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    Can someone help me with the following question:
    I've done the parts before this however I'm struggling on this bit.

    The coordinates of four points are P(-2,-1) Q(6,3) R(9,2) & S(1,-2)

    (i) Show that the equation of SR is 2y = x - 5 and find the equation of the line L through Q perpendicular to SR

    (ii) calculate the coordinates of the point T where the line L meets SR

    (iii) calculate the area of the quadrilateral PQRS


    Some parts earlier asked to work out the gradients. The gradient of SR is 1/2
    I'm not sure where I go from here for (i) ?? Do I find the mid point of SR?
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    (Original post by Aty100)
    Can someone help me with the following question:
    I've done the parts before this however I'm struggling on this bit.

    The coordinates of four points are P(-2,-1) Q(6,3) R(9,2) & S(1,-2)

    (i) Show that the equation of SR is 2y = x - 5 and find the equation of the line L through Q perpendicular to SR

    (ii) calculate the coordinates of the point T where the line L meets SR

    (iii) calculate the area of the quadrilateral PQRS


    Some parts earlier asked to work out the gradients. The gradient of SR is 1/2
    I'm not sure where I go from here for (i) ?? Do I find the mid point of SR?
    Use y-y1 = 1/2(x-x1) to get the equation of SR.

    Then the gradient of the perpendicular will be -2 and then use y-y1 = -2(x-x1) with (x1, y1) being Q.
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    If you know the gradient of SR then you know the gradient a perpendicular to SR. You also know that it passes through Q so you can find the equation of that line. You then have 2 equations. Solve simultaneously for co-ords of T.
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    (Original post by maggiehodgson)
    If you know the gradient of SR then you know the gradient a perpendicular to SR. You also know that it passes through Q so you can find the equation of that line. You then have 2 equations. Solve simultaneously for co-ords of T.
    Do I just write the equation of SR as
    Y= 1/2x + C ?
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    (Original post by Aty100)
    Do I just write the equation of SR as
    Y= 1/2x + C ?
    Perpendicular lines have gradients whose product is -1 (minus 1). So if the gradient of one line was -3/4 the gradient of the perpendicular would be +4/3.

    Does that help?
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    To find the equation of a line:
    1) Find the gradient

    2) Find the co-ordinates of a point on the line

    3) Substitute the gradient and co-ordinates into your y-y1=m(x-x1) formula

    https://youtu.be/pQ0IKUM3qLs?t=144

    Hope this helps,

    Tom (IDK-tuition.com: online maths, economics and ACT tuition/resources)
 
 
 
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