Question

Find the value of y for which the distance between the points P(2, -3) and Q (10, y) is 10
units.
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Answer 1

Given :-

Distance between (2, – 3) and (10, y) = 10

To Find :-

The value of y.

Solution :-

Given that,

Distance between (2, – 3) and (10, y) = 10

Using distance formula,

$\underline{\boxed{\sf PQ=\sqrt{(10-2)^2+(y+3)^2} }}$

Substituting their values,

$\sf=\sqrt{(8)^2+(y+3)^2}$

Given that, PQ = 10

$\sf 10=\sqrt{(8)^2+(y+3)^2}$

Squaring both sides,

64 + (y+3)² = 100

(y+3)² = 36

y + 3 = ±6

y + 3 = +6 or y + 3 = −6

Therefore, y = 3 or -9.

Answer 2

Answer:

y= 3 or -9

Step-by-step explanation:

Using distance formula

D =√{(x2-x1)²+(y2-y1)²}

Squaring both sides and putting respective values, we get

10²= (10-2)²+(y-[-3])²

100 = 64 + (y+3)²

(y+3)²= 36

Taking Square root both sides,

y+3= |6|

→y = 3 or y=-9

As |6| =+-6