Coordinate geometry — common mistakes

For perpendicular lines, remember the phrase 'negative reciprocal'. To find the perpendicular gradient, you flip the fraction and change the sign. For example, if $m_1 = \frac{2}{5}$, the perpendicular gradient $m_2$ is $-\frac{5}{2}$. If $m_1 = -3$, then $m_2 = -\frac{1}{-3} = +\frac{1}{3}$.

$y=mx+c$ is useful when you know the gradient and the y-intercept. $y-y_1=m(x-x_1)$ is more general and powerful; you can use it whenever you know the gradient and any point on the line, not just the y-intercept. Most problems are solved more quickly starting with the point-gradient form.

Not necessarily. Your answer is mathematically correct, but exam questions often specify the format of the final answer, typically $ax+by+c=0$ where $a, b, c$ are integers. To convert your answer, multiply everything by 2 to remove the decimals: $2y = x + 5$. Then rearrange to get everything on one side: $x - 2y + 5 = 0$. Always check the required format.

Think about the geometric property you need to prove. If it involves lengths (e.g., 'show the triangle is isosceles', 'show it is a rhombus'), use the distance formula. If it involves angles, especially right angles (e.g., 'show it is a right-angled triangle', 'show it is a rectangle'), use the gradient formula to check for perpendicular lines ($m_1m_2 = -1$).

An undefined gradient occurs when you try to calculate the gradient of a vertical line. For points $(x_1, y_1)$ and $(x_1, y_2)$, the formula becomes $\frac{y_2-y_1}{x_1-x_1} = \frac{y_2-y_1}{0}$. Since division by zero is undefined, so is the gradient. The equation of such a line is simply $x = k$, where $k$ is the constant x-coordinate.