document.write( "Question 2690: Question: F'ind the equation of the line that passes through the point (6,2) and is perpendicular to the line y=(-1/3)x + 6
\n" );
document.write( "Write the equation of the line in standard form with integral coefficients
\n" );
document.write( "I know ax + by = c
\n" );
document.write( " \n" );
document.write( "
Algebra.Com's Answer #1246 by xcentaur(357)![]() ![]() ![]() You can put this solution on YOUR website! Given: \n" ); document.write( "P=(6,2) \n" ); document.write( "y=(-1/3)x+6 ..... [line 1] \n" ); document.write( "[Line 2](eqn unknown) is perpendicular to [Line 1] \n" ); document.write( " \n" ); document.write( "We know that if two lines are perpendicular,the product of their slopes is -1. \n" ); document.write( "ie. m1*m2=-1 \n" ); document.write( "Let slope of line 1 be m1. \n" ); document.write( "Then, \n" ); document.write( "-1/3*m2=-1 \n" ); document.write( "m2=(-1*3)/-1 \n" ); document.write( " =3 \n" ); document.write( " \n" ); document.write( "Then slope of required line is 3. \n" ); document.write( " \n" ); document.write( "We have one point P(6,2) on this line of slope 3. \n" ); document.write( "Then using point-slope form,\r \n" ); document.write( "\n" ); document.write( "y-y1=m(x-x1) \n" ); document.write( "y-2=3(x-6) \n" ); document.write( "y-2=3x-18 \n" ); document.write( "y=3x-18+2 \n" ); document.write( "y=3x-16 \n" ); document.write( " \n" ); document.write( "Equation of the line in standard form: y = 3x - 16 \n" ); document.write( "
|