Find the angle the normal makes with the positive x–axis.
1. 3x – 2y – 1 = 0
2.5x + y – 12 = 0
3. 4x + 3y – 4 = 0
I cant Figure any of these out or how to do them..
You should have learned that the slope of a line is equal to tanØ, where Ø is the angle that the line makes with the positive x-axis
for #1
slope of that line is 3/2
so tanØ = 3/2
Ø = tan^-1 (3/2)
my calculator gave me appr. 56.3°
do the others the same way
1. sqrt2x + sqrt2y + 36 = 0
2. Normal form: 2x/sqrt53 + 7y/sqrt53 - 4/sqrt53 = 0 Angle:74
3. 3x/srt13 + 2y/sqrt13 - 5/sqr13 = 0 Angle:34
4. 326
5. 11
6. 37
To find the angle the normal makes with the positive x-axis, we need to determine the slope of the given line. The slope of the line is related to the coefficients of x and y in the equation.
The general form of a line equation is Ax + By + C = 0, where A, B, and C are constants.
Let's solve each equation to find their slopes:
1. 3x - 2y - 1 = 0:
Rearrange the equation to get it in the form y = mx + c:
3x - 2y = 1
-2y = -3x + 1
y = (3/2)x - 1/2
The slope (m) in this case is (3/2).
2.5x + y - 12 = 0:
Rearrange the equation to get y = mx + c form:
y = -5x + 12
The slope (m) in this case is (-5).
3. 4x + 3y - 4 = 0:
Rearrange the equation to get y = mx + c form:
3y = -4x + 4
y = (-4/3)x + 4/3
The slope (m) in this case is (-4/3).
Now that we have the slopes, we can find the angles the normals make with the positive x-axis using trigonometry.
The angle (θ) that a line makes with the positive x-axis can be found using the formula: θ = tan^(-1)(m), where m is the slope of the line.
Let's calculate the angles for each equation:
1. For the equation 3x - 2y - 1 = 0:
θ = tan^(-1)(3/2)
θ ≈ 56.31 degrees
2. For the equation 5x + y - 12 = 0:
θ = tan^(-1)(-5)
θ ≈ -78.69 degrees
3. For the equation 4x + 3y - 4 = 0:
θ = tan^(-1)(-4/3)
θ ≈ -53.13 degrees
So, the angles the normals make with the positive x-axis are approximately:
1. 56.31 degrees
2. -78.69 degrees
3. -53.13 degrees