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    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.'
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    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
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    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
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    (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
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    (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?
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    (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  y = 3x - 4 . We know parallel lines have the same gradient so what is the gradient of L? Then use  y = mx + c to find the equation of L.

    Second part, L is perpendicular to  y = 3x + 5 . 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?
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    (Original post by BTAnonymous)
    First part, L is parallel to  y = 3x - 4 . We know parallel lines have the same gradient so what is the gradient of L? Then use  y = mx + c to find the equation of L.

    Second part, L is perpendicular to  y = 3x + 5 . 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.
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    (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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