If the two lines are perpendicular, then the product of their slopes is equal to - 1
m1 x m2 = -1
To find the equation of a perpendicular line, we are using the Point Slope Form.
y - y1 = m(x - x1)
Write down the equation of each of the following lines.
Example 1 :
Perpendicular to y = 2x + 4 and passing through (0, 3)
Solution :
y = 2x + 4 ----(1)
Comparing slope intercept form y = mx + c in (1), we get
Slope (m1) = 2
Since the two lines are perpendicular, then the product of their slopes is equal to - 1.
m1 x m2 = -1
2 x m2 = -1
m2 = -1/2
Using Point Slope Form :
y - y1 = m(x - x1)
We have,
Slope (m) = -1/2, Point (x1, y1) = (0, 3)
y - 3 = -1/2(x - 0)
y - 3 = -1/2x
2(y - 3) = -x
2y - 6 = -x
x + 2y - 6 = 0
So, the equation of the line is x + 2y - 6 = 0.
Example 2 :
Perpendicular to y = -3x - 8 and passing through (0, -2)
Solution :
y = -3x - 8 ----(1)
Comparing slope intercept form y = mx + c in (1), we get
Slope (m1) = -3
Since the two lines are perpendicular, then the product of their slopes is equal to - 1.
m1 x m2 = -1
-3 x m2 = -1
m2 = 1/3
Using Point Slope Form :
y - y1 = m(x - x1)
We have,
Slope (m) = 1/3, Point (x1, y1) = (0, -2)
y + 2 = 1/3(x - 0)
y + 2 = 1/3x
3(y + 2) = x
3y + 6 = x
x - 3y - 6 = 0
So, the equation of the line is x - 3y - 6 = 0.
Example 3 :
Perpendicular to x + y = 6 and passing through (0, 1)
Solution :
x + y = 6
y = -x + 6 ----(1)
Comparing slope intercept form y = mx + c in (1), we get
Slope (m1) = -1
m1 x m2 = -1
-1 x m2 = -1
m2 = 1
Using Point Slope Form :
y - y1 = m(x - x1)
We have,
Slope (m) = 1, Point (x1, y1) = (0, 1)
y - 1 = 1(x - 0)
y - 1 = x
x - y + 1 = 0
So, the equation of the line is x - y + 1 = 0.
Example 4 :
Perpendicular to y = (1/3)x - 2 and passing through the origin
Solution :
y = (1/3)x - 2 ----(1)
Comparing slope intercept form y = mx + c in (1), we get
Slope (m1) = 1/3
m1 x m2 = -1
1/3 x m2 = -1
m2 = -3
Using Point Slope Form :
y - y1 = m(x - x1)
We have,
Slope (m) = -3, Point (x1, y1) = (0, 0)
y - 0 = -3(x - 0)
y - 0 = -3x
3x + y - 0 = 0
So, the equation of the line is 3x + y - 0 = 0.
Example 5 :
Perpendicular to y = (-1/5)x + 8 and passing through (0, -2)
Solution :
y = (-1/5)x + 8 ----(1)
Comparing slope intercept form y = mx + c in (1), we get
Slope (m1) = -1/5
m1 x m2 = -1
-1/5 x m2 = -1
m2 = 5
Using Point Slope Form :
y - y1 = m(x - x1)
We have,
Slope (m) = 5, Point (x1, y1) = (0, -2)
y + 2 = 5(x - 0)
y + 2 = 5x
5x - y - 2 = 0
So, the equation of the line is 5x - y - 2 = 0.
Example 6 :
Perpendicular to y = (-2/9)x - 10 and passing through (0, 6)
Solution :
y = (-2/9)x - 10 ----(1)
Comparing slope intercept form y = mx + c with (1), we get
Slope (m1) = -2/9
m1 x m2 = -1
-2/9 x m2 = -1
m2 = 9/2
Using Point Slope Form :
y - y1 = m(x - x1)
We have,
Slope (m) = 9/2, Point (x1, y1) = (0, 6)
y - 6 = (9/2)(x - 0)
y - 6 = (9/2)x
2(y - 6) = 9x
2y - 12 = 9x
9x - 2y + 12 = 0
So, the equation of the line is 9x - 2y + 12 = 0.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM