Coordinate geometry

1. Find the gradient of the line through (-3, 7) and (5, -1).

Answer A: -1

The gradient is (-1 - 7)/(5 - (-3)) = -8/8 = -1. Reversing both subtraction orders gives the same result.

2. Which equation is the line of gradient 3 passing through (2, -4)?

Answer B: y = 3x - 10

Point-gradient form gives y + 4 = 3(x - 2), hence y = 3x - 10. Substitution of x = 2 gives y = -4.

3. What is the gradient of the line 6x + 3y - 12 = 0?

Answer C: -2

3y = -6x + 12, so y = -2x + 4. The coefficient of x is the gradient, which is -2.

4. Which equation represents the vertical line through (4, -7)?

Answer D: x = 4

The x-coordinate is fixed at 4 while y may vary, so the equation is x = 4. Its gradient is undefined.

5. Find p if 2x - py + 5 = 0 is parallel to y = (1/2)x - 3.

Answer A: p = 4

The first line is y = (2/p)x + 5/p. Parallel gradients require 2/p = 1/2, giving p = 4.

6. What is the gradient of a line perpendicular to 3x - 2y + 8 = 0?

Answer B: -2/3

Rearranging gives y = (3/2)x + 4, so the given gradient is 3/2. The perpendicular gradient is -2/3.

7. Find the midpoint of the segment joining (-5, 4) and (7, -2).

Answer C: (1, 1)

Average corresponding coordinates: (-5 + 7)/2 = 1 and (4 + (-2))/2 = 1. Therefore the midpoint is (1, 1).

8. Find the exact distance between (1, -2) and (7, 6).

Answer D: 10

The distance is sqrt(6^2 + 8^2) = sqrt(100) = 10. Distance is reported as the non-negative value.

9. Find the intersection of 2x + y = 11 and x - y = 1.

Answer A: (4, 3)

Adding gives 3x = 12, so x = 4. Then 4 - y = 1 gives y = 3. The point (4, 3) satisfies both equations.

10. State the centre and radius of (x + 2)^2 + (y - 5)^2 = 49.

Answer B: Centre (-2, 5), radius 7

The bracket signs give centre (-2, 5), and r^2 = 49 gives the positive radius r = 7, not 49.

11. Which standard form is equivalent to x^2 + y^2 - 8x + 6y = 0?

Answer C: (x - 4)^2 + (y + 3)^2 = 25

Completing both squares adds 16 and 9 to both sides, giving (x - 4)^2 + (y + 3)^2 = 25.

12. Relative to x^2 + y^2 = 20, where is the point (2, 3)?

Answer D: Inside the circle

The point's squared distance from the centre is 4 + 9 = 13, which is less than 20, so it is inside the circle.

13. A circle has diameter endpoints (-2, 1) and (4, 5). What is its equation?

Answer A: (x - 1)^2 + (y - 3)^2 = 13

The centre is (1, 3). The diameter squared is 6^2 + 4^2 = 52, so r^2 = 52/4 = 13 and the stated equation follows.

14. Find the tangent to x^2 + y^2 = 25 at the point (3, 4).

Answer B: 3x + 4y = 25

The tangent gradient is -3/4. Thus y - 4 = (-3/4)(x - 3), which rearranges to 3x + 4y = 25.

15. How many points of intersection do y = 6 and x^2 + y^2 = 25 have?

Answer C: None

Substitution gives x^2 + 36 = 25, so x^2 = -11. There is no real x, hence no intersection.

16. Find the intersections of y = x - 1 with x^2 + y^2 = 5.

Answer D: (-1, -2) and (2, 1)

x^2 + (x - 1)^2 = 5 gives x^2 - x - 2 = 0, so x = -1 or 2. The paired y-values are -2 and 1, giving (-1, -2) and (2, 1).

17. For which values of k is y = x + k tangent to x^2 + y^2 = 10?

Answer A: k = 2sqrt(5) or k = -2sqrt(5)

Substitution gives 2x^2 + 2kx + k^2 - 10 = 0. Tangency requires 4k^2 - 8(k^2 - 10) = 0, so k^2 = 20 and k = +/-2sqrt(5).

18. For which values of k does y = k not meet (x - 2)^2 + (y + 1)^2 = 16?

Answer B: k < -5 or k > 3

No intersection requires |k + 1| > 4. Therefore k + 1 > 4 or k + 1 < -4, giving k > 3 or k < -5. Equality would be tangency.

19. The chord x = 3 cuts x^2 + y^2 = 25 at two points. What is the chord's midpoint?

Answer C: (3, 0)

With x = 3, y^2 = 16, so the endpoints are (3, 4) and (3, -4). Their midpoint is (3, 0), consistent with symmetry about the x-axis.

20. A(0, 0) and B(6, 0) are endpoints of a diameter, and P is any other point on the circle. What is angle APB?

Answer D: 90 degrees

AB is explicitly a diameter and P is another point on the circle, so the angle in the semicircle, angle APB, is 90 degrees.

21. What line contains the common points of x^2 + y^2 = 25 and (x - 4)^2 + y^2 = 9?

Answer A: x = 4

Subtracting x^2 - 8x + 16 + y^2 = 9 from x^2 + y^2 = 25 gives 8x - 16 = 16, hence x = 4.

22. A circle's centre lies on x + y = 4 and x - y = 2, and the circle passes through (6, 5). What is its equation?

Answer B: (x - 3)^2 + (y - 1)^2 = 25

The centre equations give (3, 1). The squared radius to (6, 5) is 3^2 + 4^2 = 25, so the circle is (x - 3)^2 + (y - 1)^2 = 25.

23. How many intersections are there between y = 2x + 1 and y = x^2 - 2?

Answer C: Two

Equating gives x^2 - 2x - 3 = 0 = (x - 3)(x + 1), which has two distinct real roots. Therefore the graphs have two intersections.

24. Find k if the line y = kx is tangent at the origin to (x - 3)^2 + (y - 4)^2 = 25.

Answer D: k = -3/4

The radius gradient is (0 - 4)/(0 - 3) = 4/3. Its perpendicular tangent gradient is -3/4, so k = -3/4.