Find the distance between the lines with equations 3x-y=9 and y=3x-4
a. 5/4
b. (5√10)/2
c. √10/2
d. (13√10)/2
not too sure on this one!
If you know how to calculate the perpendicular distance of a point P(x0,y0) from a given line put in the form
Ax+By+C=0
then all you need to do is to find a point on one line, calculate the distance from the other.
The distance is
D=(Ax0+By0+C)/sqrt(A^2+B^2)
To find a point on the second line, put x=0 and solve for y.
y=3(0)-4=-4
Therefore
P(0,-4) is a point on the second line.
Now put the first line in canonical form
3x-y-9=0
Calculate the distance
D=(3(xp)-yp-9)/sqrt(3^2+1^2)
=(3(0)-(-4)-9)/sqrt(10)
You can carry on from here.
Let's first rwrite them in y = mx + b form and see if they are parallel.
y = 3x -9
y = 3x -4
The slope of both lines is 3 (so they are parallel) and the difference of the y-intercepts is 5. That is NOT the minimum distance between the lines, however. You need to figure that out from the slope and the y-intercept difference, using the Pythagorean theorem.
If d is the separation between the lines, d^2 + 9 d^2 = 25, the square of the y-intercept difference. See if you can derive the same equation.
Solving for d will result in one of the available choices
I did the theorem and got
16/√10
?
I get 5/sqrt10 = (sqrt10)/2
The triangle in question has the two parallel lines intercepting the y-axis.
The line joining the two intercepts has a length of 5, as explained above.
The other two sides have lengths of d, distance between the two lines, and 3d (because of the slope m=3).
Thus you need to solve for a right-triangle with a diagonal of 5, the sides adjacent to the right-angle being d and 3d.
Using Pythagoras' theorem, we get
d^2+(3d)^2=5^2 (as above)
10d^2=25
can you now solve for d?
Remember also, when you get your answer, you'd have to eliminate the square-root radicals in the denominator by multiplying both the numerator and the denominator by the conjugate of the term containing the square-root.
For example:
2/sqrt(5)
=2sqrt(5)/(sqrt(5)^2)
=2sqrt(5)/5
Yes thanks! I got sqrt10/2 also!!
To find the distance between the two lines, we need to find the shortest distance between any two points on the lines.
First, we'll rewrite the equations of the lines in the form y = mx + b, where m is the slope and b is the y-intercept.
For the equation 3x - y = 9, we rewrite it as:
y = 3x - 9
For the equation y = 3x - 4, it is already in the form y = mx + b.
Now, we can find two points on the lines. Let's choose points on each line such that their x-coordinates are the same.
For the first line, when x = 0:
y = 3(0) - 9 = -9
So we have the point (0, -9) on the first line.
For the second line, when x = 0:
y = 3(0) - 4 = -4
So we have the point (0, -4) on the second line.
Now we can use the distance formula to find the distance between these two points:
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Let's substitute the coordinates of the two points into the formula:
Distance = sqrt((0 - 0)^2 + (-4 - (-9))^2)
Simplifying this expression:
Distance = sqrt(0 + 25)
Distance = sqrt(25)
Distance = 5
So the distance between the lines is 5.
Among the given options, the correct answer is: a. 5/4