Gradient = $\frac{\mathrm{Change\; in\; Y}}{\mathrm{Change\; in\; X}}$ = $\frac{{\mathrm{Y}}_2-{\mathrm{Y}}_1}{{\mathrm{X}}_2-{\mathrm{X}}_1}$

Y₂ = 4
Y₁ = 6
X₂ = x
X₁ = 3

Gradient = $\frac{\mathrm{4\; -\; 6}}{\mathrm{x\; -\; 3}}$

Gradient = $\frac{-2}{\mathrm{x\; -\; 3}}$

But gradient = - $\frac{2}{5}$ = $\frac{-2}{5}$

$\frac{-2}{\mathrm{x\; -\; 3}}$ = $\frac{-2}{5}$

Since the numerator at the left and right are the same, the denominators are also the same.

x - 3 = 5

x = 5 + 3

x = 8

Rotation through 90^o anticlockwise

When rotating a point 90^o anticlockwise(counterclockwise) about the origin our point A(x,y) becomes A'(-y,x).

In other words, switch x and y and make y negative

Rotation through 270^o anticlockwise

When rotating a point 270^o anticlockwise(counterclockwise) about the origin our point A(x,y) becomes A'(y,-x).

This means, we switch x and y and make x negative

Rotation through 180^o anticlockwise

When rotating a point 180^o anticlockwise(counterclockwise) about the origin our point A(x,y) becomes A'(-x,-y).

So all we do is make both x and y negative

Perimeter is the sum of all the sides of a shape.

The total perimeter before the sector was cut is the circumference of the circular plate.

Circumference of a circle = 2πr

The remaining perimeter will be the circumference - the length of the arc of the sector cut + the two radii.

Remaining perimeter = Circumference - length of cut arc + the two radii

Length of an arc = $\frac{\mathrm{Angle}}{360}$ x circumference of the circle

Length of an arc = $\frac{\mathrm{Angle}}{360}$ x 2 x π x radius

Angle = 150°

π = $\frac{22}{7}$

Radius = 14 cm

Length of cut arc = $\frac{\mathrm{150°}}{\mathrm{360°}}$ x 2 x $\frac{22}{7}$ x 14 cm

Length of cut arc = 36.666666667 ≈ 36.7 cm

Circumference = 2 x x $\frac{22}{7}$ x 14 cm

Circumference = 88 cm

Remaining perimeter = 88 cm - 36.7 cm + 14 cm + 14 cm = 79.3 cm

Alternatively

The total angle for the circle = 360°

Remaining angle for the sector left = 360° - 150°

Remaining angle for the sector left = 210°

Remaining perimeter = Length of arc left + the two radii

Length of arc left = $\frac{\mathrm{210°}}{\mathrm{360°}}$ x 2 x $\frac{22}{7}$ x 14 cm = 51.3 cm

Remaining perimeter = 51.3 cm + 14 cm + 14 cm

Remaining perimeter = 79.3 cm

Concept

Parallel lines have the same gradient.

The general equation of a line is given as y = mx + b
m = slope/gradient
b = y - intercept

Determine the gradient/slope in the given equation.

2x + 5y = -10

Make y the subject

5y = -2x - 10

Divide both sides by 5

y = -$\frac{2}{5}$x - 2

Compare the equation with the general equation for a line (y = mx + b).

Gradient/slope = -$\frac{2}{5}$

Since the line is parallel to the line with gradient -$\frac{2}{5}$, that line has the same gradient.

Gradient = $\frac{\mathrm{Change\; in\; Y}}{\mathrm{Change\; in\; X}}$ = $\frac{{\mathrm{Y}}_2-{\mathrm{Y}}_1}{{\mathrm{X}}_2-{\mathrm{X}}_1}$

Calculate the gradient of the line using the given point (4,1) and (x,y), equate it to the known gradient and simplify.

Cross multiply.

5(y - 1) = -2(x - 4)
5y - 5 = -2x + 8
5y + 2x = 8 + 5
2x + 5y = 13

For a midpoint between two points A(x₁, y₁) and B(x₂, y₂),

Midpoint x coordinate = $\frac{{\mathrm{x}}_1+{\mathrm{x}}_2}{2}$

Midpoint y coordinate = $\frac{{\mathrm{y}}_1+{\mathrm{y}}_2}{2}$

Calculating the midpoint of X(-8, -12) and Y(p, q)

Midpoint of XY x coordinate = $\frac{\mathrm{-8\; +\; p}}{2}$

Midpoint of XY y coordinate = $\frac{\mathrm{-12\; +\; q}}{2}$

Midpoint of XY = (-4, -2)

But Midpoint of XY x coordinate = -4

$\frac{\mathrm{-8\; +\; p}}{2}$ = -4

Get rid of the fraction by multiplying both sides by the denominator (2)

-8 + p = -4 x 2

-8 + p = -8

p = -8 + 8

p = 0

But Midpoint of XY y coordinate = -2

$\frac{\mathrm{-12\; +\; q}}{2}$ = -2

Get rid of the fraction by multiplying both sides by the denominator (2)

-12 + q = -2 x 2

-12 + q = -4

q = -4 + 12

q = 8

∴ the coordinates of Y = (0, 8)

The general equation of a line is y = mx + c

Where m is the gradient or slope and c is the y-intercept.

Express the given equation to be in the form of the general equation of a line and compare to find m.

3x - 5y = 7

3x - 7 = 5y

Divide both sides by 5

$\frac{3}{5}$x - $\frac{7}{5}$ = y

Comparing the above with the general equation of a line, m = $\frac{3}{5}$ hence gradient (slope) = $\frac{3}{5}$

To be parallel, two lines must have the same gradient.

For a general equation of a line, y = mx + c

m = gradient and c = y intercept.

Express the given equation to be in the form y = mx + c and compare to find the value of m, the gradient.

2y = 3(x - 2)

2y = 3x - 6

Divide both sides by 2

y = $\frac{\mathrm{3x}}{2}$ - $\frac{6}{2}$

y = $\frac{\mathrm{3x}}{2}$ - 3

Comparing the equation with the general equation y = mx + c, the gradient is $\frac{3}{2}$ and the y intercept is -3

Use the given point to calculate the gradient with the point (x, y) and make y the subject.

Gradient = $\frac{\mathrm{Change\; in\; Y}}{\mathrm{Change\; in\; X}}$ = $\frac{{\mathrm{Y}}_2-{\mathrm{Y}}_1}{{\mathrm{X}}_2-{\mathrm{X}}_1}$

Gradient using (x, y) and (2, 3)

Gradient = $\frac{\mathrm{y\; -\; 3}}{\mathrm{x\; -\; 2}}$

But the gradient = $\frac{3}{2}$

$\frac{\mathrm{y\; -\; 3}}{\mathrm{x\; -\; 2}}$ = $\frac{3}{2}$

Cross multiply.

2 x (y - 3) = 3 x (x - 2)

2y - 6 = 3x - 6

2y = 3x - 6 + 6

2y = 3x

Divide both sides by 2.

y = $\frac{3}{2}$x

Gradient (m) = $\frac{3}{4}$

Point (1, -5) → x₁ = 1 and y₁ = -5

y -(-5) = $\frac{3}{4}$(x - 1)

Note: -- = + or -(-) = +

-(-5) = + 5

y + 5 = $\frac{3}{4}$(x - 1)

Get rid of the fraction by multiplying both sides by the denominator 4

4 x y + 5 x 4 = 3(x - 1)

4y + 20 = 3x - 3

0 = 3x - 3 - 20 - 4y

0 = 3x - 4y - 23

3x - 4y - 23 = 0

Gradient = $\frac{\mathrm{Change\; in\; Y}}{\mathrm{Change\; in\; X}}$ = $\frac{{\mathrm{Y}}_2-{\mathrm{Y}}_1}{{\mathrm{X}}_2-{\mathrm{X}}_1}$

Y₂ = 6
Y₁ = 4
X₂ = -2
X₁ = 1

Gradient = $\frac{\mathrm{6\; -\; 4}}{\mathrm{-2\; -\; 1}}$ = $\frac{2}{-3}$ = -$\frac{2}{3}$

Distance = $\sqrt{{\left(\mathrm{8\; -\; 3}\right)}^2+{\left(\mathrm{10\; --\; 2}\right)}^2}$

Note: - - = +

Distance = $\sqrt{{\left(5\right)}^2+{\left(\mathrm{10\; +\; 2}\right)}^2}$

Distance = $\sqrt{5^2+{12}^2}$

Distance = $\sqrt{\mathrm{25\; +\; 144}}$

Distance = $\sqrt{169}$ = 13 units

The angle between a tangent and the radius is 90° and not complementary.

Two angles are complementary if the sum of the angles is equal to 90º (ninety degrees). For example, an angle that is 30º and an angle that is 60º are two complementary angles.

|OP| = $\sqrt{{(4-0)}^2+{(-1-0)}^2}$

|OP| = $\sqrt{\mathrm{17}}$

|OQ| = $\sqrt{{(1-0)}^2+{(-4-0)}^2}$

|OQ| = $\sqrt{\mathrm{17}}$

|PQ| = $\sqrt{{(4-1)}^2+{(-1--4)}^2}$

Note: -- = +. --4 = + 4

|OP| = $\sqrt{\mathrm{18}}$

|OP| = |OQ| hence ∆OPQ has two sides equal and therefore is an isosceles triangle.

Reflection in the line x - 2 = 0, x = 2

k = 2

(-7, 8) → (2 x 2 - (-7), 8)

(-7, 8) → (4 + 7, 8)

(-7, 8) → (11, 8)

Note:
1. -- = + or -(-) = +
2. -(-7) = + 7

Gradient = $\frac{\mathrm{Change\; in\; Y}}{\mathrm{Change\; in\; X}}$ = $\frac{{\mathrm{Y}}_2-{\mathrm{Y}}_1}{{\mathrm{X}}_2-{\mathrm{X}}_1}$

Pick any two points from the table to calculate the gradient.

Using the first two points, (0, 1) and (2, 2).

Y₂ = 2
Y₁ = 1
X₂ = 2
X₁ = 0

Gradient = $\frac{\mathrm{2\; -\; 1}}{\mathrm{2\; -\; 0}}$ = $\frac{1}{2}$

Reflection in the line x - 2 = 0, x = 2

k = 2

(-7, 8) → (2 x 2 - (-7), 8)

(-7, 8) → (4 + 7, 8)

(-7, 8) → (11, 8)

Note:
1. -- = + or -(-) = +
2. -(-7) = + 7

Gradient = $\frac{\mathrm{Change\; in\; Y}}{\mathrm{Change\; in\; X}}$ = $\frac{{\mathrm{Y}}_2-{\mathrm{Y}}_1}{{\mathrm{X}}_2-{\mathrm{X}}_1}$

Y₂ = 4
Y₁ = 6
X₂ = x
X₁ = 3

Gradient = $\frac{\mathrm{4\; -\; 6}}{\mathrm{x\; -\; 3}}$

Gradient = $\frac{-2}{\mathrm{x\; -\; 3}}$

But gradient = - $\frac{2}{5}$ = $\frac{-2}{5}$

$\frac{-2}{\mathrm{x\; -\; 3}}$ = $\frac{-2}{5}$

Since the numerator at the left and right are the same, the denominators are also the same.

x - 3 = 5

x = 5 + 3

x = 8

At the point of intersection, the y and x values are the same.

3x + 2y = 4

2y = -3x + 4

Divide both sides by 2

y = $\frac{\mathrm{-3x}}{2}$ + $\frac{4}{2}$

y = $\frac{\mathrm{-3x}}{2}$ + 2 ---- eq. 1

y = 2x - 5 ---- eq. 2

eq. 1 = eq. 2

$\frac{\mathrm{-3x}}{2}$ + 2 = 2x - 5

Get rid of the fraction by multiplying both sides by the denominator, 2

2 x $\frac{\mathrm{-3x}}{2}$ + 2 x 2 = 2x x 2 - 5 x 2

-3x + 4 = 4x - 10

4 + 10 = 4x + 3x

14 = 7x

Divide both sides by 7

x = $\frac{14}{7}$ = 2

Substitute the value of x into --- eq. 2 and solve for the value of y.

y = 2x - 5 ---- eq. 2

y = 2 x 2 - 5

y = 4 - 5

y = -1

∴ P = (2, -1)

Concept

Parallel lines have the same gradient.

The general equation of a line is given as y = mx + b
m = slope/gradient
b = y - intercept

Determine the gradient/slope in the given equation.

2x + 5y = -10

Make y the subject

5y = -2x - 10

Divide both sides by 5

y = -$\frac{2}{5}$x - 2

Compare the equation with the general equation for a line (y = mx + b).

Gradient/slope = -$\frac{2}{5}$

Since the line is parallel to the line with gradient -$\frac{2}{5}$, that line has the same gradient.

Gradient = $\frac{\mathrm{Change\; in\; Y}}{\mathrm{Change\; in\; X}}$ = $\frac{{\mathrm{Y}}_2-{\mathrm{Y}}_1}{{\mathrm{X}}_2-{\mathrm{X}}_1}$

Calculate the gradient of the line using the given point (4,1) and (x,y), equate it to the known gradient and simplify.

Cross multiply.

5(y - 1) = -2(x - 4)
5y - 5 = -2x + 8
5y + 2x = 8 + 5
2x + 5y = 13

For a midpoint between two points A(x₁, y₁) and B(x₂, y₂),

Midpoint x coordinate = $\frac{{\mathrm{x}}_1+{\mathrm{x}}_2}{2}$

Midpoint y coordinate = $\frac{{\mathrm{y}}_1+{\mathrm{y}}_2}{2}$

Calculating the midpoint of X(-8, -12) and Y(p, q)

Midpoint of XY x coordinate = $\frac{\mathrm{-8\; +\; p}}{2}$

Midpoint of XY y coordinate = $\frac{\mathrm{-12\; +\; q}}{2}$

Midpoint of XY = (-4, -2)

But Midpoint of XY x coordinate = -4

$\frac{\mathrm{-8\; +\; p}}{2}$ = -4

Get rid of the fraction by multiplying both sides by the denominator (2)

-8 + p = -4 x 2

-8 + p = -8

p = -8 + 8

p = 0

But Midpoint of XY y coordinate = -2

$\frac{\mathrm{-12\; +\; q}}{2}$ = -2

Get rid of the fraction by multiplying both sides by the denominator (2)

-12 + q = -2 x 2

-12 + q = -4

q = -4 + 12

q = 8

∴ the coordinates of Y = (0, 8)

To be parallel, two lines must have the same gradient.

For a general equation of a line, y = mx + c

m = gradient and c = y intercept.

Express the given equation to be in the form y = mx + c and compare to find the value of m, the gradient.

2y = 3(x - 2)

2y = 3x - 6

Divide both sides by 2

y = $\frac{\mathrm{3x}}{2}$ - $\frac{6}{2}$

y = $\frac{\mathrm{3x}}{2}$ - 3

Comparing the equation with the general equation y = mx + c, the gradient is $\frac{3}{2}$ and the y intercept is -3

Use the given point to calculate the gradient with the point (x, y) and make y the subject.

Gradient = $\frac{\mathrm{Change\; in\; Y}}{\mathrm{Change\; in\; X}}$ = $\frac{{\mathrm{Y}}_2-{\mathrm{Y}}_1}{{\mathrm{X}}_2-{\mathrm{X}}_1}$

Gradient using (x, y) and (2, 3)

Gradient = $\frac{\mathrm{y\; -\; 3}}{\mathrm{x\; -\; 2}}$

But the gradient = $\frac{3}{2}$

$\frac{\mathrm{y\; -\; 3}}{\mathrm{x\; -\; 2}}$ = $\frac{3}{2}$

Cross multiply.

2 x (y - 3) = 3 x (x - 2)

2y - 6 = 3x - 6

2y = 3x - 6 + 6

2y = 3x

Divide both sides by 2.

y = $\frac{3}{2}$x

Gradient = $\frac{\mathrm{Change\; in\; Y}}{\mathrm{Change\; in\; X}}$ = $\frac{{\mathrm{Y}}_2-{\mathrm{Y}}_1}{{\mathrm{X}}_2-{\mathrm{X}}_1}$

Choose any two points from the table.

P₁(6.20, 3.90) and P₂(6.85, 5.20)

Gradient = $\frac{\mathrm{5.2-3.9}}{\mathrm{6.85-6.20}}$ = $\frac{1.3}{0.65}$ = 2

Gradient (m) = $\frac{3}{4}$

Point (1, -5) → x₁ = 1 and y₁ = -5

y -(-5) = $\frac{3}{4}$(x - 1)

Note: -- = + or -(-) = +

-(-5) = + 5

y + 5 = $\frac{3}{4}$(x - 1)

Get rid of the fraction by multiplying both sides by the denominator 4

4 x y + 5 x 4 = 3(x - 1)

4y + 20 = 3x - 3

0 = 3x - 3 - 20 - 4y

0 = 3x - 4y - 23

3x - 4y - 23 = 0

Gradient = $\frac{\mathrm{Change\; in\; Y}}{\mathrm{Change\; in\; X}}$ = $\frac{{\mathrm{Y}}_2-{\mathrm{Y}}_1}{{\mathrm{X}}_2-{\mathrm{X}}_1}$

Y₂ = 6
Y₁ = 4
X₂ = -2
X₁ = 1

Gradient = $\frac{\mathrm{6\; -\; 4}}{\mathrm{-2\; -\; 1}}$ = $\frac{2}{-3}$ = -$\frac{2}{3}$

Distance = $\sqrt{{\left(\mathrm{8\; -\; 3}\right)}^2+{\left(\mathrm{10\; --\; 2}\right)}^2}$

Note: - - = +

Distance = $\sqrt{{\left(5\right)}^2+{\left(\mathrm{10\; +\; 2}\right)}^2}$

Distance = $\sqrt{5^2+{12}^2}$

Distance = $\sqrt{\mathrm{25\; +\; 144}}$

Distance = $\sqrt{169}$ = 13 units

If a and b are the roots, (x - a)(x - b) = 0

Change the mixed fraction to improper fraction.

1$\frac{1}{2}$ = $\frac{\mathrm{2\; x\; 1\; +\; 1}}{2}$ = $\frac{\mathrm{2\; +\; 1}}{2}$ = $\frac{3}{2}$

For the roots -$\frac{1}{2}$ and $\frac{3}{2}$,

(x --$\frac{1}{2}$)(x - $\frac{3}{2}$) = 0

-- = +

(x + $\frac{1}{2}$)(x - $\frac{3}{2}$) = 0

x(x - $\frac{3}{2}$) + $\frac{1}{2}$(x - $\frac{3}{2}$) = 0

x² - $\frac{\mathrm{3x}}{2}$ + $\frac{\mathrm{x}}{2}$ - $\frac{\mathrm{1\; x\; 3}}{\mathrm{2\; x\; 2}}$ = 0

x² + $\frac{\mathrm{x\; -\; 3x}}{2}$ - $\frac{3}{4}$ = 0

x² - $\frac{\mathrm{2x}}{2}$ - $\frac{3}{4}$ = 0

Get rid of the fraction by multiplying both sides by the L.C.M of the denominators 2 and 4.

The L.C.M is 4.

4 x x² - $\frac{\mathrm{2x}}{2}$ x 4 - 4 x $\frac{3}{4}$ = 0 x 4

4x² - 4x - 3 = 0