Question

(c) Given the points (-4,8) and (6, -12):

(i) Determine the midpoint of the line segment connecting the points.

(ii) Determine the distance separating the two points.

Answer 1

Step-by-step explanation:

I ) mid point = (-1,-2)

ii.) distance = √20

Answer 2

(i) The midpoint of the line segment connecting the points is (1, - 2)

(ii) The distance separating the two points = 10√5 unit

Given :

The points (-4,8) and (6, -12)

To find :

(i) The midpoint of the line segment connecting the points

(ii) The distance separating the two points

Formula :

1. For the given two points ( x₁ , y₁) & (x₂ , y₂)

The midpoint of the line AB is

$\displaystyle \sf{ \bigg( \frac{x_1 + x_2}{2} , \frac{y_1 + y_2}{2} \bigg)}$

2. For the given two points ( x₁ , y₁) & (x₂ , y₂) the distance between the points

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

Solution :

Step 1 of 3 :

Write down the given points

Here the given points are (-4,8) and (6, -12)

Step 2 of 3 :

Find the midpoint of the line segment connecting the points

The midpoint of the line segment connecting the points (-4,8) and (6, -12)

$\displaystyle \sf{ = \bigg( \frac{ - 4 + 6}{2} , \frac{8 - 12}{2} \bigg)}$

$\displaystyle \sf{ = \bigg( \frac{ 2}{2} , \frac{ - 4}{2} \bigg)}$

$\displaystyle \sf{ = ( 1 , - 2)}$

Step 3 of 3 :

Find the distance separating the two points

The distance separating the two points (-4,8) and (6, -12)

$\displaystyle \sf{ = \sqrt{ {( - 4 - 6)}^{2} + {(8 + 12)}^{2} } \: \: unit }$

$\displaystyle \sf{ = \sqrt{ {( - 10)}^{2} + {(20)}^{2} } \: \: unit }$

$\displaystyle \sf{ = \sqrt{ 100 + 400 } \: \: unit }$

$\displaystyle \sf{ = \sqrt{ 500 } \: \: unit }$

$\displaystyle \sf{ = \sqrt{ 10 \times 10 \times 5 } \: \: unit }$

$\displaystyle \sf{ = 10\sqrt{5 } \: \: unit }$

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