document.write( "Question 929360: Can someone help me please?? thank you.\r
\n" ); document.write( "\n" ); document.write( "Consider the points A(-2, 5), B(2, -3), C(8, 0), and P(4, 3). P is on the bisector of \n" ); document.write( "\n" ); document.write( "thanks !!
\n" ); document.write( "

Algebra.Com's Answer #564267 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
A(-2, 5), B(2, -3), C(8, 0), and P(4, 3).
\n" ); document.write( "
\n" ); document.write( "Find the slope of AB and BC.
\n" ); document.write( "AB:
\n" ); document.write( "\"m=%28-3-5%29%2F%282-%28-2%29%29=-8%2F4=-2\"
\n" ); document.write( "BC:
\n" ); document.write( "\"m=%280-%28-3%29%29%2F%288-2%29=3%2F6=1%2F2\"
\n" ); document.write( "AB and BC are perpendicular since their slopes are negative reciprocals.
\n" ); document.write( "\"%28-2%29%281%2F2%29=-1\"
\n" ); document.write( "The angle between the two lines is then 90 so a bisector would split the angle into 2 45 degree segments.
\n" ); document.write( "The line BC makes an angle with the x-axis determined by the slope.
\n" ); document.write( "\"tan%28alpha%29=1%2F2\"
\n" ); document.write( "\"alpha=26.6\" \"degrees\"
\n" ); document.write( "Adding \"45\" to this and then taking the tangent to find the slope,
\n" ); document.write( "\"m=tan%28alpha%2B45%29=tan%2871.6%29=3\"
\n" ); document.write( "Using the point slope form of a line with point B,
\n" ); document.write( "\"y-%28-3%29=3%28x-2%29\"
\n" ); document.write( "\"y%2B3=3x-6\"
\n" ); document.write( "\"highlight%28y=3x-9%29\"
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( ".\r
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "You could have also used point P and found the slope of BP.
\n" ); document.write( "\"m=%283-%28-3%29%29%2F%284-2%29=6%2F2=3\"
\n" ); document.write( "Continuing the same way as shown previously.
\n" ); document.write( "
\n" );