Question

Given: A(-3,8) and B(-5,8)
i) Find the co-ordinates of mid-point of segment AB.
ii) What is y co-ordinate on point A & B ?
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Answer 1

Given Co-ordinates of A( -3 ,8) and B (-5 ,8)

Question-1

We have to find mid-point of AB

$\bf{Mid point formula:-} = \bf\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}$

A( -3 ,8) and B (-5 ,8)

$\bf{x_1 = -3}$

$\bf{x_2 = -5}$

$\bf{y_1 = 8}$

$\bf{y_2=8}$

$\bf{Mid point :-} = \bf\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}$

$\bf{Mid point :-} = \bf\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}$

$\bf{Mid point } = \bf\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}$

$\bf{Mid point } = \bf\dfrac{-3-5}{2}, \dfrac{8+8}{2}$

$\bf{Mid point } = \bf\dfrac{-8}{2}, \dfrac{16}{2}$

$\bf{Mid point } = \bf{(-4,8)}$

So, the co-ordinates mid point are (-4,8)

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Question-2 :-

We have to find the y co-ordinates on point (A,B)

A(-3,8) and B(-5,8)

y co-ordinate on point A is 8

y co-ordinate on point B is 8

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Know more :-

Distance formula:-

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

Centroid formula:-

$\bf\dfrac{x_1+x_2+x_3}{3},\bf\dfrac{y_1+y_2+y_3}{3}$

Section formula Internal

division

$\bf\dfrac{mx_2+nx_1}{m+n}, \bf\dfrac{my_2+ny_1}{m+n}$

Section formula External division

$\bf\dfrac{mx_2-nx_1}{m-n}, \bf\dfrac{my_2-ny_1}{m-n}$