How to find the equation of a line with one point

Question

If we have a point,		, and a slope, m, here's the formula we
use to find the equation of a line:

It's called the point-slope formula
(Duh!)

You are going to use this a LOT!

Luckily, it's pretty easy -- let's just do one:

Let's find the equation of the line that passes through the point
( 4 , -3 ) with a slope of -2:

Just stick the stuff in a clean it up!

TRY IT:

Find the equation of the line that passes through the point ( -2 , 6 ) with a slope of 3.

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Delete this to disable. //listMenu.animations[listMenu.animations.length] = animFade; //listMenu.animations[listMenu.animations.length] = animClipDown; //listMenu.animSpeed = 20; var arrow = null; if (document.createElement && document.documentElement) { arrow = document.createElement('span'); arrow.appendChild(document.createTextNode('-->')); arrow.className = 'subind'; } addEvent(window, 'load', new Function('listMenu.activateMenu("listMenuRoot", arrow)')); addEvent(window, 'load', new Function('listMenu2.activateMenu("listMenuRoot2", arrow)')); addEvent(window, 'load', new Function('listMenu3.activateMenu("listMenuRoot3", arrow)')); <p> </p> </td><td width="10"></td> <td valign="top"> <h2 class="linkedH1"><a target="_blank" href="http://mathsfirst.massey.ac.nz/Algebra/StraightLinesin2D.htm">Linear Equations and Graphs</a></h2> <h2>Background</h2> <p>We assume that you are familiar with the <a target="_blank" href="http://mathsfirst.massey.ac.nz/Algebra/CoordSystems/Coordinates2D.htm">Cartesian coordinate system</a> and <a target="_blank" href="http://mathsfirst.massey.ac.nz/Algebra/SystemsofLinEq/SystemsofLinEq.htm">solving linear equations</a>.</p> <p>Functions which have straight line graphs are called <em><strong>linear functions</strong></em> (for the obvious reason). These functions can always be written in the form</p> <p align="center"><img src="http://mathsfirst.massey.ac.nz/Algebra/StraightLinesin2D/images/ymxb.gif" alt="y=m*x+b " width="141" height="50"></p> <p> (To see why <a target="_blank" href="http://mathsfirst.massey.ac.nz/Algebra/StraightLinesin2D/Points.htm#" onclick="window.open('equation.htm','exercise1','width=500,height=400,toolbar=yes,location=yes,status=yes,menubar=yes,scrollbars=yes,resizable=yes')">click here</a>.) The set of points given by the <a target="_blank" href="http://mathsfirst.massey.ac.nz/Algebra/CoordSystems/Coordinates2D.htm">ordered pairs</a> that satisfies the above equation is a straight line. </p> <p> </p> <h2>Finding Points on a Line</h2> <p> To find points on the line <em>y</em> = <em>mx</em> + <em>b</em>, </p> <ul> <li>choose <em>x</em> and solve the equation for <em>y</em>, or</li> <li>choose <em>y</em> and solve for <em>x</em>.</li> </ul> <h3>Example 1.</h3> <p>Find two points on the line <em>y</em> = 2<em>x </em>+ 1:</p> <p><strong>Point 1 - Choose <em>x</em> and solve for <em>y</em>:</strong></p> <p>Let <em>x</em> =1. Substitute <em>x</em> = 1 into <em>y</em> = 2<em>x </em>+ 1 and solve for <em>y</em>.</p> <p align="center"><font size="3">y = 2(1) + 1 = 2 + 1 = 3</font></p> <p>Hence (1,3) is one point on the line <em>y</em> = 2<em>x </em>+ 1. </p> <br> <p><strong>Point 2 - Choose <em>y</em> and solve for <em>x</em>:</strong></p> <p>Let <em>y</em> = -3. Substitute<em> y</em> = -3 into <em>y</em> = 2<em>x </em>+ 1 and solve for <em>x</em>.</p> <table width="65%" border="0" align="center" cellpadding="0" cellspacing="0"> <tr> <td width="7%"><div align="right"><font size="3">-3</font></div></td> <td width="6%"><div align="center"><font size="3"> = </font></div></td> <td width="22%"><font size="3">2x + 1</font></td> <td width="65%"><font size="3">(Subtract 1 from both sides)</font></td> </tr> <tr> <td><div align="right"><font size="3">-4</font></div></td> <td><div align="center"><font size="3">=</font></div></td> <td><font size="3">2x</font></td> <td><font size="3">(Divide both sides by 2)</font></td> </tr> <tr> <td><div align="right"><font size="3">-2</font></div></td> <td><div align="center"><font size="3">=</font></div></td> <td><font size="3">x</font></td> <td><font size="3"> </font></td> </tr> </table> <p>Hence (-2,-3) is another point on the line <em>y</em> =2<em>x</em>+1.</p> <p> </p> <h3>Exercise 1.</h3> <p>Follow the above example and try some of the following exercises:</p> <form> <table align="CENTER" cellpadding="5" cellspacing="0" bgcolor="#003366"> <tr> <td align="CENTER"><div align="left"> <p align="center"><font face="Arial" color="#FFFFFF"><b> <input name="button23" type="button" onclick="setup1(this.form);" value="Get New Exercise"> </b></font></p> <p><font color="#FFFFFF" face="Arial, Helvetica, sans-serif"><strong>Locate points on the line defined by</strong></font><font face="Arial" color="#FFFFFF"><b><em> y</em> = <input type="text" size="3" name="a"> <em>x</em> + <input type="text" size="3" name="b"> </b></font><br> </p> </div></td> </tr> <tr> <td align="CENTER" bgcolor="#669900"><p> </p> <table width="218" border="1" align="left" cellpadding="0"> <tr> <td width="88"><div align="center"><font face="Arial" color="#FFFFFF"><b><em>x</em></b></font></div></td> <td width="118"><div align="center"><font face="Arial" color="#FFFFFF"><b><em>y</em></b></font></div></td> </tr> <tr> <td><div align="center"><font face="Arial" color="#FFFFFF"><b> <input type="text" size="3" name="x1"> </b></font></div></td> <td><div align="center"><font face="Arial" color="#FFFFFF"><b> <input type="text" size="3" name="y1"> </b></font></div></td> </tr> <tr> <td><div align="center"><font face="Arial" color="#FFFFFF"><b> <input type="text" size="3" name="x2"> </b></font></div></td> <td><div align="center"><font face="Arial" color="#FFFFFF"><b> <input type="text" size="3" name="y2"> </b></font></div></td> </tr> </table> <table width="92" height="84" border="0" align="left" cellpadding="0"> <tr> <td width="88"><div align="center"><font face="Arial" color="#FFFFFF"><b></b></font></div></td> </tr> <tr> <td><div align="center"><font face="Arial" color="#FFFFFF"><b> <input name="button2" type="button" onclick="checky(this.form);" value="Check"> </b></font></div></td> </tr> <tr> <td><div align="center"><font face="Arial" color="#FFFFFF"><b> <input name="button3" type="button" onclick="checkx(this.form);" value="Check"> </b></font></div></td> </tr> </table> <div align="right"> </div> <p>  </p> <p> </p> <p> </p></td> </tr> </table> </form> <p align="Center"><a target="_blank" href="http://mathsfirst.massey.ac.nz/Algebra/StraightLinesin2D.htm">Linear Equations and Graphs Index</a> | <a target="_blank" href="http://mathsfirst.massey.ac.nz/Algebra/StraightLinesin2D/GrLinFun.htm">The Graph of a Linear Function (Drawing a Graph Using 2 Points)</a> >></p> <p> </p> </td> </tr> </table> <table width="100%" border="0" cellpadding="0" cellspacing="2"> <tr bgcolor="#003366"> <td height="30">    <span class="nav"><a target="_blank" href="http://aboutmassey.massey.ac.nz/massey/about-massey/contacts.cfm">Contact Us</a></span> <span class="nav"> | <a target="_blank" href="http://aboutmassey.massey.ac.nz/">About Massey University</a> | <a target="_blank" href="http://sitemap.massey.ac.nz/">Sitemap</a> | <a target="_blank" href="http://www.massey.ac.nz/disclaim.htm">Disclaimer</a> | Last updated: November 21, 2012     © Massey University 2003</span> </td> </tr> </table> </body> </html>

How do you find the slope of a line with one point?

To derive the point slope formula, we consider a line of slope 'm' with a point (x1,y1) ( x 1 , y 1 ) . If (x, y) is any general point on the line, then the slope of the line is, m = (y - y1 1 )/(x - x1 1 . From this, we can get the point slope formula y − y1 1 = m (x − x1 1 ).

find equation line point