Determine the ratio in which the line $3 x + y - 9 = 0$ divides the line segment joining the points (1, 3) and (2, 7).
[4 MARKS]

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Solution

Formula: 1 Mark
Steps: 2 Marks
Answer: 1 Mark

Let the required ratio be k :1
Using section formula, we have
$x = \frac{1 \times 1 + 2 \times k}{k + 1} = \frac{2 k + 1}{k + 1}$
and $y = \frac{1 \times 3 + 7 \times k}{k + 1} = \frac{7 k + 3}{k + 1}$

Point P (x, y) lies on the line $3 x + y - 9 = 0$ , so
$3 (\frac{2 k + 1}{k + 1}) + (\frac{7 k + 3}{k + 1}) - 9 = 0 \Rightarrow \frac{3 (2 k + 1) + (7 k + 3) - 9 (k + 1)}{k + 1} = 0 \Rightarrow 6 k + 3 + 7 k + 3 - 9 k - 9 = 0 \Rightarrow 4 k - 3 = 0 \Rightarrow 4 k = 3$
So, $k = \frac{3}{4}$
$∴$ Required ratio is 3 : 4

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Determine the ratio in which 3 + y – 9 = 0 divides the line segment joining the points (1 , 3) & (2 , 7).

Determine the ratio in which the graph of the equation 3x + y = 9 divides line segment joining the points A (2,7) and B (1,3).

3 x + y = 9

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(2, 7)

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