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Exercise 6.3 · Q20

Q.Find all the equations of the straight lines in the family of the lines y=mx−3y = mx - 3, for which mm and the xx-coordinate of the point of intersection of the lines with x−y=6x - y = 6 are integers.

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Substitute y=mx−3y=mx-3 into x−y=6x-y=6 to get x(m−1)=−3x(m-1)=-3, i.e. x=−3m−1x=-\dfrac{3}{m-1}. Since m,xm,x must both be integers, m−1m-1 must be an integer divisor of 33: m−1∈{1,−1,3,−3}m-1\in\{1,-1,3,-3\}.

Since the question asks to find all such lines, every integer divisor of the fixed number on the right must be checked — not just the ones that happen to look 'nice'.

Step 1. Substitute the family into x−y=6x-y=6 (i.e. y=x−6y=x-6).

mx−3=x−6mx-3=x-6

mx−x=−6+3mx-x=-6+3

x(m−1)=−3x(m-1)=-3

x=−3m−1(m≠1)x=-\frac{3}{m-1}\qquad(m\ne1)

Step 2. Impose that both mm and xx are integers.

xx is an integer exactly when (m−1)(m-1) divides 33 exactly, i.e. (m−1)(m-1) is one of the four integer divisors of 33: {1,−1,3,−3}\{1,-1,3,-3\}. Each gives one value of mm (already an integer by construction) and one value of xx:

m−1m-1mmx=−3/(m−1)x=-3/(m-1)y=mx−3y=mx-3Line
1122−3-3−9-92x−y−3=02x-y-3=0
−1-10033−3-3y+3=0y+3=0
3344−1-1−7-74x−y−3=04x-y-3=0
−3-3−2-211−5-52x+y+3=02x+y+3=0

Step 3. Verify each row on x−y=6x-y=6.

(−3,−9)(-3,-9): −3−(−9)=6-3-(-9)=6 ✓ (3,−3)(3,-3): 3−(−3)=63-(-3)=6 ✓ (−1,−7)(-1,-7): −1−(−7)=6-1-(-7)=6 ✓ (1,−5)(1,-5): 1−(−5)=61-(-5)=6 ✓ — all four intersection points genuinely lie on x−y=6x-y=6, and each came from an integer mm. …

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