plse friends help me,
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)
okay
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.
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.
I hope that helps.
okay ., i wil try
i m not getting it
@msingh Do you still need help with this?
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:
\[y=4x-20\]
We see that the slope of this line and all lines parallel to it is 4.
Solve x/c+y/3=1 for y to get: \[y=\frac{-3x}{c}+3\]
But -3/c=4 So c = -3/4 And \[y=4x+3\]
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
Now put the two lines into the form : \[ax+by+c _{1}=0 \] and \[ax+by+c _{2}=0\]
The distance between those two lines is: \[d=\frac{|c _{2}-c _{1}|}{\sqrt{a^2+b^2}}\]
I got: \[\frac{23}{\sqrt{17}} = \frac{23\sqrt{17}}{17}=5.58\]
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