EQUATION OF A LINE IN DEFFERENT FORMS

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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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