Resources | Subject Notes | Mathematics
The general equation of a straight line is given by:
$y = mx + c$
where m is the gradient and c is the y-intercept.
Alternatively, the equation can be written in the form:
$ax + by + d = 0$
where a, b, and d are constants.
The gradient m can be calculated using the two-point form:
$m = \frac{y_2 - y_1}{x_2 - x_1}$
where $(x_1, y_1)$ and $(x_2, y_2)$ are two distinct points on the line.
To find the point of intersection of two straight lines, we need to solve the system of equations:
Setting the two expressions for y equal to each other gives:
$m_1x + c_1 = m_2x + c_2$
$m_1x - m_2x = c_2 - c_1$
$x(m_1 - m_2) = c_2 - c_1$
$x = \frac{c_2 - c_1}{m_1 - m_2}$
Substitute the value of x into either of the original equations to find the corresponding value of y.
To find the points of intersection between a straight line and a circle, substitute the equation of the line into the equation of the circle. This will result in a quadratic equation in terms of x. Solve the quadratic equation to find the values of x, and then substitute these values back into the equation of the line to find the corresponding values of y.
The midpoint of a line segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:
$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$
The gradient of a straight line represents its steepness and direction. It is calculated as the change in the y-coordinate divided by the change in the x-coordinate between any two points on the line.
A positive gradient indicates a line that slopes upwards from left to right.
A negative gradient indicates a line that slopes downwards from left to right.
A gradient of zero indicates a horizontal line.
An undefined gradient indicates a vertical line.
The equation of a circle with centre $(a, b)$ and radius r is given by:
$(x - a)^2 + (y - b)^2 = r^2$
The centre of the circle is $(a, b)$ and the radius is r.
The equation of a circle can also be written in general form:
$x^2 + y^2 + 2ax + 2by + c = 0$
where the centre is $(-a, -b)$ and the radius is $\sqrt{a^2 + b^2 - c}$.
As mentioned earlier, to find the intersection points of a line and a circle, substitute the equation of the line into the equation of the circle and solve the resulting quadratic equation.
Topic | Formula | Description |
---|---|---|
Gradient | $m = \frac{y_2 - y_1}{x_2 - x_1}$ | Change in y divided by change in x between two points. |
Midpoint | $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$ | The point halfway between two given points. |
Equation of a Line (Slope-intercept) | $y = mx + c$ | The standard form of a linear equation. |
Equation of a Circle | $(x - a)^2 + (y - b)^2 = r^2$ | The standard form of a circle equation. |