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
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