Straight Lines
Equation of a line:
y = mx + b
m = slope b = y- intercept (where the line crosses on the y-axis)
How to find the slope given the coordinates of 2 points on that line:
m = y2 – y1
x2 – x1
Example: (3,5) (1,-4)
m = -4 – 5
1 – 3
m = 9
-2
How to find an equation of a line given its slope (m) and its y-intercept:
y = mx + b
Example: m = 1 , b = 5
2
Plug-in what you know: y = 1x + 5
2
Formula to find the midpoint of a given line:
P = (x1 + x2) , (y1 + y2)
2 2
Example: End points of a line segment (0, 4) (3, -2)
P = (0+ 3) , (4 + (-2))
2 2
P = (1.5, 1)
How to find an equation of a line given its slope (m) and a point on that line:
Example: m = 2 (4, 4)
3
Plug-in what we know: y = mx + b
Y = 2x + b
3
Now plug-in the coordinate: (4, 4)
4 = 2(4) + b
3
4 = 8 + b
3
4 – 8 = b *find common denominator (= 3)
3
12 – 8 = b
3 3
4 = b
3
The equation is: y = 2x + 4
3 3
How to find the equation of a line, given 2 points:
Example: (-2, 6) (3, -5)
Find the slope of the line by using m = y2 – y1
x2 – x1
-5 – 6 = -11 = m
3- (-2) 5
Now use the slope and one of the given points and plug in to m = y – y1
Use: M = -11 (-2, 6) x – x1
5
-11 = y – 6
5 X – (-2) *now cross-multiply
-11(x+2) = 5(y-6)
-11x – 22 = 5y – 30
CONVERT TO y = mx + b : -5y = 11x - 8
-5 -5
y = 11x + 8
-5 -5
In order for two lines to be perpendicular, the product of their slopes has to be -1. If a line has a slope of 1/2, the slope of the perpendicular line is its negative reciprocal = -2/1
Therefore: 1 x -2 = -2 = -1
2 1 2
Parallel lines are lines that have the exact same slope. m1 = m2
Two lines are coincident if they have the same slope and the same y-intercept.
How to find the equation of a line given a point on that line and the equation of a line parallel to it:
1. Find the slope of the known equation
2. Let m1 = m2 (since all parallel lines have the same slope)
3. Find the equation using the slope and the coordinates of a given point
Example: Find the equation of a line (l2) if it passes through (2, -1) and is parallel to line (l1) whose equation is 2x – 3y + 6 = 0
-3y = -2x – 6
-3y = -2x – 6
-3 -3 -3
y = 2x + 2
3
m = 2
3 (since we are working with parallel lines, the slopes are the same) m1 = m2
Now plug-in to m = y2 – y1
x2 – x1
2 = y – (-1)
3 x – 2
(Cross-multiply)
2(x – 2) = 3 (y +1)
2x – 4 = 3y + 3
-3y = -2x + 7
3y = 2x – 7
y = 2x – 7
3 3
How to find the equation of a line given a point on that line and the equation of a line perpendicular to it:
1. Determine the slope of known equation
2. Take the found slope and find its negative reciprocal (since they are perpendicular lines)
3. Find the equation of the line using the value of the slope and the coordinate of a given point
Example: Find the equation of a line that passes through the point (7, 4) and that is perpendicular to a line whose equation is 2x + y – 9 = 0 3
y = -2x + 9 m1 = -2 then m2 = 3
3 3 2
Now plug-in to m = y2 – y1
x2 – x1
3 = y – 4
2 x – 7
3(x – 7) = 2(y – 4)
3x – 21 = 2y – 8
-2y = -3x + 13
-2 -2 -2
y = 3x - 13
2 2
Distance between two points on a Cartesian plane (result is always in units):
Formula: √ (x2 – x1)² + (y2 – y1)²
Example: Given two points, find the distance of the line.
P1 (3 , 4) P2 (6, 8)
√ (x2 – x1)² + (y2 – y1)²
√(6-3)² + (8-4)²
√(3² ) +(4² )
√ (9) + (16)
√ 25
= 5 units
Equation of a line:
y = mx + b
m = slope b = y- intercept (where the line crosses on the y-axis)
How to find the slope given the coordinates of 2 points on that line:
m = y2 – y1
x2 – x1
Example: (3,5) (1,-4)
m = -4 – 5
1 – 3
m = 9
-2
How to find an equation of a line given its slope (m) and its y-intercept:
y = mx + b
Example: m = 1 , b = 5
2
Plug-in what you know: y = 1x + 5
2
Formula to find the midpoint of a given line:
P = (x1 + x2) , (y1 + y2)
2 2
Example: End points of a line segment (0, 4) (3, -2)
P = (0+ 3) , (4 + (-2))
2 2
P = (1.5, 1)
How to find an equation of a line given its slope (m) and a point on that line:
Example: m = 2 (4, 4)
3
Plug-in what we know: y = mx + b
Y = 2x + b
3
Now plug-in the coordinate: (4, 4)
4 = 2(4) + b
3
4 = 8 + b
3
4 – 8 = b *find common denominator (= 3)
3
12 – 8 = b
3 3
4 = b
3
The equation is: y = 2x + 4
3 3
How to find the equation of a line, given 2 points:
Example: (-2, 6) (3, -5)
Find the slope of the line by using m = y2 – y1
x2 – x1
-5 – 6 = -11 = m
3- (-2) 5
Now use the slope and one of the given points and plug in to m = y – y1
Use: M = -11 (-2, 6) x – x1
5
-11 = y – 6
5 X – (-2) *now cross-multiply
-11(x+2) = 5(y-6)
-11x – 22 = 5y – 30
CONVERT TO y = mx + b : -5y = 11x - 8
-5 -5
y = 11x + 8
-5 -5
In order for two lines to be perpendicular, the product of their slopes has to be -1. If a line has a slope of 1/2, the slope of the perpendicular line is its negative reciprocal = -2/1
Therefore: 1 x -2 = -2 = -1
2 1 2
Parallel lines are lines that have the exact same slope. m1 = m2
Two lines are coincident if they have the same slope and the same y-intercept.
How to find the equation of a line given a point on that line and the equation of a line parallel to it:
1. Find the slope of the known equation
2. Let m1 = m2 (since all parallel lines have the same slope)
3. Find the equation using the slope and the coordinates of a given point
Example: Find the equation of a line (l2) if it passes through (2, -1) and is parallel to line (l1) whose equation is 2x – 3y + 6 = 0
-3y = -2x – 6
-3y = -2x – 6
-3 -3 -3
y = 2x + 2
3
m = 2
3 (since we are working with parallel lines, the slopes are the same) m1 = m2
Now plug-in to m = y2 – y1
x2 – x1
2 = y – (-1)
3 x – 2
(Cross-multiply)
2(x – 2) = 3 (y +1)
2x – 4 = 3y + 3
-3y = -2x + 7
3y = 2x – 7
y = 2x – 7
3 3
How to find the equation of a line given a point on that line and the equation of a line perpendicular to it:
1. Determine the slope of known equation
2. Take the found slope and find its negative reciprocal (since they are perpendicular lines)
3. Find the equation of the line using the value of the slope and the coordinate of a given point
Example: Find the equation of a line that passes through the point (7, 4) and that is perpendicular to a line whose equation is 2x + y – 9 = 0 3
y = -2x + 9 m1 = -2 then m2 = 3
3 3 2
Now plug-in to m = y2 – y1
x2 – x1
3 = y – 4
2 x – 7
3(x – 7) = 2(y – 4)
3x – 21 = 2y – 8
-2y = -3x + 13
-2 -2 -2
y = 3x - 13
2 2
Distance between two points on a Cartesian plane (result is always in units):
Formula: √ (x2 – x1)² + (y2 – y1)²
Example: Given two points, find the distance of the line.
P1 (3 , 4) P2 (6, 8)
√ (x2 – x1)² + (y2 – y1)²
√(6-3)² + (8-4)²
√(3² ) +(4² )
√ (9) + (16)
√ 25
= 5 units