write the equation of the line a) parallel to b)perpendicular to 4x+3y-10=0 and is 3 units from the point (2,-1)
given: 4x+3y - 10 = 0
a) easiest way:
since the new line is parallel it must look like
4x+3y + c = 0
using your distance formula from a point to a line
|4(2) + 3(-1) + c|/√(4^2 + 3^2) = 3
8 - 3 + c = 15
c = 10
4x + 3y + 10 = 0
or
-8 + 3 - c = 15
-c = 20
c = -20
4x + 3y - 20 = 0
We have two equations, since one is 3 units above the given point, and the other is 3 units below the given point (2,-1)
b)for perpendicular, the equation must look like
3x - 4y + k = 0
repeat as above
To find the equation of a line parallel or perpendicular to a given line, we need to determine the slope of the given line.
First, let's rearrange the equation of the given line, 4x + 3y - 10 = 0, into slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept.
4x + 3y - 10 = 0
3y = -4x + 10
y = (-4/3)x + 10/3
From this equation, we can determine that the slope of the given line is -4/3.
a) To find the equation of a line parallel to the given line, it must have the same slope (-4/3).
Using the point-slope form of a linear equation (y - y1 = m(x - x1)), where (x1, y1) is a point on the line, and m is the slope, we can substitute the given point (2, -1) and the slope (-4/3) into the equation.
y - (-1) = (-4/3)(x - 2)
y + 1 = (-4/3)x + 8/3
y = (-4/3)x + 8/3 - 1
y = (-4/3)x + 8/3 - 3/3
y = (-4/3)x + 5/3
Therefore, the equation of the line parallel to 4x + 3y - 10 = 0 and passing through the point (2, -1) is y = (-4/3)x + 5/3.
b) To find the equation of a line perpendicular to the given line, we need to find the negative reciprocal of the slope (-4/3).
The negative reciprocal of -4/3 is 3/4.
Using the same point-slope form, we can substitute the given point (2, -1) and the negative reciprocal slope (3/4) into the equation.
y - (-1) = (3/4)(x - 2)
y + 1 = (3/4)x - 3/2
y = (3/4)x - 3/2 - 2/2
y = (3/4)x - 3/2 - 4/2
y = (3/4)x - 7/2
Therefore, the equation of the line perpendicular to 4x + 3y - 10 = 0 and passing through the point (2, -1) is y = (3/4)x - 7/2.
To find the equation of a line parallel or perpendicular to a given line and passing through a specific point, you need to follow a few steps:
Step 1: Find the slope of the given line.
Step 2: Determine the slope of the line parallel or perpendicular to the given line.
Step 3: Use the point-slope form of a line to find the equation.
Let's start with the given line:
4x + 3y - 10 = 0
Step 1: Find the slope of the given line.
For that, we need to rewrite the equation in slope-intercept form (y = mx + b), where m is the slope:
4x + 3y = 10
3y = -4x + 10
y = (-4/3)x + 10/3
From this equation, we see that the slope of the given line is -4/3.
Step 2: Determine the slope of the line parallel or perpendicular to the given line.
a) Parallel Line:
A parallel line will have the same slope as the given line (-4/3). Therefore, the slope of the line parallel to the given line is also -4/3.
b) Perpendicular Line:
A perpendicular line will have a negative reciprocal slope to the given line. The negative reciprocal of -4/3 is 3/4. Therefore, the slope of the line perpendicular to the given line is 3/4.
Step 3: Use the point-slope form of a line to find the equation.
a) Equation of the line parallel to the given line:
Using the point-slope form (y - y1) = m(x - x1), where (x1, y1) is the point (2, -1):
(y - (-1)) = (-4/3)(x - 2)
y + 1 = (-4/3)(x - 2)
y + 1 = (-4/3)x + 8/3
y = (-4/3)x + 8/3 - 1
y = (-4/3)x + 8/3 - 3/3
y = (-4/3)x + 5/3
Therefore, the equation of the line parallel to the given line and passing through the point (2, -1) is y = (-4/3)x + 5/3.
b) Equation of the line perpendicular to the given line:
Using the point-slope form (y - y1) = m(x - x1), where (x1, y1) is the point (2, -1):
(y - (-1)) = (3/4)(x - 2)
y + 1 = (3/4)(x - 2)
y + 1 = (3/4)x - 3/2
y = (3/4)x - 3/2 - 1
y = (3/4)x - 3/2 - 2/2
y = (3/4)x - 3/2 - 4/2
y = (3/4)x - 7/2
Therefore, the equation of the line perpendicular to the given line and passing through the point (2, -1) is y = (3/4)x - 7/2.