Turn on thread page Beta
 You are Here: Home >< Maths

# Parallel and perpendicular lines watch

1. How do you solve these two equations?
'A straight line L is parallel to y = 3x - 4 and passes through the point(4,5).
Find the equation of line L.'

'A straight line L passes through the point with coordinates(3,7) and is perpendicular to the line with equation y = 3x +5. Find the equation of the line L.'

'The point P has coordinates(2,1) and the point Q has coordinates(-2,-1).
Find the equation of the perpendicular bisector of PQ.'
2. if parallel, they have the same gradient.
so for the first one you keep the value in front of the x and swap x and y for the coordinates.

For perpendicular, the gradient is 1/gradient of the line. in your example the new gradient is 1/3.
gradient of the line times the gradient of the perpendicular line equals negative one.

I'm on the app so can't input maths symbols
Posted on the TSR App. Download from Apple or Google Play
3. just to add the gradient of a perpendicular line is the reciprocal of the gradient of the line it's perpendicular to.

I couldn't remember the correct wording last night
Posted on the TSR App. Download from Apple or Google Play
4. (Original post by BT144)
How do you solve these two equations?
'A straight line L is parallel to y = 3x - 4 and passes through the point(4,5).
Find the equation of line L.'

'A straight line L passes through the point with coordinates(3,7) and is perpendicular to the line with equation y = 3x +5. Find the equation of the line L.'

'The point P has coordinates(2,1) and the point Q has coordinates(-2,-1).
Find the equation of the perpendicular bisector of PQ.'
Moved to Maths
5. (Original post by RuthieG101)
if parallel, they have the same gradient.
so for the first one you keep the value in front of the x and swap x and y for the coordinates.

For perpendicular, the gradient is 1/gradient of the line. in your example the new gradient is 1/3.
gradient of the line times the gradient of the perpendicular line equals negative one.

I'm on the app so can't input maths symbols
The answer to the first part is y=3x - 7 and the last is y=-2x. How?
6. (Original post by BT144)
The answer to the first part is y=3x - 7 and the last is y=-2x. How?

First part, L is parallel to . We know parallel lines have the same gradient so what is the gradient of L? Then use to find the equation of L.

Second part, L is perpendicular to . We know that the gradients of perpendicular lines are exactly negative reciprocals of each other (in other words, you flip the fraction and stick a negative sign in front - note that if the fraction already has a negative sign, the new gradient is then positive). So what is the gradient of L and hence the equation of the line of L?
7. (Original post by BTAnonymous)
First part, L is parallel to . We know parallel lines have the same gradient so what is the gradient of L? Then use to find the equation of L.

Second part, L is perpendicular to . We know that the gradients of perpendicular lines are exactly negative reciprocals of each other (in other words, you flip the fraction and stick a negative sign in front - note that if the fraction already has a negative sign, the new gradient is then positive). So what is the gradient of L and hence the equation of the line of L?
Sorry, yeah. I meant that the answer to the third one is y= -2x(don't know how to solve. For the second, the answer is y=-1/3x + 8(how?). I'm not very familiar with graphs.
8. (Original post by BT144)
Sorry, yeah. I meant that the answer to the third one is y= -2x(don't know how to solve. For the second, the answer is y=-1/3x + 8(how?). I'm not very familiar with graphs.
I've told you how to do the second one.

1. What is the gradient of L?
2. Write out the equation of L once you have found it's gradient using y = mx + c OR y - y1 = m(x + x1)
3. Substitute the co-ordinates for which line L satisfies (the points which line L passes through) into either of the two equations above

Turn on thread page Beta
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: February 22, 2018
Today on TSR

### Uni league tables

Do they actually matter?

### University open days

• Staffordshire University
Everything except: Midwifery, Operating Department Practice, Paramedic Undergraduate
Sun, 21 Oct '18
• University of Exeter
Undergraduate Open Days - Exeter Campus Undergraduate
Wed, 24 Oct '18
• University of Bradford
Faculty of Health Studies Postgraduate
Wed, 24 Oct '18
Poll
Useful resources

## Make your revision easier

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

Can you help? Study help unanswered threads

## Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE