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प्रश्न
A point P (2, -1) is equidistant from the points (a, 7) and (-3, a). Find a.
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उत्तर
We know that the distance between the two points (x1, y1) and (x2, y2) is
d = `sqrt((x_2−x_1)^2+(y_2−y_1)^2)`
Let the given points be A = (a, 7) and B = (−3, a) and the third point given is P(2, −1).
We first find the distance between P(2, −1) and A =(a, 7) as follows:
PA = `sqrt((x_2−x_1)^2+(y_2−y_1)^2)`
= `sqrt((a−2)^2+(7−(−1))^2)`
= `sqrt((a−2)^2+(7+1)^2)`
= `sqrt((a−2)^2+8^2)`
= `sqrt((a−2)^2+64)`
Similarly, the distance between P(2,−1) and B = (−3, a) is:
PB = `sqrt((x_2−x_1)^2+(y_2−y_1)^2)`
= `sqrt((−3−2)^2+(a−(−1))^2)`
= `sqrt((−5)^2+(a+1)^2)`
= `sqrt(25+(a+1)^2)`
Since the point P(2,−1) is equidistant from the points A(a, 7) and B = (−3, a), therefore, PA = PB that is:
⇒ a2 − 4a + 4 + 64 = 25 + a2 + 2a + 1
⇒ a2 − 4a + 68 = a2 + 2a + 26
⇒ −4a − 2a = 26 − 68
⇒ −6a = −42
⇒ a =`(−42)/(−6)` =7
Hence, a = 7.
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Case Study -2
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