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.
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!
6 answers
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
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
?
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
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!!
1 answer