Ask your own question, for FREE!
Mathematics 11 Online
OpenStudy (anonymous):

plse friends help me,

OpenStudy (allank):

Okay, so you have two parallel lines with the equations: x/5+y/b=1 (passes through 13,32) x/c+y/3=1 We want the distance between the two lines. Step 1: Find the equation of the perpendicular line joining the two lines and passing through point (13,32)

OpenStudy (anonymous):

okay

OpenStudy (allank):

That shouldn't be hard if you're comfortable with perpendicular line qualities, particularly how the slopes of two perpendicular lines relate to each other.

OpenStudy (allank):

Step 2: Get the co-ordinates of the point where the perpendicular joining line meets line K. Step 3: Use the equation for finding the distance between two points on a Cartesian plane.

OpenStudy (allank):

I hope that helps.

OpenStudy (anonymous):

okay ., i wil try

OpenStudy (anonymous):

i m not getting it

OpenStudy (mertsj):

@msingh Do you still need help with this?

OpenStudy (mertsj):

Plug (13,32) into x/5+y/b=1 to find that b = -20 Now we have: \[\frac{x}{5}+\frac{y}{-20}=1\] Solving for y:

OpenStudy (mertsj):

\[y=4x-20\]

OpenStudy (mertsj):

We see that the slope of this line and all lines parallel to it is 4.

OpenStudy (mertsj):

Solve x/c+y/3=1 for y to get: \[y=\frac{-3x}{c}+3\]

OpenStudy (mertsj):

But -3/c=4 So c = -3/4 And \[y=4x+3\]

OpenStudy (amistre64):

an idea i have would be something like this given a line: ax+by=c the intercepts can be determined by zeroing out x or y, when x=0, y = c/b when y=0, x = c/a the intercept form of a line can then be expressed as:\[\frac{x}{c/a}+\frac{y}{c/b}=1\] your first equation has an x intercept at (5,0), the slope of the line can be determined from the given point (13,32) and the x intercept (5,0), simply by subtracting 5 from the x values and stacking y/x (13-5,32) = 32/8 = 4 the other line therefore needs a slope of 4, and runs thru the y intercept of (0,3): y = 4x+3 , the same as Mertz came to

OpenStudy (mertsj):

Now put the two lines into the form : \[ax+by+c _{1}=0 \] and \[ax+by+c _{2}=0\]

OpenStudy (mertsj):

The distance between those two lines is: \[d=\frac{|c _{2}-c _{1}|}{\sqrt{a^2+b^2}}\]

OpenStudy (mertsj):

I got: \[\frac{23}{\sqrt{17}} = \frac{23\sqrt{17}}{17}=5.58\]

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!