(D) Let the ratio in which the line $2x + 3y + 7 = 0$ divides the segment joining $A(3, 4)$ and $B(7, 8)$ be $k:1$.Using the section formula,the coord

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$5 : 9$ externally

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(D) Let the ratio in which the line $2x + 3y + 7 = 0$ divides the segment joining $A(3, 4)$ and $B(7, 8)$ be $k:1$.Using the section formula,the coordinates of the point of division are $(\frac{7k + 3}{k + 1}, \frac{8k + 4}{k + 1})$.Since this point lies on the line $2x + 3y + 7 = 0$,we substitute these coordinates into the equation:$2(\frac{7k + 3}{k + 1}) + 3(\frac{8k + 4}{k + 1}) + 7 = 0$Multiply by $(k + 1)$:$2(7k + 3) + 3(8k + 4) + 7(k + 1) = 0$$14k + 6 + 24k + 12 + 7k + 7 = 0$$45k + 25 = 0$$45k = -25$$k = -\frac{25}{45} = -\frac{5}{9}$The negative sign indicates external division in the ratio $5:9$.

In what ratio does the line $2x + 3y + 7 = 0$ divide the line segment joining the points $(3, 4)$ and $(7, 8)$?

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