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प्रश्न
If A(-1, 1) and B(2, 3) are two fixed points, then find the locus of a point P so that the area of triangle APB = 8 sq.units.
उत्तर
Let the point P(x1, y1).
Fixed points are A(-1, 1) and B(2, 3).
Given area (formed by these points) of the triangle APB = 8
⇒ `1/2` [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 8
⇒ `1/2` [x1(1 – 3) + (-1) (3 – y1) + 2(y1 – 1)] = 8
⇒ `1/2` [-2x1 – 3 + y1 + 2y1 – 2] = 8
⇒ `1/2` [-2x1 + 3y1 – 5] = 8
⇒ -2x1 + 3y1 – 5 = 16
⇒ -2x1 + 3y1 – 21 = 0
⇒ 2x1 – 3y1 + 21 = 0
∴ The locus of the point P(x1, y1) is 2x – 3y + 21 = 0.
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