Figures and equations (Mathematics II)

The formula for an internal division point gives x=(1·1+2·7)/3=5 and y=(1·2+2·5)/3=4, so the point is (5,4). The option (3,3) takes the ratio as 1:2, (13,8) computes an external division point, and (4,7/2) is the midpoint.

AB=√((2−(−1))²+(−1−3)²)=√(9+16)=√25=5. The option √7 takes the square root of 3+4 without squaring the differences, and 25 forgets the final square root. Square each difference first, then add.

From y−(−1)=3(x−2) we get y+1=3x−6, so y=3x−7. The option y=3x−5 comes from the sign error y−1=3(x−2). Substituting x=2 to check that y=−1 prevents this.

The slope is (8−2)/(3−1)=3. From y−2=3(x−1) we get y=3x−1. You can also check that x=3 gives y=8. The option (1/3)x+5/3 inverts the slope fraction, and y=3x+1 has the wrong sign on the intercept.

For a parallel line the coefficients of x and y stay the same, so write 2x+3y+c=0. Substituting (3,−1) gives 6−3+c=0, so c=−3 and the line is 2x+3y−3=0. The option 3x−2y−11=0 is the perpendicular line instead.

Perpendicularity means the product of the slopes is −1, so the required slope is −1/2. From y−1=−(1/2)(x−4) we get y=−(1/2)x+3. The option y=−2x+9 only changes the sign, and y=(1/2)x−1 only takes the reciprocal.

d=|3·1+4·2−1|/√(3²+4²)=|10|/5=2. The option 10 forgets to divide by √(a²+b²)=5, and 12/5 substitutes the constant as +1 instead of −1.

The external division point is ((−1·2+3·5)/(3−1), (−1·1+3·7)/(3−1)) = (13/2, 10). The option (17/4, 11/2) is the internal division point in the ratio 3:1, and (1/2, −2) divides externally in the ratio 1:3. Check both the type of division and the order of the ratio.

The centroid is the average of the three vertices: ((2−1+5)/3, (3+4−1)/3) = (6/3, 6/3) = (2,2). The option (6,6) gives the sums without dividing by 3, (3,3) divides by 2 instead of 3, and (2,3) simply repeats vertex A.

For parallel lines ax+by+c₁=0 and ax+by+c₂=0 the distance is |c₁−c₂|/√(a²+b²), so it is |2−(−8)|/√(3²+(−4)²) = 10/5 = 2. Equivalently, the distance from a point on one line, say (−2/3, 0), to the other line 3x−4y−8=0 is |−2−8|/5 = 2. The option 10 forgets to divide by √(a²+b²)=5, and 6/5 mishandles the sign as |2+(−8)|.

Substituting a=2, b=−3 and r=4 into the standard form (x−a)²+(y−b)²=r² gives (x−2)²+(y+3)²=16. The signs inside the brackets are opposite to the coordinates of the centre. The option with =4 puts the radius on the right side without squaring it.

Completing the square gives (x+1)²+(y−4)² = −8+1+16 = 9, so the centre is (−1,4) and the radius is √9=3. The centre (1,−4) misreads the signs, the radius 9 reports r² as the radius, and √8 takes the constant term 8 as r².

The centre is the midpoint of AB, namely (1,3). The radius is the distance from the centre to A, √((1−(−1))²+(3−2)²)=√5, so r²=5 and the circle is (x−1)²+(y−3)²=5. The option =20 puts the square of the diameter, AB²=20, on the right side, and =√5 fails to square the radius.

Write the line as 2x−y+5=0; the distance from the centre (0,0) is d=|5|/√(4+1)=5/√5=√5. This equals the radius r=√5, so they are tangent. The point is to compare d with r, both to the first power. You can also solve simultaneously and check that the discriminant is D=0.

The distance from the centre (0,0) to x+y−3=0 is d=|−3|/√2=3/√2≈2.12. Since the radius is r=2 we have d>r, so there is no common point. Taking d as 3 by forgetting to divide by √2 leads to the same conclusion here, but it is important to compute d=3/√2 correctly before comparing.

The tangent to x²+y²=r² at the point (x₁,y₁) is x₁x+y₁y=r², so it is 3x+4y=25. The option 4x+3y=25 swaps x₁ and y₁, and 3x+4y=5 leaves r on the right side instead of r².

Write the tangent as y=m(x−3)+1 and set the distance from the centre (0,0) equal to √2: |1−3m|/√(m²+1)=√2. Squaring gives 7m²−6m−1=0, so m=1 or −1/7. With m=1 the line is y=x−2. The line y=x+2 is indeed tangent to this circle but does not pass through (3,1), since 3+2=5≠1, so it is wrong.

Tangency means the distance from the centre to the line equals the radius. Here d=|3·1+4·2+9|/√(3²+4²)=|20|/5=4, so r=4. The option 20 forgets to divide by 5, and 2/5 substitutes −9 instead of +9.

The distance from the centre (0,0) to the line x=3 is d=3. The chord length is 2√(r²−d²) = 2√(25−9) = 2×4 = 8. The option 4 gives only half, one side of the chord, and 10 confuses it with the diameter.

The distance from the centre (0,0) to x−y+k=0 is d=|k|/√2, and it must be less than the radius 2. From |k|<2√2 we get −2√2<k<2√2. The option −2<k<2 forgets to multiply by √2, and including the equalities would also admit the tangent case, with only one common point.

Writing P(x,y), the condition AP²=BP² gives (x+2)²+y²=(x−4)²+y². Rearranging gives 12x=12, so x=1. This is the perpendicular bisector of the segment AB and passes through its midpoint (1,0). You can check that it agrees with the midpoint x-coordinate (−2+4)/2=1.

From AP²=4PB² we get x²+y²=4{(x−3)²+y²}. Rearranging gives x²+y²−8x+12=0, that is (x−4)²+y²=4, the circle of Apollonius with centre (4,0) and radius 2. When the ratio is not 1:1 the locus is a circle rather than a line.

Writing P(x,y) and Q(s,t), we have x=(s+6)/2 and y=t/2, so s=2x−6 and t=2y. Substituting into s²+t²=9 gives (2x−6)²+4y²=9, that is (x−3)²+y²=9/4, the circle with centre (3,0) and radius 3/2. The radius 9/4 reads r² as the radius, and the radius 3 forgets the factor of one half.

The condition y > f(x) means the region above the curve or line. Since there is no equality, the boundary line is not included. For instance the point (0,3) satisfies 3>1 and lies above the line, which confirms it.

The condition x²+y² ≦ r² is the interior of the circle, and since equality is allowed the boundary circle itself is included. Substituting the origin gives 0≦9, which holds, so the region is the side containing the origin, that is the interior.

Test each point by substitution. For (2,1): 1≦3·2−4=2 holds. For (0,0): 0≦−4 fails; for (1,0): 0≦−1 fails; for (0,3): 3≦−4 fails. Testing points by substitution is a reliable method that works for any region problem.

The region is the right triangle with vertices (0,0), (4,0) and (0,4). Its area is 4×4÷2 = 8. The option 16 treats it as a square of side 4, forgetting the division by 2. Sketch the common part of the three inequalities before computing.

The region is the quadrilateral with vertices (0,0), (3,0), (2,2) and (0,3). Sliding the line x+y=k, the value of k is largest when it passes through the vertex (2,2), giving k=4. The option 3 is the value at (3,0) or (0,3), and 6 simply repeats the right side of a constraint.

Setting 3x+4y=k gives a line. It suffices that the distance from the origin to the line is at most the radius 5, so |k|/√(3²+4²) ≦ 5, that is |k| ≦ 25. The maximum is 25, attained at the point of tangency (3,4). The option 35 comes from 5×(3+4), 15 is the value at (5,0), and 7 is the sum of the coefficients.

Setting x+y=k, the line x+y−k=0 meets the circle when |−k|/√2 ≦ 2, that is |k| ≦ 2√2. The maximum is 2√2, attained at the point of tangency (√2,√2). The option 2 simply repeats the radius, and 4 wrongly assumes x and y can each reach 2 independently, which is impossible on the circle.