A line Q has an equation of x+5y+5=0. Find the equation of the line that is "perpendicular" to Q an passes through (3,1). Find the equation of the line that is "parallel" to Q and passes to through (3,1). Find the equation of the line that passes through (3,1) that makes an "angle of 45 degrees clockwise" from the line that is perpendicular to line Q.
Line Q has slope -1/5; so do all parallel lines
Perpendicular lines P therefore have slope 5
So, just use the point-slope form of a line to write the needed equations.
Now, what is the slope of a line 45° clockwise from P? If P makes an angle Ø with the x-axis, then we want θ such that
tanθ = tan(Ø-45) = (tanØ-1)/(1+tanØ)
= (5-1)/(1+5) = 2/3
Now plug that into your point-slope form.
Well, well, well! Looks like we have quite a few line-related questions here. Let's tackle them one by one, shall we?
First off, we need to find the equation of the line that is perpendicular to line Q and passes through (3,1). To do that, we'll need to find the slope of line Q.
The equation of line Q is x + 5y + 5 = 0, right? Let me rearrange that for you: 5y = -x - 5. Now we can isolate y by dividing everything by 5: y = (-1/5)x - 1.
Now, the slope of line Q is the coefficient of x, which in this case is -1/5. Since lines that are perpendicular to each other have slopes that are negative reciprocals, the slope of our new line will be the negative reciprocal of -1/5. That's 5/1, or simply 5!
So, we have the slope of the new line, and we also have a point it passes through, which is (3,1). Using the point-slope form of a line, we can write the equation of the line that is perpendicular to line Q and passes through (3,1): y - 1 = 5(x - 3).
Now, onto the line that is parallel to line Q and passes through (3,1). Parallel lines have the same slope, so we can use the slope of line Q, which is -1/5. Again using the point-slope form, we can write the equation: y - 1 = (-1/5)(x - 3).
Now, for the third one. We want a line that makes an angle of 45 degrees clockwise from the line that is perpendicular to line Q and passes through (3,1). To find the slope of this line, we'll use a little trigonometry. The slope of the line perpendicular to line Q is 5, so the slope of the line that makes a 45 degree angle clockwise from it will be -5.
Using the point-slope form again, we can write the equation of this line: y - 1 = -5(x - 3).
Remember, these are just the equations for the lines! If you need the actual jokes, well, we might have to work on that one!
To find the equation of the line that is perpendicular to line Q and passes through (3,1), we need to determine the slope of line Q first.
Step 1: Rewrite the equation of line Q in slope-intercept form (y = mx + b).
x + 5y + 5 = 0
5y = -x - 5
y = -1/5x - 1
From this equation, we can see that the slope of line Q is -1/5.
Step 2: Determine the slope of the line that is perpendicular to line Q.
The slopes of perpendicular lines are negative reciprocals of each other.
Therefore, the slope of the line perpendicular to line Q is 5/1, which can be simplified to 5.
Step 3: Use the point-slope form of a line (y - y1 = m(x - x1)), where (x1, y1) is the given point (3,1) and m is the slope of the line.
y - 1 = 5(x - 3)
y - 1 = 5x - 15
y = 5x - 14
So, the equation of the line that is perpendicular to line Q and passes through (3,1) is y = 5x - 14.
To find the equation of the line that is parallel to line Q and passes through (3,1), we use the same slope as line Q since parallel lines have the same slope.
Step 4: Use the point-slope form of a line (y - y1 = m(x - x1)), where (x1, y1) is the given point (3,1) and m is the slope of line Q (-1/5).
y - 1 = -1/5(x - 3)
y - 1 = -1/5x + 3/5
y = -1/5x + 8/5
So, the equation of the line that is parallel to line Q and passes through (3,1) is y = -1/5x + 8/5.
To find the equation of the line that passes through (3,1) and makes an angle of 45 degrees clockwise from the line perpendicular to line Q, we need to determine the slope and use the angle rotation formula.
Step 5: Calculate the slope of the line perpendicular to line Q (same as Step 3) which is 5.
Step 6: Calculate the slope of the line that makes an angle of 45 degrees clockwise from the line perpendicular to line Q.
We know that the slope of a line with a 45-degree clockwise angle from another line is the negative reciprocal of 1.
Therefore, the slope of the line we are looking for is -1.
Step 7: Use the point-slope form of a line (y - y1 = m(x - x1)), where (x1, y1) is the given point (3,1) and m is the slope of the line (-1).
y - 1 = -1(x - 3)
y - 1 = -x + 3
y = -x + 4
So, the equation of the line that passes through (3,1) and makes an angle of 45 degrees clockwise from the line perpendicular to line Q is y = -x + 4.
To find the equation of a line that is perpendicular to a given line, we need to determine the slope of the given line and then find the negative reciprocal of the slope.
1. Perpendicular line: Line Q has the equation x + 5y + 5 = 0.
Let's rearrange the equation in slope-intercept form (y = mx + c):
x + 5y + 5 = 0
5y = -x - 5
y = -x/5 - 1
The slope of line Q is -1/5.
The negative reciprocal of -1/5 is 5. So, the slope of the perpendicular line is 5.
Now, we can use the slope-intercept form to find the equation of the perpendicular line passing through (3,1):
y - y1 = m(x - x1), where (x1, y1) is the point (3,1) and m is the slope of the perpendicular line.
Plugging in the values, we get:
y - 1 = 5(x - 3)
y - 1 = 5x - 15
y = 5x - 14
Therefore, the equation of the line perpendicular to Q and passing through (3,1) is y = 5x - 14.
2. Parallel line: To find the equation of a line parallel to a given line, we need to use the same slope as the given line.
The slope of line Q is -1/5. So, the slope of the parallel line will also be -1/5.
Using the point-slope form of a line, we can write the equation:
y - y1 = m(x - x1), where (x1, y1) is the point (3,1) and m is the slope of the parallel line.
Plugging in the values, we get:
y - 1 = -1/5(x - 3)
y - 1 = -1/5x + 3/5
y = -1/5x + 8/5
Therefore, the equation of the line parallel to Q and passing through (3,1) is y = -1/5x + 8/5.
3. Line at a 45-degree clockwise angle from the perpendicular line:
To find the equation of a line at a specific angle from a given line, we can use either trigonometry or algebraic manipulation. Here, let's use algebraic manipulation.
The angle between the perpendicular line and the desired line is 45 degrees clockwise.
For any line, if we rotate its slope by 90 degrees clockwise, we obtain the negative reciprocal of the original slope.
So, the slope of the desired line will be the negative reciprocal of the slope of the perpendicular line, which is -1/5.
Using the point-slope form of a line, we can write the equation:
y - y1 = m(x - x1), where (x1, y1) is the point (3,1) and m is the slope of the desired line.
Plugging in the values, we get:
y - 1 = -1/5(x - 3)
y - 1 = -1/5x + 3/5
y = -1/5x + 8/5
Therefore, the equation of the line that passes through (3,1) and makes an angle of 45 degrees clockwise from the line perpendicular to Q is y = -1/5x + 8/5, which is the same as the equation of the line parallel to Q.