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. Do it fɑst✔︎ About the author Hailey
[tex]\mathbb{Question:} [/tex] 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. [tex] \\ [/tex] [tex]\huge \fbox \pink{Answer:}[/tex] (a,b) is the mid-point of Line segment joining the points A(10,−6) and B(k,4) So, [tex]⇒a = \frac{10 + k}{2} \: \textbf{and \: } b = \frac{ – 6 + 4}{2} = – 1[/tex] [tex] \textbf{it is given that,} [/tex] [tex]⇒a – 2b = 18[/tex] [tex]⇒ \textbf{put} \: b = – 1[/tex] [tex]⇒a – 2( – 1 ) = 18[/tex] [tex]⇒a = 18 – 2 = 16[/tex] Now, [tex]⇒a = \frac{10 + k}{2} [/tex] [tex]⇒16 = \frac{10 + k}{2} [/tex] [tex]⇒k + 10 = 32[/tex] [tex]⇒k = 22[/tex] More Answer: [tex]\text{Distance between the Points} \: (x_{1},y_{1}\text{and} \: (x_{2},y_{2}) \: \text{is}[/tex] [tex] = \sqrt{(x_{2} – x_{1} {)}^{2} + (y_{2} – y_{1}} {)}^{2} [/tex] [tex]\text{Distance between the Points} \: (10 , – 6) \: \text{and} \: (22,4) \: \text{is}[/tex] [tex]⇒ \sqrt{(22 – 10 {)}^{2} + (4 + 6 {)}^{2} } [/tex] [tex] = \sqrt{244} [/tex] [tex] = 2 \sqrt{61} [/tex] [tex] \\ \\ \\ [/tex] btw aapka dil se sukriya thanks dene ke liya [tex] \\ \\ \\ \\ \sf \colorbox{gold} {\red(ANSWER ᵇʸ ⁿᵃʷᵃᵇ⁰⁰⁰⁸}[/tex] Reply
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: Hope it’s helpful… pls mark me brainliest… Reply
[tex]\mathbb{Question:} [/tex]
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.
[tex] \\ [/tex]
[tex]\huge \fbox \pink{Answer:}[/tex]
(a,b) is the mid-point of Line segment joining the points A(10,−6) and B(k,4)
So,
[tex]⇒a = \frac{10 + k}{2} \: \textbf{and \: } b = \frac{ – 6 + 4}{2} = – 1[/tex]
[tex] \textbf{it is given that,} [/tex]
[tex]⇒a – 2b = 18[/tex]
[tex]⇒ \textbf{put} \: b = – 1[/tex]
[tex]⇒a – 2( – 1 ) = 18[/tex]
[tex]⇒a = 18 – 2 = 16[/tex]
Now,
[tex]⇒a = \frac{10 + k}{2} [/tex]
[tex]⇒16 = \frac{10 + k}{2} [/tex]
[tex]⇒k + 10 = 32[/tex]
[tex]⇒k = 22[/tex]
More Answer:
[tex]\text{Distance between the Points} \: (x_{1},y_{1}\text{and} \: (x_{2},y_{2}) \: \text{is}[/tex]
[tex] = \sqrt{(x_{2} – x_{1} {)}^{2} + (y_{2} – y_{1}} {)}^{2} [/tex]
[tex]\text{Distance between the Points} \: (10 , – 6) \: \text{and} \: (22,4) \: \text{is}[/tex]
[tex]⇒ \sqrt{(22 – 10 {)}^{2} + (4 + 6 {)}^{2} } [/tex]
[tex] = \sqrt{244} [/tex]
[tex] = 2 \sqrt{61} [/tex]
[tex] \\ \\ \\ [/tex]
btw aapka dil se sukriya thanks dene ke liya
[tex] \\ \\ \\ \\ \sf \colorbox{gold} {\red(ANSWER ᵇʸ ⁿᵃʷᵃᵇ⁰⁰⁰⁸}[/tex]
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:
Hope it’s helpful…
pls mark me brainliest…