Find the equation of a straight line that is equidistant from the points (-2, -1)and (4, 5) expressing it in the form ax+by=c, where a, b and c are integers
3 answers
The equation of the line is x + 3y = 7
Mr. Bot fails again!!!
The equation you are looking for must be the right bisector of the
segment joining the two given points.
slope for the given points = (5+1)/(4+2) = 6/6 = 1
so the slope of the required line is -1
it is y = -x + b, but it must also go through the midpoint of the above segment
which would be (1,2) , so
2 = -1 + b, or b = 3
new equation: y = -x + 3 or
x + y = 3
The equation you are looking for must be the right bisector of the
segment joining the two given points.
slope for the given points = (5+1)/(4+2) = 6/6 = 1
so the slope of the required line is -1
it is y = -x + b, but it must also go through the midpoint of the above segment
which would be (1,2) , so
2 = -1 + b, or b = 3
new equation: y = -x + 3 or
x + y = 3
or, use the point-slope form of the line to start with
slope = -1
it goes through the midpoint of AB: (1,2)
so the equation is
y-2 = -1(x-1)
which resolves to
x+y = 3
slope = -1
it goes through the midpoint of AB: (1,2)
so the equation is
y-2 = -1(x-1)
which resolves to
x+y = 3
1 answer