Expressing a straight line in algebraic form expression when the straight line passes through the origin with some slope is defined equation of a straight line, passes through the origin with slope.

In very special cases, straight lines are intersected the origin of the Cartesian coordinate system with some slope and these types of straight lines can be expressed in terms of an algebraic expression. It is very easier to represent it in mathematics by a single expression which reveals how exactly the straight line is in the geometric system. Actually, the straight line is expressed in terms of the slope and any one of the points on the straight line.

Assume,

$\overleftrightarrow{AB}$is a straight line on a plane and also assume it intersects the origin of the Cartesian coordinate geometric system with some slope.

The straight line

$\overleftrightarrow{AB}$intersects the origin at a point

$A$which is origin as well as one of the points of the straight line

$\overleftrightarrow{AB}$. Therefore, the coordinates of the point

$A$is

$(0,0)$geometrically. Assume, the point

$B$is located at

$(x,y)$in the Cartesian coordinate geometric system. The assumed angle made by the straight line

$\overleftrightarrow{AB}$with the horizontal x-axis is theta

$(\theta )$.

Draw a perpendicular line from point

$B$towards the x-axis and it intersects the horizontal x-axis at a point which is assumed to call point

$C$. Thus, it forms a right angled triangle

$\Delta BAC$. The angle made the straight line

$\overleftrightarrow{AB}$with horizontal x-axis is theta

$(\theta )$but it is also an angle of the right angled triangle

$\Delta BAC$which means

$\angle BAC=\theta $.

As per this right angled triangle,

- The line segment

$\stackrel{\u203e}{AB}$becomes the hypotenuse of the right angled triangle

$\Delta BAC$.

- The line segment

$\stackrel{\u203e}{AC}$becomes the adjacent side of the right angled triangle

$\Delta BAC$.

- The line segment

$\stackrel{\u203e}{BC}$becomes the opposite side of the right angled triangle

$\Delta BAC$.

As per the principle definition of slope of the straight line, the slope of the straight line is mathematically expressed as

$m=tan\theta $.

But, as per the right angled triangle

$\Delta BAC,tan\theta =\frac{BC}{AC}$As per the right angled triangle

$\Delta BAC$,

The length of the opposite side

$\left(\stackrel{\u203e}{BC}\right)$of the right angled triangle

$\Delta BAC$is

$BC=y$The length of the adjacent side

$\left(\stackrel{\u203e}{AC}\right)$of the right angled triangle

$\Delta BAC$is

$AC=x$Therefore,

$tan\theta =\frac{BC}{AC}=\frac{y}{x}$$\Rightarrow tan\theta =\frac{y}{x}$

Replace the value of

$tan\theta $in slope form.

$\Rightarrow m=\frac{y}{x}$

$\Rightarrow mx=y$

$\Rightarrow y=mx$

This is an algebraic expression which represents a straight line when the straight line passes through the origin of the Cartesian coordinate geometric system.

List of most recently solved mathematics problems.

Jul 04, 2018

Limit (Calculus)

Evaluate $\displaystyle \large \lim_{x \,\to\, \tan^{-1}{3}} \normalsize {\dfrac{\tan^2{x}-2\tan{x}-3}{\tan^2{x}-4\tan{x}+3}}$

Jun 23, 2018

Limit (Calculus)

Evaluate $\displaystyle \large \lim_{x \to 0} \normalsize \dfrac{e^{x^2}-\cos{x}}{x^2}$

Jun 22, 2018

Integral Calculus

Evaluate $\displaystyle \int \dfrac{1+\cos{4x}}{\cot{x}-\tan{x}} dx$

Jun 21, 2018

Limit

Evaluate $\displaystyle \large \lim_{x \to \infty} \normalsize {\sqrt{x^2+x+1}-\sqrt{x^2+1}}$

Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers.
Know more

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.