Hi, I need some help with these, I am really tired b/c of my surgery, an explanation would be enough, thank you ever so much.
1) Given the following line and point not on the line, find the distance between them.
y = 7x + 5
(-8,-1)
2) Find the distance between the lines.
y=32x+32
y=32x−5
3) Find the distance between the lines.
y=3x+10
y=3x−20
4) Given the following line and point not on the line, find the distance between them.
y=−23x+2
(-2,-1)
I will not do them all for you, they are all virtually the same.
1) Given the following line and point not on the line, find the distance between them.
y = 7x + 5
(-8,-1)
==============================
find a line perpendicular to that one through that point
slope = m' = -1/m = -1/7
y = -x/7 + b
-1 = 8/7 + b
b = -15/7
so the line we want is y = -x/7 - 15/7
or 7 y = -x -15
now where does that hit the first line?
7 y = - 1 x - 15
y = 7 x + 5 multiply this by 7
7 y = - x - 15
7 y = 49 x + 35
-----------------------subtract
0 = -50 x - 50
x = -1
then y = 7x+5 = -2 so that point is at (-1 , -2)
NOW
I suspect you can find the distance between
(-1,-2) and (-8,-1)
sqrt (7^2+1^1) = sqrt 50 = 5 sqrt 2
by the way lines in #2 and #3 are parallel, so pick any old point on one of them
thank you so
much
You are welcome.
Of course, I'm here to help! Let's tackle each problem one by one:
1) To find the distance between a line and a point, you can use the formula for the distance between a point and a line. In this case, we have the line equation y = 7x + 5 and the point (-8, -1). The formula for the distance D between a point (x1, y1) and a line Ax + By + C = 0 is:
D = |Ax1 + By1 + C| / √(A^2 + B^2)
In our case, A = 7, B = -1, and C = -5. Plugging these values into the formula, we get:
D = |7*(-8) + (-1)*(-1) + (-5)| / √(7^2 + (-1)^2)
D = |-56 + 1 - 5| / √(49 + 1)
D = |-60| / √50
D = 60 / √50
The distance between the line y = 7x + 5 and the point (-8, -1) is 60 / √50.
2) To find the distance between two parallel lines, you can find the distance between any point on one line and the other line. In this case, we have the equations y = 32x + 32 and y = 32x - 5.
Let's pick a point on one line, say (0, 32x + 32), and find the distance to the other line. Using the distance formula as mentioned in problem 1, we get:
D = |32*0 + (32x + 32)*1 + (-5)| / √(32^2 + 1^2)
D = |32x + 32 - 5| / √(1024 + 1)
D = |32x + 27| / √1025
Therefore, the distance between the lines y = 32x + 32 and y = 32x - 5 is |32x + 27| / √1025.
3) Similar to problem 2, the distance between two parallel lines can be found by selecting a point on one line and calculating the distance to the other line. In this case, we have the equations y = 3x + 10 and y = 3x - 20.
Choosing a point on the first line, let's say (0, 3x + 10), we can use the same distance formula as before:
D = |3*0 + (3x + 10)*1 + (-20)| / √(3^2 + 1^2)
D = |3x - 10| / √(9 + 1)
D = |3x - 10| / √10
Therefore, the distance between the lines y = 3x + 10 and y = 3x - 20 is |3x - 10| / √10.
4) Applying the same concept as in problem 1, we have the line equation y = -23x + 2 and the point (-2, -1). Plugging these values into the distance formula, we get:
D = |-23*(-2) + (-1)*(-1) + 2| / √((-23)^2 + (-1)^2)
D = |-46 + 1 + 2| / √(529 + 1)
D = |-43| / √530
The distance between the line y = -23x + 2 and the point (-2, -1) is 43 / √530.
I hope the explanations provided above are helpful. If you have any further questions or need additional assistance, feel free to ask!