Exam practice

Call the points (x₁, y₁) and (x₂, y₂). First find the slope, $m = \dfrac{y_2 - y_1}{x_2 - x_1}$. Any other point (x, y) on the line must give the same slope with (x₁, y₁), so $\dfrac{y - y_1}{x - x_1} = m$, that is y − y₁ = m(x − x₁). Put in the numbers and simplify to the form ax + by + c = 0. Check the answer with the second point. If x₁ = x₂, the line is upright: x = x₁.

From the grid, P(−5, 0) and Q(4, 3). Slope $\dfrac{3 - 0}{4 + 5} = \dfrac{1}{3}$.

$$\begin{aligned} y - 0 &= \frac{1}{3}(x + 5) \\ 3y &= x + 5 \end{aligned}$$

x − 3y + 5 = 0

P(5, 0) and Q(−1, −2). Slope $\dfrac{-2 - 0}{-1 - 5} = \dfrac{1}{3}$.

$$\begin{aligned} y - 0 &= \frac{1}{3}(x - 5) \\ 3y &= x - 5 \end{aligned}$$

x − 3y − 5 = 0

P(−5, 4) and Q(4, 0). Slope $\dfrac{0 - 4}{4 + 5} = -\dfrac{4}{9}$.

$$\begin{aligned} y - 0 &= -\frac{4}{9}(x - 4) \\ 9y &= -4x + 16 \end{aligned}$$

4x + 9y − 16 = 0

P(−3, 0) and Q(5, −4). Slope $\dfrac{-4 - 0}{5 + 3} = -\dfrac{1}{2}$.

$$\begin{aligned} y - 0 &= -\frac{1}{2}(x + 3) \\ 2y &= -x - 3 \end{aligned}$$

x + 2y + 3 = 0

Slope $\dfrac{0 - 3}{-1 - 2} = 1$; y − 3 = x − 2. x − y + 1 = 0

Slope $\dfrac{4 - 3}{-2 - 4} = -\dfrac{1}{6}$.

$$\begin{aligned} y - 3 &= -\frac{1}{6}(x - 4) \\ 6y - 18 &= -x + 4 \end{aligned}$$

x + 6y − 22 = 0

Slope $\dfrac{-5 + 3}{-1 - 6} = \dfrac{2}{7}$.

$$\begin{aligned} y + 3 &= \frac{2}{7}(x - 6) \\ 7y + 21 &= 2x - 12 \end{aligned}$$

2x − 7y − 33 = 0

Slope $\dfrac{0 + a}{b - 0} = \dfrac{a}{b}$.

$$\begin{aligned} y + a &= \frac{a}{b}(x - 0) \\ by + ab &= ax \end{aligned}$$

ax − by − ab = 0

Slope $\dfrac{q - 0}{0 - p} = -\dfrac{q}{p}$.

$$\begin{aligned} y - 0 &= -\frac{q}{p}(x - p) \\ py &= -qx + pq \end{aligned}$$

qx + py − pq = 0, which is $\dfrac{x}{p} + \dfrac{y}{q} = 1$ divided through by pq.

Slope $\dfrac{6 - 3}{5 - 4} = 3$; y − 3 = 3(x − 4), so y − 3 = 3x − 12. 3x − y − 9 = 0

Line through the first two: slope $\dfrac{2 + 1}{5 - 1} = \dfrac{3}{4}$, $y + 1 = \dfrac{3}{4}(x - 1)$, so 4y + 4 = 3x − 3, that is 3x − 4y − 7 = 0.
(9, 5): 27 − 20 − 7 = 0. True, so the three points are collinear.

Line through the first two: slope $\dfrac{-6 - 0}{0 + 3} = -2$, y = −2(x + 3), that is 2x + y + 6 = 0.
(−1, −4): −2 − 4 + 6 = 0. True: collinear.

Slope $\dfrac{6 - 4}{4 - 2} = 1$, y − 4 = x − 2, that is x − y + 2 = 0.
(6, 8): 6 − 8 + 2 = 0. True: collinear.

Slope $\dfrac{1 + 1}{2 - 1} = 2$, y + 1 = 2(x − 1), that is 2x − y − 3 = 0.
(4, 5): 8 − 5 − 3 = 0. True: collinear.

Slope $\dfrac{3 + 2}{2 - 7} = -1$, y + 2 = −(x − 7), that is x + y − 5 = 0.
(−1, 6): −1 + 6 − 5 = 0. True: collinear.

Slope $\dfrac{-2 + 4}{3 - 0} = \dfrac{2}{3}$, $y + 4 = \dfrac{2}{3}x$, so 3y + 12 = 2x, that is 2x − 3y − 12 = 0.
(6, 0): 12 − 0 − 12 = 0. True: collinear.

Not in general. Collinear only means the three points are on one line; B could be anywhere on it. B is the midpoint of AC only if $\left(\dfrac{b + 3}{2}, \dfrac{0 + 3}{2}\right) = (0, c)$, that is only when b = −3 and c = 3/2: the points (−3, 0), (0, 3/2) and (3, 3).

Line AB through (b, 0) and (0, c): slope $-\dfrac{c}{b}$.

$$\begin{aligned} y - 0 &= -\frac{c}{b}(x - b) \\ by &= -cx + bc \\ cx + by &= bc \end{aligned}$$

C(3, 3) is on it: 3c + 3b = bc. Divide both sides by 3bc:

$$\frac{1}{b} + \frac{1}{c} = \frac{1}{3}$$

All three lie on one straight line: the slope between any two of them is the same, and the third point satisfies the equation of the line through the other two.

Line AB: slope $-\dfrac{q}{p}$, so $y = -\dfrac{q}{p}(x - p)$, that is qx + py = pq.
C(5, 5) is on it: 5q + 5p = pq. Divide by 5pq: $\dfrac{1}{p} + \dfrac{1}{q} = \dfrac{1}{5}$.

$M = \left(\dfrac{0 + 2}{2}, \dfrac{1 + 3}{2}\right) =$ (1, 2).

P(1, 0) and M(1, 2) have the same x, so the median is an upright line: x = 1 (the two-point form can’t be used, since x₂ − x₁ = 0).

$M = \left(\dfrac{2 - 6}{2}, \dfrac{8 + 2}{2}\right) =$ (−2, 5).

$AM = \sqrt{(-2 - 2)^2 + (5 - 2)^2} = \sqrt{16 + 9} =$ 5 units.

Through A(2, 2) and M(−2, 5): slope $\dfrac{5 - 2}{-2 - 2} = -\dfrac{3}{4}$.

$$\begin{aligned} y - 2 &= -\frac{3}{4}(x - 2) \\ 4y - 8 &= -3x + 6 \\ 3x + 4y &= 14 \end{aligned}$$

D, the midpoint of YZ: $\left(\dfrac{1 + 5}{2}, \dfrac{-1 + 1}{2}\right) = (3, 0)$.
$XD = \sqrt{(3 + 1)^2 + (0 - 3)^2} = \sqrt{25} =$ 5 units.

Through X(−1, 3) and D(3, 0): slope $\dfrac{0 - 3}{3 + 1} = -\dfrac{3}{4}$.

$$\begin{aligned} y - 0 &= -\frac{3}{4}(x - 3) \\ 4y &= -3x + 9 \end{aligned}$$

3x + 4y − 9 = 0

A on the first line: 6a − b = 1, so b = 6a − 1. B on the second: 2b − 5a = 5.

$$\begin{aligned} 2(6a - 1) - 5a &= 5 \\ 7a &= 7 \end{aligned}$$

a = 1, b = 5.

A(1, 5) and B(5, 1).

$AB = \sqrt{(5 - 1)^2 + (1 - 5)^2} = \sqrt{32} =$ $4\sqrt{2}$ units (about 5.66).

Slope $\dfrac{1 - 5}{5 - 1} = -1$; y − 5 = −(x − 1). x + y − 6 = 0

Slopes 3 and −2: $\dfrac{3 + 2}{1 - 6} = -1$. The acute angle has tan θ = 1: 45∘°.

Slopes 3 and −1/2. Times 2 top and bottom:

$$\frac{3 + \dfrac{1}{2}}{1 - \dfrac{3}{2}} = \frac{6 + 1}{2 - 3} = -7$$

The acute angle: $\theta = \tan^{-1} 7 \approx$ 81.87∘°.

As in 2a, $\dfrac{m_1 - m_2}{1 + m_1m_2} = -1$: the angles are 45∘° and 135∘° (obtuse).

Slopes −2/3 and −1/2. Times 6 top and bottom:

$$\frac{-\dfrac{2}{3} + \dfrac{1}{2}}{1 + \dfrac{1}{3}} = \frac{-4 + 3}{6 + 2} = -\frac{1}{8}$$

The acute angle is $\tan^{-1}\dfrac{1}{8} \approx 7.13^\circ$, so the obtuse angle is 180∘° − 7.13∘° = 172.87∘°.

Slopes −3 and −1/3. Times 3 top and bottom:

$$\frac{-3 + \dfrac{1}{3}}{1 + 1} = \frac{-9 + 1}{6} = -\frac{4}{3}$$

$\tan^{-1}\dfrac{4}{3} \approx 53.13^\circ$: the angles are 53.13∘° and 126.87∘°.

Slopes 2 and −1/2: $2 \times \left(-\dfrac{1}{2}\right) = -1$. The lines are perpendicular: 90∘°.

Both slopes are $-\dfrac{2}{-3} = \dfrac{2}{3}$. Equal slopes: parallel. (The equations differ only in the constant.)

Slopes $\dfrac{1}{5}$ and $-\dfrac{3}{-15} = \dfrac{1}{5}$. Equal: parallel.

Slopes: $y = -\sqrt{3}x - 4$ gives $-\sqrt{3}$; $y = \dfrac{x - 5}{\sqrt{3}}$ gives $\dfrac{1}{\sqrt{3}}$. Product: $-\sqrt{3} \times \dfrac{1}{\sqrt{3}} = -1$. Perpendicular.

Slopes $-\dfrac{a}{b}$ and $-\dfrac{b}{-a} = \dfrac{b}{a}$. Product: $-\dfrac{a}{b} \times \dfrac{b}{a} = -1$. Perpendicular.

Slopes $-\dfrac{p}{3}$ and $\dfrac{3}{4}$. $-\dfrac{p}{3} = \dfrac{3}{4}$, so $p = -\dfrac{9}{4}$.

Slopes $\dfrac{2}{8} = \dfrac{1}{4}$ and $\dfrac{q}{12}$. $\dfrac{q}{12} = \dfrac{1}{4}$, so q = 3.

Slopes $-\dfrac{a}{3}$ and $-\dfrac{3}{9} = -\dfrac{1}{3}$. So a = 1.

The second is 3x + 2y = 6, slope $-\dfrac{3}{2}$. The first has slope $-\dfrac{3}{m}$. Equal when m = 2.

$-\dfrac{p}{3} \times \dfrac{3}{4} = -1$, so $-\dfrac{p}{4} = -1$: p = 4.

Slopes $-\dfrac{2}{3}$ and $\dfrac{m}{5}$. $-\dfrac{2m}{15} = -1$, so $m = \dfrac{15}{2}$.

Slopes $-\dfrac{4}{3}$ and $-\dfrac{3}{q}$. $\dfrac{4}{q} = -1$, so q = −4.

Slopes $-\dfrac{a}{5}$ and $-\dfrac{6}{10} = -\dfrac{3}{5}$. $\dfrac{3a}{25} = -1$, so $a = -\dfrac{25}{3}$.

$-\dfrac{p}{q}$

The second line’s slope is $-\dfrac{l}{-m} = \dfrac{l}{m}$. Perpendicular:

$$\begin{aligned} -\frac{p}{q} \times \frac{l}{m} &= -1 \\ pl &= qm \end{aligned}$$

3x + 4y + k = 0 through (3, 4): 9 + 16 + k = 0, k = −25. 3x + 4y − 25 = 0

Slope of the join: $\dfrac{-1 - 3}{3 - 2} = -4$. Through (3, 1): y − 1 = −4(x − 3), so y − 1 = −4x + 12. 4x + y − 13 = 0

5x − 2y + k = 0 through (2, 5): 10 − 10 + k = 0, k = 0. 5x − 2y = 0

Slope of the join: $\dfrac{-3 - 7}{2 - 4} = 5$, so the required slope is $-\dfrac{1}{5}$.

$$\begin{aligned} y - 1 &= -\frac{1}{5}(x - 3) \\ 5y - 5 &= -x + 3 \end{aligned}$$

x + 5y − 8 = 0

(i) Twice the second is 10x − 4y = 6; add it to 3x + 4y = 7: 13x = 13, so x = 1, and 4y = 7 − 3, y = 1. Z(1, 1)

(ii) Perpendicular to 2x + 3y = 5: 3x − 2y + k = 0. Through Z: 3 − 2 + k = 0, k = −1. 3x − 2y − 1 = 0

Meeting point: x = 3 − y, so 2(3 − y) − 3y = 1, 5y = 5: y = 1, x = 2. The point is (2, 1).

Parallel to x − y − 6 = 0 (slope 1): the slope of $\dfrac{x}{a} + \dfrac{y}{b} = 1$ is $-\dfrac{b}{a}$, so $-\dfrac{b}{a} = 1$ and b = −a.

Through (2, 1): $\dfrac{2}{a} + \dfrac{1}{b} = 1$, so $\dfrac{2}{a} - \dfrac{1}{a} = 1$: a = 1.

a = 1, b = −1: the line x − y = 1.

Midpoint (1, 5); slope of the segment $\dfrac{4}{6} = \dfrac{2}{3}$, so the bisector’s slope is $-\dfrac{3}{2}$.

$$\begin{aligned} y - 5 &= -\frac{3}{2}(x - 1) \\ 2y - 10 &= -3x + 3 \end{aligned}$$

3x + 2y − 13 = 0

Midpoint $\left(-2, \dfrac{7}{2}\right)$; slope of the segment $\dfrac{9 + 2}{-8 - 4} = -\dfrac{11}{12}$, so the bisector’s slope is $\dfrac{12}{11}$.

$$\begin{aligned} y - \frac{7}{2} &= \frac{12}{11}(x + 2) \\ 22y - 77 &= 24x + 48 \end{aligned}$$

24x − 22y + 125 = 0

Midpoint (−1, 2); slope of the segment $\dfrac{14}{-14} = -1$, so the bisector’s slope is 1. y − 2 = x + 1: x − y + 3 = 0

The given slope is $\dfrac{2}{3}$. With slope m, times 3 top and bottom:

$$\begin{aligned} 1 &= \pm\frac{m - \dfrac{2}{3}}{1 + \dfrac{2m}{3}} \\ 3 + 2m &= \pm(3m - 2) \end{aligned}$$

With +: m = 5, and y + 2 = 5(x − 3) gives 5x − y − 17 = 0.
With −: 3 + 2m = −3m + 2, $m = -\dfrac{1}{5}$, and 5(y + 2) = −(x − 3) gives x + 5y + 7 = 0.

5x − y − 17 = 0 and x + 5y + 7 = 0

The given slope is −1; $\tan 60^\circ = \sqrt{3}$.

$$\begin{aligned} \sqrt{3} &= \pm\frac{m + 1}{1 - m} \\ m + 1 &= \pm\sqrt{3}(1 - m) \end{aligned}$$

With +: $m(1 + \sqrt{3}) = \sqrt{3} - 1$, so $m = \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1} = \dfrac{(\sqrt{3} - 1)^2}{2} = 2 - \sqrt{3}$.
With −: $m(1 - \sqrt{3}) = -\sqrt{3} - 1$, so $m = \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1} = 2 + \sqrt{3}$.

Through the origin, y = mx: $y = (2 - \sqrt{3})x$ and $y = (2 + \sqrt{3})x$, that is $(2 \mp \sqrt{3})x - y = 0$.

• 1.vertex A
• 2.semi-vertical angle α = ∠BAC
• 3.generator or slant height AC
• 4.axis or height AB
• 5.radius BC of the base
• 6.the circular base

• a.circle
• b.ellipse
• c.parabola
• d.hyperbola (two branches)

The curve in which a plane parallel to a generator of a cone (θ = α) cuts it. It doesn’t close.

The closed curve in which a plane at an angle θ to the axis, with $\alpha < \theta < 90^\circ$, cuts one nappe of a cone.

The curve in which a plane at right angles to the axis (parallel to the base) cuts a cone; every point of it is the same distance from the center.

The curve of two separate branches in which a plane at an angle θ < α to the axis cuts both nappes of a double-napped cone.

$\theta = 90^\circ$

$\alpha < \theta < 90^\circ$

θ = α

θ < α

$(x + 4)^2 + (y - 1)^2 = 16$, so $x^2 + 8x + 16 + y^2 - 2y + 1 = 16$. $x^2 + y^2 + 8x - 2y + 1 = 0$

$(x + 3)^2 + (y + 4)^2 = 49$, so $x^2 + 6x + 9 + y^2 + 8y + 16 = 49$. $x^2 + y^2 + 6x + 8y - 24 = 0$

$(x - 2)^2 + (y + 1)^2 = 9$, so $x^2 - 4x + 4 + y^2 + 2y + 1 = 9$. $x^2 + y^2 - 4x + 2y - 4 = 0$

$x^2 + (y - 10)^2 = 25$, so $x^2 + y^2 - 20y + 100 = 25$. $x^2 + y^2 - 20y + 75 = 0$

$(x + a)^2 + (y + a)^2 = 2a^2$, so $x^2 + 2ax + a^2 + y^2 + 2ay + a^2 = 2a^2$. $x^2 + y^2 + 2ax + 2ay = 0$: the circle passes through the origin, since c = 0.

$(x - a)^2 + (y - b)^2 = b^2$, so $x^2 - 2ax + a^2 + y^2 - 2by = 0$. $x^2 + y^2 - 2ax - 2by + a^2 = 0$ (it touches the X-axis, since r = k).

Center (0, 0), radius 5.

Center (0, 0), radius 6.

Center (0, −2), radius 6.

$x^2 + y^2 - 2y - 24 = 0$: g = 0, f = −1, c = −24. Center (0, 1), radius $\sqrt{0 + 1 + 24} = 5$.

g = −5, f = 2, c = 13. Center (5, −2), radius $\sqrt{25 + 4 - 13} = 4$.

g = 3, f = −4, c = −11. Center (−3, 4), radius $\sqrt{9 + 16 + 11} = 6$.

Divide by 2: $x^2 + y^2 - 4x - 6y + \dfrac{1}{2} = 0$. g = −2, f = −3, $c = \dfrac{1}{2}$.

$$r = \sqrt{4 + 9 - \frac{1}{2}} = \sqrt{\frac{25}{2}} = \frac{5}{\sqrt{2}}$$

Center (2, 3), radius $\dfrac{5}{\sqrt{2}} = \dfrac{5\sqrt{2}}{2} \approx 3.54$.

This is the circle of Example 3: center $\left(2, -\dfrac{1}{3}\right)$, radius 4.

$$\begin{aligned} (x - a)(x + a) + (y + 1)(y - 2) &= 0 \\ x^2 - a^2 + y^2 - y - 2 &= 0 \end{aligned}$$

$x^2 + y^2 - y - a^2 - 2 = 0$

$$\begin{aligned} (x - 2)(x - 3) + (y + 6)(y + 7) &= 0 \\ x^2 - 5x + 6 + y^2 + 13y + 42 &= 0 \end{aligned}$$

$x^2 + y^2 - 5x + 13y + 48 = 0$

$$\begin{aligned} x(x + 3) + (y + 7)(y - 2) &= 0 \\ x^2 + 3x + y^2 + 5y - 14 &= 0 \end{aligned}$$

$x^2 + y^2 + 3x + 5y - 14 = 0$

g = −2, f = 3: center (2, −3). (Its radius is $\sqrt{4 + 9 + 12} = 5$.)

The center is the midpoint of the diameter. If the other end is (x, y): $\dfrac{5 + x}{2} = 2$ and $\dfrac{1 + y}{2} = -3$, so x = −1, y = −7. (−1, −7)

Check: (5, 1) is on the circle, 25 + 1 − 20 + 6 − 12 = 0, and so is (−1, −7): 1 + 49 + 4 − 42 − 12 = 0.

g = −3, f = −4: center (3, 4).

$r = \sqrt{9 + 16 + 11} = \sqrt{36} =$ 6 units.

$\dfrac{3 + x}{2} = 3$ and $\dfrac{-2 + y}{2} = 4$: x = 3, y = 10. (3, 10). (The diameter is upright, 12 units long: twice the radius.)

Divide by 3: $x^2 + y^2 + 2x - y + 1 = 0$. g = 1, $f = -\dfrac{1}{2}$: center $\left(-1, \dfrac{1}{2}\right)$.

$r = \sqrt{1 + \dfrac{1}{4} - 1} = \sqrt{\dfrac{1}{4}} =$ $\dfrac{1}{2}$ unit.

$\dfrac{-1 + x}{2} = -1$ and $\dfrac{1 + y}{2} = \dfrac{1}{2}$: x = −1, y = 0. (−1, 0)

Center (−5, −5), as in Example 4: $x^2 + y^2 + 10x + 10y + 25 = 0$.

Center (6, −6): $(x - 6)^2 + (y + 6)^2 = 36$, so $x^2 - 12x + 36 + y^2 + 12y + 36 = 36$. $x^2 + y^2 - 12x + 12y + 36 = 0$

The diameter joins (4, 0) and (0, 6): x(x − 4) + y(y − 6) = 0. $x^2 + y^2 - 4x - 6y = 0$

x(x − 5) + y(y − 10) = 0. $x^2 + y^2 - 5x - 10y = 0$

As in Example 5: $x^2 + y^2 - 12x - 8y = 0$.

CM = CN: $(h + 2)^2 + k^2 = h^2 + (k + 2)^2$, so 4h + 4 = 4k + 4 and h = k.
On the line: 2h − 3h + 1 = 0, so h = 1, k = 1. $r^2 = (1 + 2)^2 + 1^2 = 10$.
$(x - 1)^2 + (y - 1)^2 = 10$: $x^2 + y^2 - 2x - 2y - 8 = 0$

$CP^2 = CQ^2$:

$$\begin{aligned} (h - 4)^2 + (k - 1)^2 &= (h - 6)^2 + (k - 5)^2 \\ -8h + 16 - 2k + 1 &= -12h + 36 - 10k + 25 \\ 4h + 8k &= 44 \\ h + 2k &= 11 \end{aligned}$$

With 4h + k = 16: h = 11 − 2k, so 44 − 8k + k = 16, k = 4, h = 3. $r^2 = (3 - 4)^2 + (4 - 1)^2 = 10$.

$(x - 3)^2 + (y - 4)^2 = 10$: $x^2 + y^2 - 6x - 8y + 15 = 0$

The angle at (0, 0) between the axes is 90∘°, so (4, 0) and (0, 2) are the ends of a diameter: x(x − 4) + y(y − 2) = 0. $x^2 + y^2 - 4x - 2y = 0$

In the same way, x(x − 5) + y(y − 5) = 0. $x^2 + y^2 - 5x - 5y = 0$

g = −4, f = 0: the center is (4, 0). The radius is the distance to (4, 2): 2.
$(x - 4)^2 + y^2 = 4$, so $x^2 - 8x + 16 + y^2 = 4$. $x^2 + y^2 - 8x + 12 = 0$