Question

8. Find the values of y for which the distance between the point P(2, -3) and Q(10, y) is 10 units.​

Answer 1

Answer:-

y = 6, -9

Given :-

Distance between P( 2, -3) and Q(10, y) is 10 units

To find :-

Value of y

Solution:-

If (x_1, y_1) and (x_2 , y_2) are the points distance between them is

Distance formula :-

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

So,

$(x_1 ,\: y_1) = (2 ,\: - 3) \\ (x_2 ,\: y_2) = (10, \: y)$

Substituting the values ,

$\: 10 = \sqrt{(2 - 10) {}^{2} + ( - 3 - y) {}^{2} }$

Squaring on both sides,

$(10) {}^{2} = \bigg( \sqrt{(2 - 10) {}^{2} + ( - 3 - y) {}^{2} } \bigg) {}^{2}$

$100 = (2 - 10) {}^{2} + ( - 3 - y) {}^{2}$

$100 = ( - 8) {}^{2} + ( - 3 - y) {}^{2}$

$100 = 64 + ( - ) {}^{2} (3 + y) {}^{2}$

$100 = 64 + (3 + y) {}^{2}$

$100 - 64 = (3 + y) {}^{2}$

$36 = (3 + y) {}^{2}$

$(6) {}^{2} = (3 + y) {}^{2}$

$(3 + y) = \pm \: 6$

Case - 1:-

$3 + y = + 6 \:$

$y = + 6 - 3$

$\red {y = + 3}$

Case - 2:-

$3 + y = \: - 6$

$y = - 6 - 3$

$\red{y = \: - 9}$

$\red{ \boxed{So \: the \: value \: of \: y \: are \purple{\: 3 \: and \: - 9}}}$

Know more :-

Distance formula can also be used as

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

The distance of the point (x,y) from origin (0,0) is

$\sqrt{x^2 + y^2 }$

Answer 2

Step-by-step explanation:

Answer:-

y = 6, -9

Given :-

Distance between P( 2, -3) and Q(10, y) is 10 units

To find :-

Value of y

Solution:-

If (x_1, y_1) and (x_2 , y_2) are the points distance between them is

Distance formula :-

\red {\sqrt{(x_1 - x_2) {}^{2} + (y_1 - y_2) {}^{2} } }

(x

1

−x

2

)

2

+(y

1

−y

2

)

2

So,

\begin{gathered}(x_1 ,\: y_1) = (2 ,\: - 3) \\ (x_2 ,\: y_2) = (10, \: y)\end{gathered}

(x

1

,y

1

)=(2,−3)

(x

2

,y

2

)=(10,y)

Substituting the values ,

\: 10 = \sqrt{(2 - 10) {}^{2} + ( - 3 - y) {}^{2} }10=

(2−10)

2

+(−3−y)

2

Squaring on both sides,

(10) {}^{2} = \bigg( \sqrt{(2 - 10) {}^{2} + ( - 3 - y) {}^{2} } \bigg) {}^{2}(10)

2

=(

(2−10)

2

+(−3−y)

2

)

2

100 = (2 - 10) {}^{2} + ( - 3 - y) {}^{2}100=(2−10)

2

+(−3−y)

2

100 = ( - 8) {}^{2} + ( - 3 - y) {}^{2}100=(−8)

2

+(−3−y)

2

100 = 64 + ( - ) {}^{2} (3 + y) {}^{2}100=64+(−)

2

(3+y)

2

100 = 64 + (3 + y) {}^{2}100=64+(3+y)

2

100 - 64 = (3 + y) {}^{2}100−64=(3+y)

2

36 = (3 + y) {}^{2}36=(3+y)

2

(6) {}^{2} = (3 + y) {}^{2}(6)

2

=(3+y)

2

(3 + y) = \pm \: 6(3+y)=±6

Case - 1:-

3 + y = + 6 \:3+y=+6

y = + 6 - 3y=+6−3

\red {y = + 3}y=+3

Case - 2:-

3 + y = \: - 63+y=−6

y = - 6 - 3y=−6−3

\red{y = \: - 9}y=−9

\red{ \boxed{So \: the \: value \: of \: y \: are \purple{\: 3 \: and \: - 9}}}

Sothevalueofyare3and−9

Know more :-

Distance formula can also be used as

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

(x

2

−x

1

)

2

+(y

2

−y

1

)

2

The distance of the point (x,y) from origin (0,0) is

\sqrt{x^2 + y^2 }

x

2

+y

2