find the distance between the lines with equations 3x -4y=8 and y=3x/4+4.
lines are
y = (3/4) x + 4 through (0,4)(-16/3,0)
and
y = (3/4) x - 2 through (0,-2)(8/3,0)
so they are indeed parallel (same slope)
sketch the lines
the lower line goes up (2 while it goes right (8/3)
so the tangent of its angle to the y axis is (8/3)/2 = 8/6 = 4/3
or tan T = 4/3
of course the hypotenuse of a triangle with tangent = 4/3 is 5
so sin T = 4/5
and cos T 3/5
now if we draw a line perpendicular to the upper line from point (0,4) to the lower line it is of our unknown length d.
then sin T = d/6 (the distance between lines on the y axis)
4/5 = d/6
d = 24/5
To find the distance between the lines with equations 3x - 4y = 8 and y = (3/4)x + 4, we can use the formula for the distance between a point and a line.
Step 1: Convert the equations to slope-intercept form (y = mx + b).
For the equation 3x - 4y = 8:
-4y = -3x + 8
y = (3/4)x - 2
For the equation y = (3/4)x + 4, the equation is already in slope-intercept form.
Step 2: Identify the slopes of the lines.
The slope for the equation y = (3/4)x - 2 is (3/4), and for y = (3/4)x + 4, the slope is also (3/4). Since the slopes are the same, the lines are parallel.
Step 3: Determine any point on one of the lines.
Choose a point on one of the lines. Let's take the equation y = (3/4)x + 4. We can set x = 0 to find the y-intercept. When x = 0, y = (3/4)(0) + 4, so the point (0, 4) lies on the line.
Step 4: Find the distance between the point and the other line.
We will calculate the perpendicular distance between the point (0, 4) and the line 3x - 4y = 8.
To find the distance, we use the formula:
distance = |Ax + By + C| / √(A^2 + B^2)
where A, B, and C are the coefficients of the line equation in the form Ax + By + C = 0.
In our case, A = 3, B = -4, and C = -8. Substituting these values into the formula, we get:
distance = |3(0) - 4(4) - 8| / √(3^2 + (-4)^2)
= |-16| / √(9 + 16)
= 16 / √25
= 16 / 5
Therefore, the distance between the lines 3x - 4y = 8 and y = (3/4)x + 4 is 16/5 units.