Perpendicular bisector question and general checking

[greener1993]

The main help i need is with this question:

The perpendicular bisector of a straight line joining the points (3,2) and (5,6) meets the x-axis at A and the y at B. Prove that the distance AB is equal to 6root5.

I just need to have theses checked to make sure there right, would love if you could just skim it and check :D

1)What is the gradient of the line perpendicular to y=3x-5. hence find the equation of the line which goes through (3,1) and is perpendicular to y=3x-5
a) to find the gradient I did, m2=-1/m1. m1=3 so it was -1/3 = -1/3
b) We know the coordinates (3,1) which is the y and X, so 1=(-1/3*3)+c
1=-1+c, so
2=C all together the answer is y=-1/3x+2

2)Find the equation of line B which is perpendicular to line A and goes though the mid-point.
I am given the coordinates for line A, (2,6) and (4,8). Using that i can find both the midpoint and the gradient of line A = midpoint = (3,7) Gradient = 1
So M2=1/M1 so -1/1=-1
we know from midpoint, y=7 and x=3 so
7=(-1*3)+c
7+3=c
10=c so the equation is Y=-x+10

 

[HallsofIvy]

The main help i need is with this question:

The perpendicular bisector of a straight line joining the points (3,2) and (5,6) meets the x-axis at A and the y at B. Prove that the distance AB is equal to 6root5.

I presume, from the problems below that you know that the straight line joining (3, 2) and (5, 6) has slope (6- 2)/(6-3)= 4/2= 2 and that the midpoint is ((3+ 5)/2, (2+ 6)/2)= (4, 4).
What is the line with slope -1/2 that passes through (4, 4)?

The point "A" is (x, 0) and the point "B" is at (0, y). Knowing the equation of the line, it should be easy to find x such that y= 0 and find y such that x= 0.

The distance AB is equal to $\sqrt{x^2+ y^2}$ where x and y were found above.

I just need to have theses checked to make sure there right, would love if you could just skim it and check :D

1)What is the gradient of the line perpendicular to y=3x-5. hence find the equation of the line which goes through (3,1) and is perpendicular to y=3x-5
a) to find the gradient I did, m2=-1/m1. m1=3 so it was -1/3 = -1/3
b) We know the coordinates (3,1) which is the y and X, so 1=(-1/3*3)+c
1=-1+c, so
2=C all together the answer is y=-1/3x+2

2)Find the equation of line B which is perpendicular to line A and goes though the mid-point.
I am given the coordinates for line A, (2,6) and (4,8). Using that i can find both the midpoint and the gradient of line A = midpoint = (3,7) Gradient = 1
So M2=1/M1 so -1/1=-1
we know from midpoint, y=7 and x=3 so
7=(-1*3)+c
7+3=c
10=c so the equation is Y=-x+10

Yes, both of those are correct.

 

[greener1993]

:D thank you for checking them two questions for me.

As for the main one, putting into an equation had sliped me, mainly because i don't know how to find the x intecepy from a y=mx+c graph... However i put into equation and got -1/2x +8, so the y intercept is 8. but was unsure from there for the x axis. Working back however, knowing the answer is 6root5 or root36*root5, we know the answer must be 180 before rooting. 8*8 =64 180-64 = 116 and root116 isn't a whole number, something my teacher is unlikly to do. so I am not even confident i know how to find the y :S

 

[greener1993]

ouch some very bad maths sorry, -1/2 *4 = 2 :P there for c=6 :D making what i said ilrelivant