Asked by Austin
find the equation of a straight line which is equidistant from the points(2,3) and (6,1) expressing it in the form ax+by=c where a,b and c are constants.
Answers
Answered by
Steve
The line must be the perpendicular bisector of the segment between the two points.
Since the slope of that segment is (1-3)/(6-2) = -1/2, your line must have slope 2
The midpoint of the segment is (4,2), so using the point-slope form, your line is
y-3 = 2(x-2)
now just put that into the desired form.
Since the slope of that segment is (1-3)/(6-2) = -1/2, your line must have slope 2
The midpoint of the segment is (4,2), so using the point-slope form, your line is
y-3 = 2(x-2)
now just put that into the desired form.
Answered by
Emmanuel
not understood
Answered by
samuel
Equidistant of course means that the line its perpendicular to the two lines given find the gradient first that is
Gradient=Dy/DX
1-3/6-4= -1/2
Because it is perpendicular
M1XM2 =-1
Gradient of perpendicular = 2
Equidistant now (6+2)/2. (3+1)/2
Then follow it to the end
Gradient=Dy/DX
1-3/6-4= -1/2
Because it is perpendicular
M1XM2 =-1
Gradient of perpendicular = 2
Equidistant now (6+2)/2. (3+1)/2
Then follow it to the end
Answered by
Tuney
Y=1/2x+2
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