If P(x, y) is any point on the line joining the points (a, 0) and (0, b) then the value of $\frac{x}{a} + \frac{y}{b}$

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Solution

The correct option is D 1
Since the given points are collinear, they do not form a
triangle, which means area of the triangle is Zero.

Area of a triangle with vertices $(x_{1}, y_{1})$ ; $(x_{2}, y_{2})$ and $(x_{3}, y_{3})$ is $∣ ∣ ∣ \frac{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})}{2} ∣ ∣ ∣$
Hence, substituting the points $(x_{1}, y_{1}) = (a, 0)$ ; $(x_{2}, y_{2}) = (x, y)$ and $(x_{3}, y_{3}) = (0, b)$
in the area formula, we get
$∣ ∣ ∣ \frac{a (y - b) + x (b - 0) + 0 (0 - y)}{2} ∣ ∣ ∣ = 0$
$=> a y - a b + b x = 0$
Dividing by ab both sides, we get
$=> \frac{x}{a} + \frac{y}{b} = 1$

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If P(x, y) is any point on the line joining the points (a, 0) and (0, b) then the value ofxa+yb

If P(x, y) is any point on the line joining the points (a, 0) and (0, b) then the value of $\frac{x}{a} + \frac{y}{b}$