Linear Coordinate Geometry

This section has been studied for many years now (since year 8). As such, we have simply provided the necessary formulas below. \[A = (x_1, y_1)\] \[B = (x_2, y_2)\]

Distance between points A and B \[AB = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\]

Midpoint of a line segment AB has coordinates: \[(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})\]

Gradient of the line AB \[m = \frac{y_2-y_1}{x_2-x_1}\] \[m = tan(\theta)\] θ is the angle that the line makes with the positive direction of the x-axis

Equations of a straight line \[y = mx + c\] (c is the y - intercept) \[y-y_1 = m(x-x_1)\]

For any two perpendicular lines \[m_1 \cdot m_2 = -1\]

Geometry of Simultaneous Equations

For any two lines, they can intersect once only, infinitely many times or not at all.
A line, most commonly represented in the form: y = mx + c is identified by its gradient (m) and y-intercept (c).

Case 1: If two lines intersect once only, then, the gradients of the lines must be different. (No requirements for the y-intercept.)

Case 2: If two lines intersect infinitely many times, then, the gradients and the y-intercepts must be the same.

Case 3: If two lines do not intersect at all, then, the gradients must be the same, but the y-intercepts must be different.

\[ \text{Example 2.1: Find the value(s) of k for which the pair of equations } \\ kx+(k+1)y=8 \text{ and } 4x+3ky=4 \text{ have:}\\ \text{a. no solutions}\\ \text{b. a unique solution}\\ \text{c. infinitely many solutions}\\ \text{ }\\ \text{Step 1: Rearrange the equations into the standard form } y = mx + c\\ \begin{aligned} kx + (k+1)y &=8\\ (k+1)y &= 8 - kx\\ y&=-\frac{kx}{k+1}+\frac{8}{k+1}...\boxed{1}\\ 4x+3ky&=4\\ 3ky &= 4-4x\\ y&=-\frac{4x}{3k}+\frac{4}{3k}...\boxed{2}\\ \end{aligned}\\ \text{Step 2: Equate the gradients to find when they are equal}\\ \begin{aligned} -\frac{4}{3k}&=-\frac{k}{k+1}\\ -4(k+1) &= 3k \cdot (-k) \\ -4k-4 &= -3k^2 \\ 0&=3k^2-4k-4\\ &=(3k+2)(k-2)\\ k&=-\frac{2}{3},\,2\\ \end{aligned}\\ \text{Step 3: Find when the y-intercepts are not equal}\\ \begin{aligned} \frac{8}{k+1}&\neq\frac{4}{3k}\\ 24k&\neq4k+4\\ 20k &\neq 4\\ k&\neq\frac{1}{5}\\ \end{aligned}\\ \therefore \text{No solutions when } k=-\frac{2}{3}, 2\\ \text{ }\\ \text{Step 4: Find when the gradients are not equal}\\ \begin{aligned} -\frac{4}{3k}&\neq-\frac{k}{k+1}\\ -4(k+1) &\neq 3k \cdot (-k) \\ -4k-4 &\neq -3k^2 \\ 0&\neq3k^2-4k-4\\ &\neq(3k+2)(k-2)\\ k&\neq-\frac{2}{3},\,2\\ \end{aligned}\\ \therefore \text{A unique solution when } k\in \mathbb{R} \backslash\{-\frac{2}{3}, 2\}\\ \text{ }\\ \text{Step 5: Equate the y-intercepts to find when they are equal}\\ \begin{aligned} \frac{8}{k+1}&=\frac{4}{3k}\\ 24k&=4k+4\\ 20k&=4\\ k&=\frac{1}{5}\\ \end{aligned}\\ \text{Step 6: Equate the gradients to find when they are equal}\\ \begin{aligned} -\frac{4}{3k}&=-\frac{k}{k+1}\\ -4(k+1) &= 3k \cdot (-k) \\ -4k-4 &= -3k^2 \\ 0&=3k^2-4k-4\\ &=(3k+2)(k-2)\\ k&=-\frac{2}{3},\,2\\ \end{aligned}\\ \text{Thus, no value of } k \text{ will produce infinitely many solutions}\\ \]