Math, asked by sanjeevkoravi54589, 1 year ago

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

Answers

Answered by Anonymous
67

Question

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

Solution

K = 46

Given

The mid point of AB is (a,b)

'a' and 'b' are defined as : a - 2b = 18

Using Mid Point Formula,

 \sf{(a ,\: b) = ( \dfrac{10 + k}{2} \: ,  \dfrac{6 + 4}{2}  )}

Implies,

  \boxed{ \boxed{ \sf \: a =  \dfrac{10 + k}{2}}}

Also,

 \boxed{ \boxed{ \sf \: 2b = 10}}

Now,

 \sf \: a - 2b = 18 \\  \\  \leadsto \:  \sf \: a - 10 = 18 \\  \\  \leadsto \:  \sf \: a = 28

Thus,

 \sf \: 28 =  \dfrac{10 + k}{2}  \\  \\  \longmapsto \:  \sf \: 56 = 10 + k \\  \\  \\  \longmapsto \:  \sf \: k = 46

Answered by Equestriadash
72

Given: (a, b) is the mid - point of the line segment joining A(10, 6) and B(k, 4); a - 2b = 18.

To find: The value of k and the distance of AB.

Answer:

Formula to find the mid - point:

\tt P(x, y)\ =\ \bigg(\dfrac{x_1\ +\ x_2}{2},\ \dfrac{y_1\ +\ y_2}{2}\bigg)

From the given points,

\tt x_1\ =\ 10\\\\x_2\ =\ k\\\\y_1\ =\ 6\\\\y_2\ =\ 4

Using these in the formula,

\tt (a, b)\ =\ \bigg(\dfrac{10\ +\ k}{2}, \dfrac{6\ +\ 4}{2}\bigg)\\\\\\\implies\ a\ =\ \dfrac{10\ +\ k}{2}\ \ \ and\ \ \ b\ =\ \dfrac{6\ +\ 4}{2}\\\\\\\\b\ =\ \dfrac{6\ +\ 4}{2}\\\\\\2b\ =\ 10\\\\\\b\ =\ 5

Using b = 5 in a - 2b = 18,

\tt a\ -\ 2\ \times 5\ =\ 18\\\\\\a\ -\ 10\ =\ 18\\\\\\a\ =\ 18\ +\ \ 10\\\\\\a\ =\ 28

Now using a = 28 to find k,

\tt 28\ =\ \dfrac{10\ +\ k}{2}\\\\\\56\ =\ 10\ +\ k\\\\\\56\ -\ 10\ =\ k\\\\\\46\ =\ k

Therefore, the points are A(10, 6) and B(6, 4).

Formula to find the distance between two points:

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

From the points, we have,

\tt x_1\ =\ 10\\\\x_2\ =\ 46\\\\y_1\ =\ 6\\\\y_2\ =\ 4

Using them in the formula,

\tt Distance\ =\ \sqrt{\bigg(46\ -\ 10\bigg)^2\ +\ \bigg(4\ -\ 6\bigg)^2}\\\\\\\\\\Distance\ =\ \sqrt{\bigg(36\bigg)^2\ +\ \bigg(-2\bigg)^2}\\\\\\\\\\Distance\ =\ \sqrt{\bigg(1296\ +\ 4\bigg)^2}\\\\\\\\\\Distance\ =\ \sqrt{\bigg(1300\bigg)^2}\\\\\\\\\bf Distance\ =\ 1300\ units.

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