Find the condition that the point $P (x, y)$ may lie on the line joining (3, 4) and (-5, -6).

5 x + 4 y - 1 = 0

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5 x - 4 y + 1 = 0

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5 x - 4 y - 1 = 0

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Solution

The correct option is B $5 x - 4 y + 1 = 0$
The Point $P (x, y)$ lies on the line joining $A (3, 4)$ and $B (- 5, - 6)$ .

$∴ P, A & B$ are collinear points.
So area of triangle formed by these points will be equal to 0.

Area of triangle having coordinates $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ is
$\frac{1}{2} \times | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) | ∴ \frac{1}{2} \times | x (4 + 6) + 3 (- 6 - y) - 5 (y - 4) | = 0 x (4 + 6) + 3 (- 6 - y) - 5 (y - 4) = 0 10 x - 18 - 3 y - 5 y + 20 = 0 10 x - 8 y + 2 = 0 \Rightarrow 5 x - 4 y + 1 = 0$

Hence, the point $P (x, y)$ lies on the line joining (3, 4) and (-5, -6), if $5 x - 4 y + 1 = 0$

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