calculate from the coordinates of a point the equation of a line passing through the origin

Online math exercise > Middle School > 9th Grade > linear functions and affine functions, equations of lines and linear systems, equations

Determine the equation of the linear function whose graphical representation passes through the point A of coordinates (`1`,`9/8`).

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This type of exercise can be solved using the function : equation_straight_line

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An equation is an equality that involves one or more variables, solving an equation in a set is to find the value or values of the variables in that set that verify the equation, these are the solutions of the equation. The variables are often referred to as the unknown, when the equation has only one unknown, it is often referred to as x.

For example, 3x-3=0 is an equation, solving for x in ℝ this equation is to find the solutions in ℝ of this equation.

When we have to solve several equations, with several variables, we speak of a system of equations.

There are methods and formulas for solving certain types of equations such as first degree equations (linear equation), second degree equations (quadratic equation), or product equations.

Solve an equation with one unknown of the first degree

Solving an equation with one unknown x in R, it is to determine the set of real numbers x verifying the said equation. This set is called the set of solutions of the equation.

When we add or subtract the same real to both members of an equation, we obtain a new equation which has the same solutions as the previous one.
When we multiply or divide the two members of an equation by a non-zero real, we obtain a new equation that has the same solutions as the previous one.

Using these rules, it is easy to show that equations of the form ax+b=0, if a is non-zero, admit a unique solution which is `x=-b/a`.

Solving a product equation

A product of two factors is zero if and only if one of the factors is zero.

Saying that a.b = 0 is equivalent to saying that a is zero or that b is zero.
Think of using the special expansions to get back to a product of factors and a classic case of solving an equation.

Solve a second degree equation using the discriminant

We call the discriminant of the trinomial `a*x^2+b*x+c`, with a not zero, the real `Delta=b^2-4*a*c`.

When `Delta<0` the equation has no root
When `Delta=0` the equation has a root `-b/2a`
When `Delta>0` the equation has two distinct roots `(-b-sqrt(Delta))/(2a)` and `(-b+sqrt(Delta))/(2a)`

Course reminder : Equations