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See below for the distance between the points and the lines

How to determine the distance between the lines and the points?

Question 2

The line and the points are given as:

x = y

P = (4, -2)

Rewrite the equation as:

y = x

The slope of the above equation is

m = 1

The slope of a line perpendicular to it is

m = -1

A linear equation is represented as:

y = mx + b

Substitute m = -1

y = -x + b

Substitute (4, -2) in y = -x + b

-2 = -4 + b

Solve for b

b = 2

Substitute b = 2 in y = -x + b

y = -x + 2

So, we have:

x = y and y = -x + 2

Substitute x for y

x = -x + 2

Solve for x

x = 1

Substitute x = 1 in y = x

y = 1

So, we have the following points

(1, 1) and (4, -2)

The distance between the above points is

d = √(x2 - x1)² + (y2 - y1)²

So, we have:

d = √(1 - 4)² + (1 + 2)²

Evaluate

d = 3√2

Hence, the distance between x = y and P = (4, -2) is 3√2 units

Question 3

The line and the points are given as:

y = 2x + 1

Q = (2, 10)

The slope of the above equation is

m = 2

The slope of a line perpendicular to it is

m = -1/2

A linear equation is represented as:

y = mx + b

Substitute m = -1/2

y = -1/2x + b

Substitute (2, 10) in y = -1/2x + b

10 = -1/2 * 2 + b

Solve for b

b = 11

Substitute b = 11 in y = -1/2x + b

y = -1/2x + 11

So, we have:

y = 2x + 1 and y = -1/2x + 11

Substitute 2x + 1 for y

2x + 1 = -1/2x + 11

Solve for x

x = 4

Substitute x = 4 in y = 2x + 1

y = 9

So, we have the following points

(4, 9) and (2, 10)

The distance between the above points is

d = √(x2 - x1)² + (y2 - y1)²

So, we have:

d = √(4 - 2)² + (9 - 10)²

Evaluate

d = √5

Hence, the distance between the line and the point is √5 units

Question 4

The line and the points are given as:

y = -x + 3

R = (-5, 0)

The slope of the above equation is

m = -1

The slope of a line perpendicular to it is

m = 1

A linear equation is represented as:

y = mx + b

Substitute m = 1

y = x + b

Substitute (-5, 0) in y = x + b

0 = 5 + b

Solve for b

b = -5

Substitute b = 5 in y = x + b

y = x + 5

So, we have:

y = x + 5 and y = -x + 3

Substitute x + 5 for y

x + 5 = -x + 3

Solve for x

x = -1

Substitute x = -1 in y = x + 3

y = 2

So, we have the following points

(-1, 2) and (-5, 0)

The distance between the above points is

d = √(x2 - x1)² + (y2 - y1)²

So, we have:

d = √(-1 + 5)² + (2 - 0)²

Evaluate

d = 2√5

Hence, the distance between the line and the point is 2√5 units

Read more about distance at:

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