A line Q has an equation of x+5x+5=0. Find the equation of the line that is "perpendicular" to Q and passes through (3,1). Find the question 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.
it's genghis khan
To find the equation of the line perpendicular to line Q and passing through the point (3,1), we need to first find the slope of line Q.
The equation of line Q is given as x + 5x + 5 = 0. Simplifying this equation, we get 6x + 5 = 0. Then, solving for x, we have x = -5/6.
The slope of line Q can be found from its equation in the form y = mx + c, where m represents the slope. Comparing the equation with y = mx + c, we have y = (0)x + (5/6), which means the slope of line Q is 0.
To find the slope of the line perpendicular to line Q, we use the fact that the product of the slopes of two perpendicular lines is -1. Therefore, the slope of the line perpendicular to line Q is the negative reciprocal of the slope of line Q, which is undefined (since the slope of line Q is 0).
Since a vertical line passing through (3,1) has an undefined slope, the equation of the line perpendicular to Q and passing through (3,1) is x = 3.
Now let's find the equation of the line parallel to line Q and passing through (3,1).
Since line Q has a slope of 0, any line parallel to it will have the same slope of 0. Therefore, the equation of the line parallel to Q and passing through (3,1) can be written as y = 1.
Lastly, let's find the equation of the line passing through (3,1) that makes an angle of 45 degrees clockwise from the line perpendicular to line Q.
Since the line perpendicular to line Q is vertical (x = 3), a line making a 45-degree clockwise angle from it will have a slope equal to the negative reciprocal of the tangent of 45 degrees.
The tangent of 45 degrees is 1, so the negative reciprocal of 1 is -1. Therefore, the slope of the line making a 45-degree clockwise angle from the line perpendicular to line Q is -1.
Using the point-slope form of the equation of a line, y - y1 = m(x - x1), where (x1, y1) is the given point (3,1) and m is the slope (-1), the equation of the line passing through (3,1) that makes an angle of 45 degrees clockwise from the line perpendicular to line Q is y - 1 = -1(x - 3). Simplifying this equation gives us y = -x + 4.