Exam practice

A locus is the set of all the points, and only the points, that satisfy a given rule or condition. For example, the locus of a point 4 units from the origin is the circle $x^2 + y^2 = 16$.

Put each point into $x^2 + y^2$:
(3, 2): 9 + 4 = 13, not on it.
(4, 3): 16 + 9 = 25, on it.
(5, 0): 25 + 0 = 25, on it.
(0, −5): 0 + 25 = 25, on it.
(4, 3), (5, 0) and (0, −5) lie on the locus.

$k(2)^2 - 2(-1)^2 - 2(2) + 3(-1) - 3 = 0$
4k − 2 − 4 − 3 − 3 = 0, so 4k = 12.
k = 3

3(k − 1) − (k + 3) + 7 = 0
3k − 3 − k − 3 + 7 = 0, so 2k + 1 = 0.
$k = -\frac{1}{2}$ (the point is $\left(-\frac{3}{2}, \frac{5}{2}\right)$).

$(x + 2)^2 + (y - 1)^2 = (x - 4)^2 + (y - 1)^2$
The y terms are the same on both sides: $x^2 + 4x + 4 = x^2 - 8x + 16$, so 12x = 12.
x = 1: the vertical line halfway between the two points.

The distance from P(x, y) to the Y-axis is |x|, so its square is $x^2$.
$x^2 = (x - 1)^2 + (y - 2)^2 = x^2 - 2x + 1 + y^2 - 4y + 4$
$y^2 - 2x - 4y + 5 = 0$
(This curve is a parabola, which comes in Grade 10.)

$OP^2 = 4^2$: $x^2 + y^2 = 16$, or $x^2 + y^2 - 16 = 0$.

$x^2 = (x + 2)^2 + (y - 5)^2 = x^2 + 4x + 4 + y^2 - 10y + 25$
$y^2 + 4x - 10y + 29 = 0$

AB = 14, so $AB^2 = 196$.
$(x - 7)^2 + y^2 + (x + 7)^2 + y^2 = 196$
$2x^2 + 2y^2 + 98 = 196$, so $x^2 + y^2 = 49$.
What it means: $PA^2 + PB^2 = AB^2$ is Pythagoras’ rule, so the angle APB is always 90∘°. The points that see AB at a right angle make the circle with AB as its diameter: center O, radius 7.

$(x - 3)^2 + (y - 2)^2 = (x - 7)^2 + (y + 4)^2$
−6x + 9 − 4y + 4 = −14x + 49 + 8y + 16
8x − 12y − 52 = 0; divide by 4: 2x − 3y − 13 = 0
Check with the midpoint (5, −1): 10 + 3 − 13 = 0. ✓

$AP^2 = 4PB^2$:
$(x - 3)^2 + (y - 2)^2 = 4\{(x - 7)^2 + (y + 4)^2\}$
$x^2 - 6x + 9 + y^2 - 4y + 4 = 4x^2 - 56x + 196 + 4y^2 + 32y + 64$
$3x^2 + 3y^2 - 50x + 36y + 247 = 0$
Check with a point that obeys the rule: $\left(\frac{17}{3}, -2\right)$, which divides AB in 2 : 1, has AP = 2PB, and $3 \times \frac{289}{9} + 12 - \frac{850}{3} - 72 + 247 = \frac{289}{3} - \frac{850}{3} + 187 = -187 + 187 = 0$. ✓

$AB^2 = 4^2 + (-6)^2 = 52$.
$(x - 3)^2 + (y - 2)^2 = 52$
$x^2 + y^2 - 6x - 4y - 39 = 0$: the circle with center A through B.

2 × (distance to (0, 2)) = distance to (2, −3). Squaring: $4\{x^2 + (y - 2)^2\} = (x - 2)^2 + (y + 3)^2$
$4x^2 + 4y^2 - 16y + 16 = x^2 - 4x + 4 + y^2 + 6y + 9$
$3x^2 + 3y^2 + 4x - 22y + 3 = 0$

$(x - 1)^2 + y^2 = 4\{x^2 + (y + 2)^2\}$
$x^2 - 2x + 1 + y^2 = 4x^2 + 4y^2 + 16y + 16$
$3x^2 + 3y^2 + 2x + 16y + 15 = 0$

$(x - 3)^2 + y^2 = 9\{x^2 + (y - 2)^2\}$
$x^2 - 6x + 9 + y^2 = 9x^2 + 9y^2 - 36y + 36$
$8x^2 + 8y^2 + 6x - 36y + 27 = 0$

The distance from the X-axis is |y|, so: $(x - 3)^2 + (y - 4)^2 = 4y^2$
$x^2 - 6x + 9 + y^2 - 8y + 16 = 4y^2$
$x^2 - 3y^2 - 6x - 8y + 25 = 0$

The distance from the Y-axis is |x|, so: $2\sqrt{(x + 2)^2 + (y - 5)^2} = |x|$, and squaring, $4\{(x + 2)^2 + (y - 5)^2\} = x^2$
$4x^2 + 16x + 16 + 4y^2 - 40y + 100 = x^2$
$3x^2 + 4y^2 + 16x - 40y + 116 = 0$

The point dividing the segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m₁:m₂ is
$$\left(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\right)$$
and externally $\left(\frac{m_1x_2 - m_2x_1}{m_1 - m_2}, \frac{m_1y_2 - m_2y_1}{m_1 - m_2}\right)$.

The midpoint (m₁ = m₂): $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$. The midpoint of (2, 5) and (6, −1) is (4, 2).
External division (m₂ replaced by −m₂): the point dividing (1, 1) and (3, 5) externally in 3 : 1 is $\left(\frac{9 - 1}{2}, \frac{15 - 1}{2}\right) = (4, 7)$.
Also, the trisection points (ratios 1 : 2 and 2 : 1), and the origin as an end: the point dividing O(0, 0) and (x₂, y₂) in m₁:m₂ is $\left(\frac{m_1x_2}{m_1 + m_2}, \frac{m_1y_2}{m_1 + m_2}\right)$.

$\left(\frac{3(-4) + 4(6)}{7}, \frac{3(14) + 4(-10)}{7}\right) =$ $\left(\frac{12}{7}, \frac{2}{7}\right)$

$\left(\frac{3(-2) + 2(3)}{5}, \frac{3(-7) + 2(5)}{5}\right) =$ $\left(0, -\frac{11}{5}\right)$

$\left(\frac{2(6) + 5(4)}{7}, \frac{2(3) + 5(3)}{7}\right) =$ $\left(\frac{32}{7}, 3\right)$

$\left(\frac{2(6) - 1(-3)}{2 - 1}, \frac{2(5) - 1(2)}{1}\right) =$ (15, 8)

$\left(\frac{4(4) - 3(-3)}{4 - 3}, \frac{4(-4) - 3(2)}{1}\right) =$ (25, −22)

$\left(\frac{1(-3) - 2(3)}{1 - 2}, \frac{1(-4) - 2(-2)}{-1}\right) = \left(\frac{-9}{-1}, \frac{0}{-1}\right) =$ (9, 0)
Here m₁ < m₂, so the point is beyond the first end, (3, −2).

With $k = \frac{m_1}{m_2}$, the y-coordinates: $2 = \frac{k(-3) + 6}{k + 1}$, so 2k + 2 = −3k + 6, 5k = 4, $k = \frac{4}{5}$.
Check with x: $\frac{\frac{4}{5} \times \frac{1}{2} + (-4)}{\frac{9}{5}} = \frac{\frac{2}{5} - \frac{20}{5}}{\frac{9}{5}} = \frac{-18}{9} = -2$. ✓
The ratio is 4 : 5 (internally).

With $k = \frac{m_1}{m_2}$, x: $1 = \frac{3k + 4}{k + 1}$, so k + 1 = 3k + 4, $k = -\frac{3}{2}$.
Check with y: $\frac{5\left(-\frac{3}{2}\right) + 6}{-\frac{1}{2}} = \frac{-\frac{3}{2}}{-\frac{1}{2}} = 3$. ✓
The ratio is 3 : 2 externally (written −3 : 2). (1, 3) is beyond (3, 5), on the far side from (4, 6).

x: $\frac{1}{2} = \frac{-7k + 3}{k + 1}$, so k + 1 = −14k + 6, 15k = 5, $k = \frac{1}{3}$.
Check with y: $\frac{9 \times \frac{1}{3} - 5}{\frac{4}{3}} = \frac{-2}{\frac{4}{3}} = -\frac{3}{2}$. ✓
The ratio is 1 : 3 (internally).

$\left(\frac{-3 + 1}{2}, \frac{-6 - 2}{2}\right) =$ (−1, −4)

$\frac{1 + x'}{2} = -2$, so x′ = −5; $\frac{4 + y'}{2} = 2$, so y′ = 0.
(x′, y′) = (−5, 0)

$\frac{0 + x}{2} = 4$ and $\frac{2 + y}{2} = 3$, so the other end is (8, 4).
A quick way: the midpoint is 4 across and 1 up from (0, 2), so go 4 across and 1 up again.

AB: $\left(\frac{2 - 1}{2}, \frac{-1 + 4}{2}\right) =$ $\left(\frac{1}{2}, \frac{3}{2}\right)$
BC: $\left(\frac{-1 - 2}{2}, \frac{4 + 2}{2}\right) =$ $\left(-\frac{3}{2}, 3\right)$
CA: $\left(\frac{-2 + 2}{2}, \frac{2 - 1}{2}\right) =$ $\left(0, \frac{1}{2}\right)$

1 : 2: $\left(\frac{4 + 2}{3}, \frac{3 - 6}{3}\right) = (2, -1)$; 2 : 1: $\left(\frac{8 + 1}{3}, \frac{6 - 3}{3}\right) = (3, 1)$.
(2, −1) and (3, 1)

1 : 2: $\left(\frac{-3 + 2}{3}, \frac{4 - 4}{3}\right) = \left(-\frac{1}{3}, 0\right)$; 2 : 1: $\left(\frac{-6 + 1}{3}, \frac{8 - 2}{3}\right) = \left(-\frac{5}{3}, 2\right)$.
$\left(-\frac{1}{3}, 0\right)$ and $\left(-\frac{5}{3}, 2\right)$

1 : 2: $\left(\frac{25 - 10}{3}, \frac{10 - 10}{3}\right) = (5, 0)$; 2 : 1: $\left(\frac{50 - 5}{3}, \frac{20 - 5}{3}\right) = (15, 5)$.
(5, 0) and (15, 5)

Every point of the X-axis has y = 0. With $k = \frac{m_1}{m_2}$: $0 = \frac{6k + 3}{k + 1}$, so $k = -\frac{1}{2}$.
The ratio is 1 : 2 externally. Both points are above the X-axis, so the line meets it outside the segment, as the minus sign says.

On the Y-axis x = 0: $0 = \frac{3k - 4}{k + 1}$, so $k = \frac{4}{3}$.
The ratio is 4 : 3 (internally): the ends are on opposite sides of the Y-axis, 4 and 3 units from it.

First the ratio from x: $3 = \frac{-2k + 7}{k + 1}$, so 3k + 3 = −2k + 7, $k = \frac{4}{5}$.
Then $y = \frac{\frac{4}{5}(-5) + (-3)}{\frac{9}{5}} = \frac{-7}{\frac{9}{5}} =$ $-\frac{35}{9}$

The ratio from y: $-5 = \frac{5k + 3}{k + 1}$, so −5k − 5 = 5k + 3, $k = -\frac{4}{5}$ (externally, as expected).
$x = \frac{-\frac{4}{5}(-6) + 2}{-\frac{4}{5} + 1} = \frac{\frac{24}{5} + \frac{10}{5}}{\frac{1}{5}} =$ 34
Check with the slope: from (2, 3) to (−6, 5) y rises 2 while x falls 8; from (2, 3) to (34, −5) y falls 8 while x rises 32: the same steepness, $-\frac{1}{4}$. ✓

Diagonal from (1, 2) to (7, 4): midpoint (4, 3). Diagonal from (3, 0) to (5, 6): midpoint (4, 3).
The diagonals bisect each other, so the points make a parallelogram.

Midpoint of (−1, 0) and (2, 2): $\left(\frac{1}{2}, 1\right)$. Midpoint of (3, 1) and (−2, 1): $\left(\frac{1}{2}, 1\right)$.
The same, so it is a parallelogram.

Midpoint of (3, −2) and (6, −3): $\left(\frac{9}{2}, -\frac{5}{2}\right)$. Midpoint of (4, 0) and (5, −5): $\left(\frac{9}{2}, -\frac{5}{2}\right)$.
The same, so it is a parallelogram.

The diagonals AC and BD have the same midpoint: $\left(\frac{2 + 0}{2}, \frac{3 + 5}{2}\right) = (1, 4)$.
So $\frac{4 + x}{2} = 1$ and $\frac{-1 + y}{2} = 4$: D(−2, 9).
Shortcut: D = A + C − B = (2 + 0 − 4, 3 + 5 + 1).

D = A + C − B = (2 + 12 − 6, 6 + 4 − 2) = (8, 8)
(The midpoint of AC, (7, 5), is also the midpoint of BD.)

D = (1 + 5 − 3, 2 + 3 − 1) = (3, 4)

The diagonals join the first and third, and the second and fourth, and they have the same midpoint:
$\frac{1 + x}{2} = \frac{3 + 5}{2}$, so x = 7; $\frac{2 + 4}{2} = \frac{0 + y}{2}$, so y = 6.

The diagonals bisect each other at (3, 4), so each vertex has its opposite vertex as far again past (3, 4).
Opposite (3, 2): (2 × 3 − 3, 2 × 4 − 2) = (3, 6).
Opposite (5, 10): (2 × 3 − 5, 2 × 4 − 10) = (1, −2).
(This takes (3, 2) and (5, 10) as neighbors. They can’t be opposite vertices: then (3, 4) would be their midpoint, which is (4, 6).)

The diagonals of a parallelogram bisect each other, so the midpoints of AC and BD are the same point:
$$\left(\frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2}\right) = \left(\frac{x_2 + x_4}{2}, \frac{y_2 + y_4}{2}\right)$$
Equal points have equal coordinates; multiplying each by 2, x₁ + x₃ = x₂ + x₄ and y₁ + y₃ = y₂ + y₄. ∎
This is the shortcut used in Question 18: D = A + C − B.

The slope (gradient) of a line is the tangent of its inclination, the angle θ it makes with the positive direction of the X-axis: m =  tan θ. It is how far the line rises for each unit it goes to the right.

m =  tan θ

$m = \frac{y_2 - y_1}{x_2 - x_1}$ (when x₁ ≠ x₂)

If a line meets the X-axis at (a, 0) and the Y-axis at (0, b), the length with sign OA = a is its x-intercept and OB = b its y-intercept: the parts of the axes cut off between the origin and the line.

$m = \tan 60^\circ =$ $\sqrt{3}$

$m = \tan 135^\circ = \tan(180^\circ - 45^\circ) = -\tan 45^\circ =$ −1

The line makes 45∘° with OX: $m = \tan 45^\circ =$ 1

The angle from NX round to the line is 120∘°, so θ = 120∘°: $m = \tan 120^\circ = -\tan 60^\circ =$ $-\sqrt{3}$

The 30∘° is between the line and BO, the negative direction of the X-axis. The inclination, from the positive direction BX, is $180^\circ - 30^\circ = 150^\circ$: $m = \tan 150^\circ = -\tan 30^\circ =$ $-\frac{1}{\sqrt{3}}$
(The line falls to the right, so a negative slope is expected.)

The 150∘° is between the line and BX′, the negative direction. The inclination is $180^\circ - 150^\circ = 30^\circ$: $m = \tan 30^\circ =$ $\frac{1}{\sqrt{3}}$

$m = \frac{-5 - 3}{2 - (-4)} = \frac{-8}{6} =$ $-\frac{4}{3}$

$m = \frac{0 - (-5)}{3 - 0} =$ $\frac{5}{3}$

$\frac{p - 5}{6 - 2} = -\frac{1}{2}$, so p − 5 = −2, p = 3.

$\frac{4 - 8}{3 - p} = 2$, so −4 = 6 − 2p, 2p = 10, p = 5.

Slope of PQ: $\frac{2 - 6}{4 - 8} = 1$. So $\frac{3 - 9}{p - 7} = 1$, p − 7 = −6, p = 1.

Points on one line give the same slope between any two of them.
Slope from (−2, −2) to (1, 1): $\frac{3}{3} = 1$. Slope from (1, 1) to (m, 2): $\frac{1}{m - 1}$.
$\frac{1}{m - 1} = 1$, so m = 2.

y = b, where b is its distance above the X-axis (negative if below).

x = a, where a is its distance to the right of the Y-axis (negative if to the left).

y = mx + c, m the slope and c the y-intercept.

$\frac{x}{a} + \frac{y}{b} = 1$, a the x-intercept and b the y-intercept.

x cos α + y sin α = p

p is the length of the perpendicular from the origin to the line (never negative), and α is the angle that perpendicular makes with the positive direction of the X-axis.

Slope-intercept form y = mx + c; double-intercept form $\frac{x}{a} + \frac{y}{b} = 1$; normal form x cos α + y sin α = p.

Every point has x-coordinate −3: x = −3, or x + 3 = 0.

Every point has y-coordinate −3: y = −3, or y + 3 = 0.

y = 2, or y − 2 = 0.

x = 5, or x − 5 = 0.

m = 5, c = −2

m = −2, c = 4

m = 12, c = 0 (the line passes through the origin).

y = 0 · x + 6: m = 0, c = 6 (a horizontal line).

$m = \frac{1}{2}$, $c = -\frac{2}{3}$

y = −x − 1: m = −1, c = −1

2y = 10x + 8, y = 5x + 4: m = 5, c = 4

$m = \sqrt{3}$, c = 0

y = 5x + 3, or 5x − y + 3 = 0

y = −2x − 1, or 2x + y + 1 = 0

$\tan 120^\circ = -\sqrt{3}$: $y = -\sqrt{3}x - 5$, or $\sqrt{3}x + y + 5 = 0$

$\tan(-60^\circ) = -\sqrt{3}$: $y = -\sqrt{3}x + 3$, or $\sqrt{3}x + y - 3 = 0$

c = 0: y = 3x, or 3x − y = 0

(0, 1) is on the Y-axis, so c = 1: $y = \frac{1}{3}x + 1$, 3y = x + 3, or x − 3y + 3 = 0

$m = \tan 45^\circ = 1$, c = 0: y = x, or x − y = 0

$m = \tan 150^\circ = -\frac{1}{\sqrt{3}}$, c = 0: $y = -\frac{x}{\sqrt{3}}$, $\sqrt{3}y = -x$, or $x + \sqrt{3}y = 0$

$\frac{x}{3} + \frac{y}{-4} = 1$; multiply by 12: 4x − 3y = 12, or 4x − 3y − 12 = 0

$\frac{x}{-2} + \frac{y}{3} = 1$; multiply by 6: −3x + 2y = 6, or 3x − 2y + 6 = 0

$\frac{x}{3/5} + \frac{y}{9/2} = 1$, that is $\frac{5x}{3} + \frac{2y}{9} = 1$; multiply by 9: 15x + 2y = 9, or 15x + 2y − 9 = 0

y = 0: 4x = 12, x = 3. x = 0: −3y = 12, y = −4. a = 3, b = −4

y = 0: 5x = −15, x = −3. x = 0: 3y = −15, y = −5. a = −3, b = −5

$\frac{x}{a} + \frac{y}{a} = 1$, so x + y = a. Through (2, −1): a = 2 − 1 = 1.
x + y − 1 = 0

x + y = a, a = 3 + 4 = 7: x + y − 7 = 0

$\frac{x}{a} + \frac{y}{-a} = 1$, so x − y = a. Through (2, 3): a = −1.
x − y = −1, or x − y + 1 = 0

x − y = a, a = 6 + 5 = 11: x − y − 11 = 0

Let the y-intercept be b, so the x-intercept is 2b: $\frac{x}{2b} + \frac{y}{b} = 1$, so x + 2y = 2b.
Through (3, 2): 2b = 3 + 4 = 7.
x + 2y − 7 = 0

Let the intercepts be a and 15 − a: $\frac{3}{a} + \frac{4}{15 - a} = 1$.
Multiply by a(15 − a): 3(15 − a) + 4a = a(15 − a), so $45 + a = 15a - a^2$, $a^2 - 14a + 45 = 0$, (a − 5)(a − 9) = 0.
a = 5, b = 10: $\frac{x}{5} + \frac{y}{10} = 1$, 2x + y − 10 = 0.
a = 9, b = 6: $\frac{x}{9} + \frac{y}{6} = 1$, 2x + 3y − 18 = 0.
Two lines fit, one steep and one gentle: both pass through (3, 4).

The line meets the axes at (a, 0) and (0, b), whose midpoint is $\left(\frac{a}{2}, \frac{b}{2}\right) = (2, 3)$: a = 4, b = 6.
$\frac{x}{4} + \frac{y}{6} = 1$; multiply by 12: 3x + 2y − 12 = 0

6 and 8

$\frac{x}{6} + \frac{y}{8} = 1$; multiply by 24: 4x + 3y − 24 = 0

$AB = \sqrt{6^2 + 8^2} = \sqrt{100} =$ 10 units

Let the x-intercept be a, so the y-intercept is 2a: $\frac{x}{a} + \frac{y}{2a} = 1$, so 2x + y = 2a.
Through (−5, 8): 2a = −10 + 8 = −2.
2x + y + 2 = 0
Put (−1, 0) in: −2 + 0 + 2 = 0. ✓ So the line passes through (−1, 0), which is in fact its x-intercept, a = −1.

Let the line meet the X-axis at A(a, 0) and the Y-axis at B(0, b), and take the ratio from A, so AP:PB = 3:4 with P(2, 3):
$2 = \frac{3 \times 0 + 4a}{7}$, so $a = \frac{7}{2}$; $3 = \frac{3b + 4 \times 0}{7}$, so b = 7.
$\frac{2x}{7} + \frac{y}{7} = 1$: 2x + y − 7 = 0
If the ratio is taken from B instead, BP:PA = 3:4: $2 = \frac{3a}{7}$, $a = \frac{14}{3}$, and $3 = \frac{4b}{7}$, $b = \frac{21}{4}$, giving $\frac{3x}{14} + \frac{4y}{21} = 1$, or 9x + 8y − 42 = 0. Both pass through (2, 3); say which end you start from.

$x\cos 30^\circ + y\sin 30^\circ = 4$: $\frac{\sqrt{3}x}{2} + \frac{y}{2} = 4$, $\sqrt{3}x + y - 8 = 0$

x · 0 + y · 1 = 2: y − 2 = 0 (a horizontal line 2 above the origin, as a perpendicular straight up says).

$\cos(-60^\circ) = \frac{1}{2}$, $\sin(-60^\circ) = -\frac{\sqrt{3}}{2}$: $\frac{x}{2} - \frac{\sqrt{3}y}{2} = 1$, $x - \sqrt{3}y - 2 = 0$

$\cos 135^\circ = -\frac{1}{\sqrt{2}}$, $\sin 135^\circ = \frac{1}{\sqrt{2}}$: $\frac{-x + y}{\sqrt{2}} = \frac{5}{7}$, so $7(-x + y) = 5\sqrt{2}$.
$7x - 7y + 5\sqrt{2} = 0$

$\frac{x + y}{\sqrt{2}} = 13$, so $x + y = 13\sqrt{2}$: $x + y - 13\sqrt{2} = 0$

$\cos 150^\circ = -\frac{\sqrt{3}}{2}$, $\sin 150^\circ = \frac{1}{2}$: $\frac{-\sqrt{3}x + y}{2} = 2\sqrt{2}$, so $-\sqrt{3}x + y = 4\sqrt{2}$.
$\sqrt{3}x - y + 4\sqrt{2} = 0$

$2y = -\sqrt{3}x + 7$: $y = -\frac{\sqrt{3}}{2}x + \frac{7}{2}$

y = x − 6

2y = 3x + 8: $y = \frac{3}{2}x + 4$

Divide by 12: $\frac{x}{4} + \frac{y}{3} = 1$

2x − 3y = 6; divide by 6: $\frac{x}{3} - \frac{y}{2} = 1$, that is $\frac{x}{3} + \frac{y}{-2} = 1$

Divide by 5: $\frac{\sqrt{3}x}{5} - \frac{7y}{5} = 1$, that is $\frac{x}{\frac{5}{\sqrt{3}}} + \frac{y}{-\frac{5}{7}} = 1$

−x + y = 4; $\sqrt{1 + 1} = \sqrt{2}$: $-\frac{1}{\sqrt{2}}x + \frac{1}{\sqrt{2}}y = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.
cos α < 0, sin α > 0: α = 135∘°. $x\cos 135^\circ + y\sin 135^\circ = 2\sqrt{2}$

$-x + y = 2\sqrt{2}$; divide by $\sqrt{2}$: $-\frac{1}{\sqrt{2}}x + \frac{1}{\sqrt{2}}y = 2$.
$x\cos 135^\circ + y\sin 135^\circ = 2$

$\sqrt{9 + 16} = 5$: $\frac{3}{5}x + \frac{4}{5}y = 3$, so $\cos\alpha = \frac{3}{5}$, $\sin\alpha = \frac{4}{5}$ (α ≈ 53.1∘°) and p = 3.

$-x - \sqrt{3}y = 4$; $\sqrt{1 + 3} = 2$: $-\frac{1}{2}x - \frac{\sqrt{3}}{2}y = 2$.
Both negative: third quadrant, α = 240∘°. $x\cos 240^\circ + y\sin 240^\circ = 2$