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New Further Maths Year 12 Edexcel TB Ex 7C question 12 watch

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    I need help with this question please... Its from the Edexcel Year 12 New Further Maths course book. Exercise 7C question 12...

    A triangle T has vertices at the points A = (k,1), B = (4,1) and C = (4,k) where k is an integer constant.

    Triangle T is formed by the matrix (4 -1)TOP ROW (k 2) BOTTOM ROW (This is my attempt at typing 2x2)


    Given that triangle T has a right-angle at B, and the area of T' is 10, find the value of k.


    Here's what I've done... I've sketched the two possible locations of A and the two possible locations of C. What I get is then four possible triangles.

    I have considered the dimensions of all four, done BxH/2 = 10/(8 + k) , since (8+k) is the determinant of the matrix and hence the scale factor of enlargement.

    In each case, I end up solving a cubic. The only cubic that gives rise to an integer value for k is when I choose the top left triangle, where k ends up = 2

    But this seemed like a lot of work for 5 marks.

    Am I missing something obvious.

    Sorry if what I'm saying makes little sense; if you attempt the question I hope you will see what I mean.

    Thank you.
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    The area scale factor of a matrix transformation is ad-bc where it was transformed by the matrix row 1: (a,b) row 2: (c,d) . I can't quite remember if this is a general rule or if there are any conditions to it.
    I'm not sure if this will help, but I hope it does.
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    (Original post by beast14)
    I need help with this question please... Its from the Edexcel Year 12 New Further Maths course book. Exercise 7C question 12...

    A triangle T has vertices at the points A = (k,1), B = (4,1) and C = (4,k) where k is an integer constant.

    Triangle T is formed by the matrix (4 -1)TOP ROW (k 2) BOTTOM ROW (This is my attempt at typing 2x2)


    Given that triangle T has a right-angle at B, and the area of T' is 10, find the value of k.


    Here's what I've done... I've sketched the two possible locations of A and the two possible locations of C. What I get is then four possible triangles.

    I have considered the dimensions of all four, done BxH/2 = 10/(8 + k) , since (8+k) is the determinant of the matrix and hence the scale factor of enlargement.

    In each case, I end up solving a cubic. The only cubic that gives rise to an integer value for k is when I choose the top left triangle, where k ends up = 2

    But this seemed like a lot of work for 5 marks.

    Am I missing something obvious.

    Sorry if what I'm saying makes little sense; if you attempt the question I hope you will see what I mean.

    Thank you.
    You can speed things up:

    Without considering all the triangles you can say that e.g. the base length is |k-4| so you can end up with the equation

    |k-4|\times |k-1| \times (8+k) = 20

    Then it's not hard to find integer solutions to this since you need a product of positive integers equal to 20. There's no need to expand and solve a cubic.

    There may be a nicer way of doing this question but I haven't spent much time on it.
 
 
 
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