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Exercise 6.5 · Q14

Q.The line (p+2q)x+(p−3q)y=p−q(p+2q)x+(p-3q)y=p-q, for different values of pp and qq, passes through the point

(1) (32,52)\left(\dfrac32,\dfrac52\right)
(2) (25,25)\left(\dfrac25,\dfrac25\right)
(3) (35,35)\left(\dfrac35,\dfrac35\right)
(4) (25,35)\left(\dfrac25,\dfrac35\right)
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Group the terms in pp and qq: p(x+y−1)+q(2x−3y+1)=0p(x+y-1)+q(2x-3y+1)=0 must hold for every p,qp,q, so both brackets vanish; solving the pair gives the fixed point.

The line (p+2q)x+(p−3q)y=p−q(p+2q)x+(p-3q)y=p-q is really a family of lines, one for each choice of pp and qq. A point that lies on every member of the family must satisfy the equation no matter what p,qp,q are, which is only possible if the coefficients of pp and qq separately vanish.

Step 1. Regroup by pp and qq. Rewrite the equation as

(p+2q)x+(p−3q)y−(p−q)=0(p+2q)x+(p-3q)y-(p-q)=0

Expand and collect terms in pp and in qq:

p(x+y−1)+q(2x−3y+1)=0p(x+y-1)+q(2x-3y+1)=0

Step 2. Force independence of p,qp,q. Since pp and qq can be chosen independently (they are not proportional to each other), the only way the sum can be zero for every p,qp,q is if each bracket is separately zero:

x+y−1=0and2x−3y+1=0x+y-1=0 \qquad\text{and}\qquad 2x-3y+1=0

Step 3. Solve the simultaneous equations. From the first equation, x=1−yx=1-y. Substitute into the second: …

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