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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.

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
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
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.

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

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