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प्रश्न
Find the values of y for which the distance between the points P (2, -3) and Q (10, y) is 10 units.
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उत्तर १
PQ = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)`
⇒ `sqrt((10 - 2)^2 + (y + 3)^2) = 10`
⇒ (8)2 + (y + 3)2 = 100
⇒ 64 + y2 + 6y + 9 = 100
⇒ y2 + 6y + 73 - 100 = 0
⇒ y2 + 6y - 27 = 0
⇒ y2 + 9y - 3y - 27 = 0
⇒ y(y + 9) - 3(y + 9) = 0
⇒ (y + 9) (y - 3) = 0
⇒ y + 9 = 0
⇒ y = -9
and y - 3 = 0
⇒ y = 3
उत्तर २
The distance d between two points `(x_1, y_1)` and `(x_2, y_2)` is given by the formula
d = `sqrt((x_1-x_2)^2 + (y_1 - y_2)^2)`
The distance between two points P(2,−3) and Q(10,y) is given as 10 units. Substituting these values in the formula for distance between two points, we have,
10 = `sqrt((2 - 10)^2 + (-3 - y)^2)`
Now, squaring the above equation on both sides of the equals sign
100 = (-8)2 + (-3 - y)2
100 = 64 + 9 + y2 + 6y
27 = y2 + 6y
Thus, we arrive at a quadratic equation. Let us solve this now.
y2 + 6y - 27 = 0
y2 + 9y - 3y - 27 = 0
y(y + 9) - 3(y + 9) = 0
(y - 3) (y + 9) = 0
The roots of the above quadratic equation are thus 3 and −9.
Thus, the value of ‘y’ could either be 3 or -9.
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