Question

If (ɑ,b) is the mid point of the line segment joining the points ɑ(10,-6) ɑnd B(k, 4) ɑnd ɑ-2b= 18, the vɑlue of k.


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Answer 1

Answer:

(a,b) is the mid-point of Line segment joining the points A(10,−6) and B(k,4)

So,

a=

2

10+k

and b=

2

−6+4

=−1

It is given that,

a−2b=18

Put b=−1

a−2(−1)=18

a=18−2=16

Now,

a=

2

10+k

16=

2

10+k

k+10=32

k=22

Distance between the points (x

1

,y

1

) and (x

2

,y

2

) is

(x

2

−x

1

)

2

+(y

2

−y

1

)

2

Distance between the points (10,−6) and (22,4) is

(22−10)

2

+(4+6)

2

=

244

=2

61

Step-by-step explanation:

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Answer 2

$\mathbb{Question:}$

If (ɑ,b) is the mid point of the line segment joining the points ɑ(10,-6) ɑnd B(k, 4) ɑnd ɑ-2b= 18, the vɑlue of k.

ㅤㅤㅤㅤㅤㅤㅤㅤOR

If (a,b) is the mid-point of the line segment joining the points A(10,−6),B(k,4) and a−2b=18, find the value of k and the distance AB.

$\huge \fbox \pink{Answer:}$

(a,b) is the mid-point of Line segment joining the points A(10,−6) and B(k,4)

So,

$⇒a = \frac{10 + k}{2} \: \textbf{and \: } b = \frac{ - 6 + 4}{2} = - 1$

$\textbf{it is given that,}$

$⇒a - 2b = 18$

$⇒ \textbf{put} \: b = - 1$

$⇒a - 2( - 1 ) = 18$

$⇒a = 18 - 2 = 16$

Now,

$⇒a = \frac{10 + k}{2}$

$⇒16 = \frac{10 + k}{2}$

$⇒k + 10 = 32$

$⇒k = 22$

More Answer:

$\text{Distance between the Points} \: (x_{1},y_{1}\text{and} \: (x_{2},y_{2}) \: \text{is}$

$= \sqrt{(x_{2} - x_{1} {)}^{2} + (y_{2} - y_{1}} {)}^{2}$

$\text{Distance between the Points} \: (10 , - 6) \: \text{and} \: (22,4) \: \text{is}$

$⇒ \sqrt{(22 - 10 {)}^{2} + (4 + 6 {)}^{2} }$

$= \sqrt{244}$

$= 2 \sqrt{61}$

btw aapka dil se sukriya thanks dene ke liya

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