Question

the ratio in which the line
$3 x + y = 9$
divides the line sequent joining the points (1;3) and (2;7) is given by
​

Answer 1

⌬ Let the line 3x + y − 9 = 0 divide the line segment joining the point (1,3) and (2,7) in the ratio K:1 at point C.

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• Using section formula,

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$\star\;{\boxed{\sf{\purple{ x\;,\;y = \bigg( \dfrac{m_2 x_1 + m_1 x_2 }{m_1 + m_2 }\;,\; \dfrac{m_2 y_1 + m_1 y_2 }{m_1 + m_2 } \bigg)}}}}$

$:\implies\sf C = \bigg( \dfrac{1 \times 1 + k \times 2}{k + 1}\;,\; \dfrac{1 \times 3 + k \times 7}{k + 1} \bigg)$

$:\implies\sf C = \bigg( \dfrac{1 + 2k}{k + 1}\;,\; \dfrac{3 + 7k}{k + 1} \bigg)$

$\sf C = \bigg( \dfrac{1 + 2k}{k + 1}\;,\; \dfrac{3 + 7k}{k + 1} \bigg)\;lies\;on\;the\;line\;3x + y - 9 = 0$

Therefore,

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$:\implies\sf 3 \bigg( \dfrac{1 + 2k}{k + 1} \bigg) + \bigg( \dfrac{3 + 7k}{k + 1} \bigg) - 9 = 0$

$:\implies\sf \dfrac{ 3(1 + 2k) + (3 + 7k) - 9(k + 1)}{(k + 1)} = 0$

$:\implies\sf 3(1 + 2k) + (3 + 7k) - 9(k + 1) = 0$

$:\implies\sf 3 + 6k + 3 + 7k - 9k - 9 = 0$

$:\implies\sf 4k - 3$

$:\implies{\boxed{\sf{\pink{ k = \dfrac{3}{4}}}}}\;\bigstar$

∴ Thus, the line 3x + y − 9 = 0 divide the line segment in the ratio 3:4.