Perpendicular Lines Help (C1 OCR)

"Find the equation of the perpendicular bisector of the line joining (2,-5) and (-4,3) give your answer in ax+by+c=0"
I had an idea of how to do it, but when I checked the answer in my textbook it didn't seem like it was correct. So here's how I did it.
1. Find the midpoint of the two lines by finding the mean of x and the mean of y which I got as (-2/2, -8/2)
2. I then found the gradient of the line by finding the differentiation of y and divided it by the differentiation of x which gave me -6/8
3. I used this gradient so that I can find the negitive reciprical of it so that I can find the perpendicular lines gradient which was 8/6
4.I substiuted my gradient and the midpoint coordinates into the formula y-y=m(x2-x)
5. I got 6y+24=8x+8
6. I minused 24 from both sides which gave me 6y=8x-16
7. I minused 8x from both sides and that gave me 6y-8x=24
8. I wasn't sure what to do on this part so I put the 8x behind 6y so then it looked like 8x-6y=16
9. I finally then subtracted both sides by sixteen, which resulted me getting 8x-6y-16=0 Which was incorrect.
According to my textbook the correct answer was 3x-4y-1=0 which I have no idea how I could of got there, any ideas?

Reply 1

TenOfThem

18

Original post by Fruitbasket786

"Find the equation of the perpendicular bisector of the line joining (2,-5) and (-4,3) give your answer in ax+by+c=0"
I had an idea of how to do it, but when I checked the answer in my textbook it didn't seem like it was correct. So here's how I did it.
1. Find the midpoint of the two lines by finding the mean of x and the mean of y which I got as (-2/2, -8/2)

The mid point of (2,-5) and (-4,3) is (-2/2,-2/2) = (-1,-1)

Reply 2

TenOfThem

18

Original post by Fruitbasket786

"Find the equation of the perpendicular bisector of the line joining (2,-5) and (-4,3) give your answer in ax+by+c=0"
I had an idea of how to do it, but when I checked the answer in my textbook it didn't seem like it was correct. So here's how I did it.

2. I then found the gradient of the line by finding the differentiation of y and divided it by the differentiation of x which gave me -6/8

the change in y divided by the change in x = -8/6

Reply 3

TenOfThem

18

Original post by Fruitbasket786

3. I used this gradient so that I can find the negitive reciprical of it so that I can find the perpendicular lines gradient which was 8/6
4.I substiuted my gradient and the midpoint coordinates into the formula y-y=m(x2-x)
5. I got 6y+24=8x+8
6. I minused 24 from both sides which gave me 6y=8x-16

if you had started correctly then this working would have been correct

Reply 4

ztibor

10

Original post by Fruitbasket786

"Find the equation of the perpendicular bisector of the line joining (2,-5) and (-4,3) give your answer in ax+by+c=0"
I had an idea of how to do it, but when I checked the answer in my textbook it didn't seem like it was correct. So here's how I did it.
1. Find the midpoint of the two lines by finding the mean of x and the mean of y which I got as (-2/2, -8/2)
2. I then found the gradient of the line by finding the differentiation of y and divided it by the differentiation of x which gave me -6/8
3. I used this gradient so that I can find the negitive reciprical of it so that I can find the perpendicular lines gradient which was 8/6
4.I substiuted my gradient and the midpoint coordinates into the formula y-y=m(x2-x)
5. I got 6y+24=8x+8
6. I minused 24 from both sides which gave me 6y=8x-16
7. I minused 8x from both sides and that gave me 6y-8x=24
8. I wasn't sure what to do on this part so I put the 8x behind 6y so then it looked like 8x-6y=16
9. I finally then subtracted both sides by sixteen, which resulted me getting 8x-6y-16=0 Which was incorrect.
According to my textbook the correct answer was 3x-4y-1=0 which I have no idea how I could of got there, any ideas?

For 1)
If the 2 point are (2,-5) and (-4,3) then the midpoint M(-2/2, -2/2)=(-1,-1)

For 2)
the gradient of the line passing throw og that 2 points is (3-(-5))/(-4-2)=8/(-6)=-8/6=-4/3

For 3)
the gradient of the perpendicular line is: 3/4

for 4)
y-(-1)=3/4(x-(-1)) -> y+1 =3/4(x+1)

You got 6y+24=8x+8 (the gradient and the midpoint was incorrect)
for 7)
subtracting 8x from both side the RHS will be -16

for 8)
from -8x will be 8x at the LHS if you multiply both side by -1
So 8x-6y=16 is right here

i should use the a(x-x0)+b(y-y0)=0 equation formula where
(a,,b) is the normal (perpendicular) vector of the line (f.e the vector between
the 2 points) and (x0,y0) is a point on the line (the midpoint here)

Arranging this eqution you get

ax+by+c=0 form

Reply 5

Fruitbasket786

OP

14

Even if I was wrong, i ended up with the answer 8x-6y-14=0 which is still wrong
y--1=-8/6(x--1)
y+1=-8/6(x+1)
6(y+1)=-8(x+1)
6y+6=-8x-8
6y=-8x-14
6y+8x=-14
8x-6y=14
8x-6y-14=0

Reply 6

Fruitbasket786

OP

14

Original post by ztibor

For 1)

i should use the a(x-x0)+b(y-y0)=0 equation formula where
(a,,b) is the normal (perpendicular) vector of the line (f.e the vector between
the 2 points) and (x0,y0) is a point on the line (the midpoint here)

Arranging this eqution you get

ax+by+c=0 form