Question

5.The distance between the points P(0, 2) and Q(6,0) is
(a)4v10
(b)2v10
(c)v10
(d)20
​

Answer 1

Answer:

We can find the distance between the points P and Q by using the distance formula.

If the coordinates of point P and Q are (x1,y1)and(x2,y2),(x1,y1)and(x2,y2), then the distance between these two points is:

(x2−x1)2+(y2−y1)2−−−−−−−−−−−−−−−−−−−√(x2−x1)2+(y2−y1)2

Here, coordinates of P and Q are (0,2) and (0,6). So, the distance between P and Q is:

= (0−0)2+(6−2)2−−−−−−−−−−−−−−−√(0−0)2+(6−2)2

= 0+42−−−−−√0+42

= 4

So, the distance between the points P and Q is 4 units.

Step-by-step explanation:

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Answer 2

Correct question: The distance between the points P(0, 2) and Q(6,0) is

(a) 4$\sqrt{10}$

(b) 2$\sqrt{10}$

(c) $\sqrt{10}$

(d) 20

Answer:

Option (b) is correct.

Step-by-step explanation:

Formula for calculating distance between the points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ . . . . . (1)

where $d=$ distance and $(x_1,y_1)$ and $(x_2,y_2)$ are the two points.

Consider the two given points as follows:

$(0,2)=(x_1,y_1)$ at point $P$

$(6,0)=(x_2,y_2)$ at point $Q$

(a) If distance $PQ=4\sqrt{10}$. Then,

Using formula (1), we have

$PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$4\sqrt{10} =\sqrt{(6-0)^2+(0-2)^2}$

$4\sqrt{10} =\sqrt{6^2+(-2)^2}$

$4\sqrt{10} =\sqrt{36+4}$

$4\sqrt{10} =\sqrt{40}$

$4\sqrt{10} \neq 2\sqrt{10}$ which is not possible.

Thus, option (a) is incorrect.

(b) If distance $PQ=2\sqrt{10}$. Then,

Using formula (1), we have

$PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$2\sqrt{10} =\sqrt{(6-0)^2+(0-2)^2}$

$2\sqrt{10} =\sqrt{6^2+(-2)^2}$

$2\sqrt{10} =\sqrt{36+4}$

$2\sqrt{10} =\sqrt{40}$

$2\sqrt{10} = 2\sqrt{10}$ which is true.

Thus, option (b) is correct.

(c) If distance $PQ=\sqrt{10}$. Then,

Using formula (1), we have

$PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\sqrt{10} =\sqrt{(6-0)^2+(0-2)^2}$

$\sqrt{10} =\sqrt{6^2+(-2)^2}$

$\sqrt{10} =\sqrt{36+4}$

$\sqrt{10} =\sqrt{40}$

$\sqrt{10} \neq 2\sqrt{10}$ which is not possible.

Thus, option (c) is incorrect.

(d) If distance $PQ=20$. Then,

Using formula (1), we have

$PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$20=\sqrt{(6-0)^2+(0-2)^2}$

$20=\sqrt{6^2+(-2)^2}$

$20=\sqrt{36+4}$

$20 =\sqrt{40}$

$20 \neq 2\sqrt{10}$ which is not possible.

Thus, option (d) is incorrect.

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