HSC Extension 1 and 2 Mathematics/Linear functions

The linear function y = mx + b and its graphEdit

The linear function y = mx + b, for numerical values of m and b, is represented by a straight line.

The linear equation mx + b = 0 (m !=­ 0) has one and only one root, which is rational if m and b are rational. This simple and apparently trivial result is worth noting; it does not extend to the quadratic equation.

The straight lineEdit

equation of a line passing through a given point with given slopeEdit

The equation of a straight line passing through a fixed point (x1, y1) and having a given slope m is y – y1 = m(x – x1).

equation of a line passing through two given pointsEdit

The equation of a straight line passing through two given points (x1, y1) and (x2, y2) may be deduced from the above by noting that the slope m must be the ratio of y2 – y1 to x2 – x1, i.e.,

m = \frac{y_2-y_1}{x_2-x_1}

Verification that the number pairs (x1, y1) and (x2, y2) do indeed satisfy the final equation.

the general equation ax + by + c = 0Edit

The linear equation ax + by + c = 0 represents a straight line provided at least one of a and b is nonzero.

parallel linesEdit

Parallel lines have the same slope

perpendicular linesEdit

two lines with gradients m and m' respectively are perpendicular if and only if mm' = –1.

Intersection of linesEdit

Intersection of two lines and the solution of two linear equations in two unknownsEdit

The equation of a line passing through the point of intersection of two given linesEdit

Any line passing through the intersection of a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 has the equation

a_1x + b_1y + c_1 + k(a_2x + b_2y + c_2) = 0

where k is a constant, whose value for any particular line may be found using the additional information given to specify that line.

Regions determined by lines: linear inequalitiesEdit

The straight line locus ax + by + c = 0 divides the rest of the number plane into two regions, satisfying the inequalities ax + by + c < 0 and ax + by + c > 0 respectively. Each region is a half-plane.

Two intersecting straight lines divide the rest of the plane into four regions, each defined by a pair of linear inequalities.

1. a1x + b1y + c1 > 0 and a2x + b2y + c2 > 0 2. a1x + b1y + c1 > 0 and a2x + b2y + c2 < 0 3. a1x + b1y + c1 < 0 and a2x + b2y + c2 > 0 4. a1x + b1y + c1 < 0 and a2x + b2y + c2 < 0

Each such region is the intersection of two half-planes. Extension to the case of three lines intersecting in pairs, thus including the description of the interior of a triangle as the intersection of three half-planes or the common solutions of three linear inequalities in x and y.

Distance between two points and the (perpendicular) distance of a point from a lineEdit

The formula for the distance between two points should be derived, as should the formula for the perpendicular distance of a point (x1, y1) from a line ax + by + c = 0.

For example, the following proof may be used.

The equation of the line through P, perpendicular to the given line, is

bx - ay = bx_1 - ay_1 \;

This meets ax + by = – c at (x0, y0), where

(a^2 + b^2)x_0 = b^2x_1 - aby_1 - ac \;

(a^2 + b^2)y_0 = -abx_1 + a^2y_1 - bc \;

thus

x_1 - x_0 = \frac{(a^2 + b^2)x_1 - (b^2x_1 - abx_1 - ac)}{a^2 + b^2}

x_1 - x_0 = \frac{a(ax_1 + by_1 + c)}{a^2 + b^2}

and

y_1-y_0 = \frac{(a^2 + b^2)y_1 - (a^2y_1 - abx_1 - bc)}{a^2 + b^2}

y_1 - y_0 = \frac{b(ax_1 + by_1 + c)}{a^2 + b^2}

hence

(x_1 - x_0)^2 + (y_1 - y_0)^2 = \frac{a^2(ax_1 + by_1 + c)^2}{(a^2 + b^2)^2} + \frac{b^2(ax_1 + by_1 + c)^2}{(a^2 + b^2)^2}

(x_1 - x_0)^2 + (y_1 - y_0)^2 = \frac{(a^2 + b^2)(ax_1 + by_1 + c)^2}{(a^2 + b^2)^2}

(x_1 - x_0)^2 + (y_1 - y_0)^2 = \frac{(ax_1 + by_1 + c)^2}{a^2 + b^2}

taking the positive square root of the numerator gives ±(ax1 + bx1 +c), but distance can only be positive, so we use the absolute value. The denominator cannot be simplified and so is left with a positive square root sign.

\mbox{distance} = \frac{|ax_1 + by_1 + c|}{ \sqrt{a^2 + b^2}}

The mid-point of an intervalEdit

A direct derivation of the coordinates of the midpoint of a given interval should be presented.

midpoint of interval with endpoints (x1,y1) and (x2,y2) is

\left ( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right )

Coordinate methods in geometryEdit

Examples, illustrating the use of coordinate methods in solving geometrical problems, are to be restricted to problems with specified data. The following are typical problems.

1. Show that the triangle whose vertices are (1, 1), (–1, 3) and (3, 5) is isosceles. 2. Show that the four points (0, 0), (2, 1), (3, –1), (1, –2) are the corners of a square. 3. Given that A, B, C are the points (–1, –2), (2, 5) and (4, 1) respectively, find D so that ABCD is a parallelogram. 4. Find the coordinates of the point A on the line x = –3 such that the line joining A to B (3, 5) is perpendicular to the line 2x + 5y = 12.

Last modified on 8 April 2012, at 00:16