EQUATION OF A LINE IN DEFFERENT FORMS

1. Slope-intercept form:

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Consider AB a straight line making an angle\displaystyle \theta with x-axis and cutting off an intercept OD=c from OY.

As the line makes intercept OD=c on the y-axis, it is called as y-intercept.

Consider the line AB intersect OX’ at T.

Take a point P (x, y) on the line AB. Draw\displaystyle PM\bot OX.

Then, \displaystyle OM=x,MP=y, draw a line DN perpendicular to MP as shown in the figure.

Now, from the right angled triangle DNP, the following result is obtained;

\displaystyle \begin{array}{l}\tan \theta =\frac{NP}{DN}\\=\frac{MP-MN}{OM}\\=\frac{y-OD}{OM}\\\tan \,\theta =\frac{y-c}{x}\end{array}

Therefore,

\displaystyle y=x\tan \theta +c

Where, \displaystyle \tan \theta =m

\displaystyle y=mx+c

Therefore, the equation of a straight line in slope intercept form is \displaystyle y=mx+c.

Example:

Find the equation of a line with slope 4 and y-intercept zero.

Take the general form of slope-intercept form

\displaystyle y=mx+c

Known that, m = 4 and c = 0, substitute the value as follows:

\displaystyle \begin{array}{l}y=\left( 4 \right)x+0\\y=4x\end{array}

Therefore, the equation is \displaystyle y=4x.

2. Point- slope form

Consider the figure as shown below:

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The equation of a line passing through the point \displaystyle A\left( {{x}_{1}},{{y}_{1}} \right) and slope m is determined as follows:

Consider \displaystyle P(x,y) be the any point on the line as shown in the figure. The slope image of the line joining \displaystyle A\left( {{x}_{1}},{{y}_{1}} \right)and\displaystyle P(x,y)is given by;

\displaystyle m=\tan \theta =\frac{y-{{y}_{1}}}{x-{{x}_{1}}}

The slope of the line AP is m, therefore,

\displaystyle m=\frac{y-{{y}_{1}}}{x-{{x}_{1}}}

Hence, the equation in the point-slope form is \displaystyle y-{{y}_{1}}=m\left( x-{{x}_{1}} \right).

Example

Determine the equation of the line passing through the point (2, 1) and having the slope \displaystyle m=\frac{2}{3}.

The general equation for the point-slope form is;

\displaystyle y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)

\displaystyle \left( {{x}_{1}},{{y}_{1}} \right)=\left( 2,1 \right);m=\frac{2}{3}, substitute the value as follows:

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3. Two point form

Consider \displaystyle A\left( {{x}_{1}},{{y}_{1}} \right) and \displaystyle B\left( {{x}_{2}},{{y}_{2}} \right) be two distinct points

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Slope of the line passing through these points written as follows:

\displaystyle m=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}

From the equation of line in the point-slope form, the following result is obtained:

\displaystyle y-{{y}_{1}}=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right)

Therefore, the equation for the two point form is \displaystyle y-{{y}_{1}}=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right).

Example

Find the equation of the line passing through \displaystyle \left( 3,-7 \right)\text{ and }\left( -2,-5 \right).

The general form for the two points is:

\displaystyle y-{{y}_{1}}=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right)

\displaystyle \left( {{x}_{1}},{{y}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}} \right)=\left( 3,-7 \right)\text{ and }\left( -2,-5 \right), substitute the value as follows:

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Therefore, the required equation is \displaystyle 2x+5y+29=0.

Question 1:

Find the equation of a line with slope 6 and y-intercept 2.

Solution 1:

Take the general form of slope-intercept form

\displaystyle y=mx+c

Known that, m=6 and c=2, substitute the value as follows:

\displaystyle \begin{array}{l}y=\left( 6 \right)x+2\\y=6x+2\end{array}

Therefore, the equation is \displaystyle y=6x+2.

Question 2:

Determine the equation of the line passing through the point (4, 5) and having the slope \displaystyle m=\frac{5}{2}.

Solution 2:

The general equation for the point-slope form is;

\displaystyle y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)

\displaystyle \left( {{x}_{1}},{{y}_{1}} \right)=\left( 4,5 \right);m=\frac{5}{2}, substitute the value as follows:

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Question 3:

Find the equation of the line passing through \displaystyle \left( 2,1 \right)\text{ and }\left( -3,4 \right).

Solution 3:

The general form for the two points is:

\displaystyle y-{{y}_{1}}=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right)

\displaystyle \left( {{x}_{1}},{{y}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}} \right)=\left( 2,1 \right)\text{ and }\left( -3,4 \right), substitute the value as follows:

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Post Author: E-Maths