Question 1Level 2 · ExamSingle correct
Let A=(2,3) and B=(6,−5). For a real constant c, suppose the line 3x−4y=c divides the segment AB internally at a point P with AP:PB=λ:1, where λ>0. The set of all values of c for which λ≥3 is ______. Exactly one option is correct.
A[27,38)
B[5,38)
C[27,38]
D(−6,5]
Show answer and solution
Answer: A
Let L(x,y)=3x−4y. Then L(A)=6−12=−6 and L(B)=18+20=38.
If P divides AB internally in the ratio λ:1, then P=λ+1λB+A, and since L is linear, L(P)=λ+138λ−6=c. Solving, λ=38−cc+6. This is positive exactly for −6<c<38; at c=38 the line passes through B, so P=B and no finite ratio λ:1 exists.
For −6<c<38: λ≥3⟺c+6≥3(38−c)⟺4c≥108⟺c≥27. Hence c∈[27,38).
Check at c=27: λ=33/11=3, and P=43B+A=(5,−3), which satisfies 15+12=27.
Difficulty: HardQuestion 2Level 3 · AdvancedSingle correct
Consider the equation 2x2+5xy+3y2+6x+7y+c=0, where c is a real constant. The origin is shifted to a point O′(h,k), the new axes being parallel to the old ones, and it is found that in the new coordinates (X,Y) the equation is homogeneous of degree two (no first-degree terms and no constant term). Then (h,k), c and the set of real points represented by the original equation are ______. Exactly one option is correct.
A(h,k)=(−1,2), c=4; a pair of lines through (−1,2)
B(h,k)=(1,−2), c=−4; a pair of lines through (1,−2)
C(h,k)=(1,−2), c=4; the pair of lines 2x+3y+4=0 and x+y+1=0
D(h,k)=(1,−2), c=4; the single point (1,−2)
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Answer: C
Put x=X+h, y=Y+k. The quadratic part 2X2+5XY+3Y2 is unchanged. The coefficients of X and Y become 4h+5k+6 and 5h+6k+7.
Both vanish: 4h+5k=−6, 5h+6k=−7. The determinant is 24−25=−1=0, so the solution is unique: h=1, k=−2.
The new constant term is the value of the left side at (1,−2): 2−10+12+6−14+c=c−4. Homogeneity forces c=4.
Then 2X2+5XY+3Y2=(2X+3Y)(X+Y)=0, and returning to x,y gives 2X+3Y=2x+3y+4 and X+Y=x+y+1. Indeed (2x+3y+4)(x+y+1)=2x2+5xy+3y2+6x+7y+4. The locus is the pair of lines 2x+3y+4=0 and x+y+1=0, meeting at (1,−2).
Difficulty: Very hardQuestion 3Level 2 · ExamSingle correct
For real t=2, let P(t)=(t−23t+1, t−2t+3). The set of all points P(t) is ______. Exactly one option is correct.
Athe whole line 5x−7y=8
Bthe line 5x−7y=8 with only the point (3,1) removed
Cthe line 5x−7y=8 with the points (3,1) and (−21,−23) removed
Dthe line 7x−5y=16 with only the point (3,1) removed
Show answer and solution
Answer: B
Write x=3+t−27 and y=1+t−25. With u=t−21, which takes every real value except 0 as t runs over R∖{2}, we get x=3+7u, y=1+5u.
So every point satisfies 5(x−3)=7(y−1), i.e. 5x−7y=8. Conversely each point (3+7u,1+5u) of this line with u=0 is attained by t=2+u1.
The only point of the line not attained is u=0, the point (3,1). The point (−21,−23) of option (C) is P(0), so it is attained. The locus is the line 5x−7y=8 with the point (3,1) removed.
Difficulty: Hard