You appear to have a typo in (9-3,5)
In any case, find the equation of the right-bisector of the line joining the two given points.
(find the midpoint, and the slope of the line through the two given points, the slope of the perpendicular is the negative reciprocal of the above slope.
Once you have the line, let x=0 to find the y-intercept.)
Find the co-ordinates of a point on the y-axis which is equidistant from (9-3,5) and (2,3).
What formula do I use to find these co-ordinates?
please and thank you
4 answers
(9-3,5) is suppose to be (-3,5)
midpoint:
((-3+2)/2,(5+3)/2)
=(-1/2,4)
correct???
slope:
m=(3-5)/(2-(-3))
m=-2/5
correct???
perpendicular slope:
m=5/2
correct???
y=(-2/5)x+b
5=(-2/5)(-3)+b
5=6/5+b
5-6/5=b
19/5=b
correct??? It doesn't look right to me but I'm not sure.
please and thank you
midpoint:
((-3+2)/2,(5+3)/2)
=(-1/2,4)
correct???
slope:
m=(3-5)/(2-(-3))
m=-2/5
correct???
perpendicular slope:
m=5/2
correct???
y=(-2/5)x+b
5=(-2/5)(-3)+b
5=6/5+b
5-6/5=b
19/5=b
correct??? It doesn't look right to me but I'm not sure.
please and thank you
P=(-3,5)
Q=(2,3)
midpoint M=(-1/2,4)
slope of PQ = (3-5)/((2-(-3)) = -2/5
slope of perpendicular is 5/2
Now you have a point M=(-1/2,4) and a slope m=5/2
so far, so good. Now, you messed up by using the slope of PQ, rather than the perpendicular bisector. (and after having gone to all the trouble of finding its slope!)
using the point-slope form of the line:
y-4 = 5/2(x+1/2) = 5/2 x + 5/4
y = 5/2 x + 21/4
y-intercept is 21/4
Q=(2,3)
midpoint M=(-1/2,4)
slope of PQ = (3-5)/((2-(-3)) = -2/5
slope of perpendicular is 5/2
Now you have a point M=(-1/2,4) and a slope m=5/2
so far, so good. Now, you messed up by using the slope of PQ, rather than the perpendicular bisector. (and after having gone to all the trouble of finding its slope!)
using the point-slope form of the line:
y-4 = 5/2(x+1/2) = 5/2 x + 5/4
y = 5/2 x + 21/4
y-intercept is 21/4
thank you
1 answer