Find the locus of points equidistant from two straight lines y-5=0 and y-3=0
y-4 = 0
make a sketch to see why.
Pls put me through this question:
find the locus of points equidistant from two straight lines y-5=0 and y-3=0
To find the locus of points equidistant from two straight lines, we need to consider the perpendicular bisector of the line segment joining any two points on the lines. Let's find the equations of the lines first.
The equation y - 5 = 0 can be written as y = 5.
The equation y - 3 = 0 can be written as y = 3.
Now, let's consider two arbitrary points on these lines. Choose two points, A(a, 5) on y = 5, and B(b, 3) on y = 3.
The midpoint(M) of the line segment AB is given by:
M( (a+b)/2, (5+3)/2 ) = M( (a+b)/2, 4 )
The slope of the line segment AB is given by:
m = (3 - 5) / (b - a) = -2 / (b - a)
The slope of the perpendicular bisector is the negative reciprocal of m.
m′ = 1/m = -1 / (-2 / (b - a)) = (b - a) / 2
Using the point-slope form, the equation of the perpendicular bisector passing through M is:
y - 4 = m′ * (x - ( (a+b)/2 ))
Simplifying this equation gives us:
y - 4 = (b - a) / 2 * (x - ( (a+b)/2 ))
Expanding the equation further:
2y - 8 = (b - a) * (x - ( (a+b)/2 ))
Simplifying, we get:
2y - 8 = b*x - a*x - (b^2 - a^2) / 2
Moving the terms to one side gives us the final equation:
a*x + 2y - b*x - 8 = (b^2 - a^2) / 2
Simplifying the right side:
a*x + 2y - b*x - 8 = ( (b - a)(b + a) ) / 2
Multiplying both sides by 2 to eliminate the fractional term gives us:
2*(a*x) + 4*y - 2*(b*x) - 16 = (b - a)(b + a)
Finally, simplifying this equation further gives us the locus of points equidistant from the lines y - 5 = 0 and y - 3 = 0:
2*(a*x) + 4*y - 2*(b*x) - 16 = b^2 - a^2