A person standing at the crossing at two straight paths represented by the equations 2x-3y-4 = 0 and 3x-4y-5 = 0, wants to reach a path represented by 6x-7y+8 = 0 in least time. Find the equations of path he should follow.
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Point of intersection P of Lines L1 2x-3y=4 and L2: 3x-4y=5 is:
P(-1,-2)
Slope of the line L3 : 6x-7y + 8 = 0 is m=6/7
Slope of the line L4 perpendicular to L3 is = -1/m = -7/6
Equation of L4 : y= -7x/6 + c
L4 passes through point P.
So -2 = 7/6 + c => c = -19/6
Equation of the path : L4 : y = -7x /6 - 19/6
or 6y + 7x + 19 = 0
Perpendicular path to the line is the least distance from the point to the straight line.
P(-1,-2)
Slope of the line L3 : 6x-7y + 8 = 0 is m=6/7
Slope of the line L4 perpendicular to L3 is = -1/m = -7/6
Equation of L4 : y= -7x/6 + c
L4 passes through point P.
So -2 = 7/6 + c => c = -19/6
Equation of the path : L4 : y = -7x /6 - 19/6
or 6y + 7x + 19 = 0
Perpendicular path to the line is the least distance from the point to the straight line.
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