|MN| = $\sqrt{{\left(\mathrm{x\; -\; (-4)}\right)}^2+{\left(\mathrm{6\; -\; 3}\right)}^2}$

|MN| = $\sqrt{{\left(\mathrm{x\; +\; 4}\right)}^2+3^2}$

|MN| = $\sqrt{{\left(\mathrm{x\; +\; 4}\right)}^2+\; 9}$

But |MN| = 3$\sqrt{\mathrm{10}}$

$\sqrt{{\left(\mathrm{x\; +\; 4}\right)}^2+\; 9}$ = 3$\sqrt{\mathrm{10}}$

Remove the square roots by squaring both sides of the equation.

(x + 4)² + 9 = (3)² x 10

(x + 4)² + 9 = 9 x 10

(x + 4)² + 9 = 90

(x + 4)² + 9 - 90 = 0

(x + 4)² - 81 = 0

Expand the bracket.

x² + 8x + 16 - 81 = 0

x² + 8x - 65 = 0

Two numbers when you multiply, you will get -65 and when you add you will get -8. The two numbers are 13 and -5. Hence split -8x into 13x - 5x.

x² + 13x - 5x - 65 = 0

Factorize out.

x(x + 13) - 5(x + 13) = 0

(x - 5)(x + 13) = 0

When x - 5 = 0, x = 5

When x + 13 = 0, x = -13

x = 5 or -13

Alternatively

(x + 4)² + 9 = 9 x 10

(x + 4)² + 9 = 90

(x + 4)² = 90 - 9

(x + 4)² = 81

Take square root of both sides.

x + 4 = $\sqrt{\mathrm{81}}$

x + 4 ± 9

Note: the result of a square root is either positive or negative.

x + 4 = 9 or x + 4 = -9

x = 9 - 4 or x = -9 - 4

x = 5 or x = -13

∴ x = 5 or -13

Distance between two points = $\sqrt{{({\mathrm{x}}_2-{\mathrm{x}}_1)}^2+{({\mathrm{y}}_2-{\mathrm{y}}_1)}^2}$

Distance between M and N = $\sqrt{{(0-5)}^2+{(5-\mathrm{(-7)})}^2}$

Note:
1. -- or -(-) = +
2. 5 - -7 or 5-(-7) = 5 + 7

Distance between M and N = $\sqrt{{\left(-5\right)}^2+{\left(\mathrm{5\; +\; 7}\right)}^2}$

Distance between M and N = $\sqrt{25+{\left(12\right)}^2}$

Distance between M and N = $\sqrt{\mathrm{25\; +\; 144}}$

Distance between M and N = $\sqrt{169}$

Distance between M and N = 13

For a general equation of a line y = mx + c, the gradient (slope) is m.

Express the given equation to be in the form of the general equation to determine for the gradient (m).

2y + x = 6

2y = -x + 6

Divide both sides by 2.

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

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

Comparing the above equation to the general equation y = mx + c, m = -$\frac{1}{2}$ hence the gradient is -$\frac{1}{2}$

Use the equation y - y₁ = m(x - x₁) with the given point (x₁, y₁) to find the equation of the line.

From the given point (-2, 3), x₁ = -2, y₁ = 3

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

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

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

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

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

2y - 6 = -x - 2

2y = -x - 2 + 6

2y = -x + 4

Divide both sides by 2.

y = -$\frac{1}{2}$x + $\frac{4}{2}$

y = -$\frac{1}{2}$x + 2