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Introduction To Trigonometry

Question
CBSEENMA10006987
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Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.

Solution

We have, P(2, -3), Q( 10, y) and PQ = 10 units.
Now,    PQ2 = (10)2 = 100
⇒ (10 - 2)2 + {y - (-3)}2 = 100
⇒    (8)2 + (y + 3)2 = 100
⇒    64 + y2 + 6y + 9 = 100
⇒    y2 + 6y - 21 = 0
⇒    y2 + 9y - 3y - 27 = 0
⇒    y(y + 9) - 3(y + 9) = 0
⇒    (y + 9) (y - 3) = 0
⇒    y + 9 = 0
or    y - 3 = 0
⇒    y = -9
or    y = 3
⇒    y = -9, 3
Hence, the required value of is -9 or 3.

Some More Questions From Introduction to Trigonometry Chapter

Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(-1, -2), (1, 0), (-1, 2), (-3, 0)

Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(–3, 5), (3, 1), (0, 3), (–1, – 4)

Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.

Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (-3, 4).

Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).

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