# can you help with this problem?Watch

#1
Find all possible values of a and b given that y=ax+14 is perpendicular bisector of the line joining (1,2) to (b,6).

I know that M1*M2=-1 because they are perpendicular.
I know the formulas for finding equation either with a point and gradient or with just two points. And here comes the difficult part as I am getting equations with three unknown things x,y,b or x,y,a. I think there is some way by simultanious equation or by forming a quadratic but I cannot find it. Please help. I know that the middlepoint is (b+1/2, 4)
0
13 years ago
#2
Gradient of line joining those points = (6-2)/(b-1) = 4/(b-1)

Gradient of perpendicular bisector = -1/[4/(b-1)] = (1-b)/4

Perpendicular bisector goes through ( (b+1)/2 , 4 ), as your rightly say, so
Equation perpendicular bisector: y-4 = [(1-b)/4][x - (b+1)/2]

y-4 = (1-b)(2x-b-1)/8
y-4 = (2x - b - 1 - 2bx + b² + b)/8
y-4 = [(2-2b)/8]x + (b²-1)/8
y = ¼(1-b)x + (b²+31)/8

This must be equivalent to y = ax+14

=> a = (1-b)/4

and
=> (b²+31)/8 = 14
=> b² = 81
=> b = ±9

=> a = 5/2 (when b = -9) or -2 (when b = 9)
0
#3
There are answers in the bak of the book

It says that a=-2,b=9 or a=5/2b=-9
0
13 years ago
#4
Look through my working and try to spot a mistake. I'll do the same.

(I might have made an incorrect assumption in the method somewhere too)
0
#5
(Original post by mockel)
Look through my working and try to spot a mistake. I'll do the same.

(I might have made an incorrect assumption in the method somewhere too)
Yes I will do it. Thanks for your effort
0
13 years ago
#6
(Original post by mockel)
Gradient of line joining those points = (6-2)/(b-1) = 4/(b-1)

Gradient of perpendicular bisector = -1/[4/(b-1)] = (1-b)/4

Perpendicular bisector goes through ( (b+1)/2 , 4 ), as your rightly say, so
Equation perpendicular bisector: y-4 = [(1-b)/4][x - (b+1)/2]

y-4 = (1-b)(2x-b-1)/8
y-4 = (2x - b - 1 - 2bx + b² + b)/8
y-4 = [(2-2b)/8]x + (b²-1)/8
y = ¼(1-b)x + (b²-31)/8

This must be equivalent to y = ax+14

=> a = (1-b)/4

and
=> (b²-31)/8 = 14
=> b² = 143
=> b = ±√143

=> a = ¼(1±√143)

Hmm, that would've been a nicer if it had been √144, but I've checked my working and can't seem to find anything wrong. Maybe you will.
But that's the method.
y-4 = [(1-b)/4][x - (b+1)/2]
4y-16 = (1-b)(x-b-1)/2
8y-32 = (1-b)(x-b-1)
8y-32 = x-b-1-bx+b²+b
8y-32 = x-1-bx+b²
y = (31+x-bx+b²)/8

This must be equivalent to y = ax+14

(31+x-bx+b²)/8 = ax+14
31/8 + (1/8-b)x + b²/8 = ax+14

=> 1/8 - b = a
=> a = (1-8b)/8

and
=> (b²+31)/8 = 14
=> b² = 81
=> b = ±√81 = ±9

=> a = 1/8 - b
=> a = 1/8 ± 9
=> a = 73/8 or -71/8

I think.
0
13 years ago
#7
Ok...I found my mistake and edited. I wrote (b²-31)/8 for the equation of y, when it should've been (b²+31)/8.

0
X

new posts
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• University of East Anglia
All Departments Open 13:00-17:00. Find out more about our diverse range of subject areas and career progression in the Arts & Humanities, Social Sciences, Medicine & Health Sciences, and the Sciences. Postgraduate
Wed, 30 Jan '19
• Aston University
Wed, 30 Jan '19
• Solent University
Sat, 2 Feb '19

### Poll

Join the discussion

Remain (1411)
79.63%
Leave (361)
20.37%