</span><span>The equation of the line through the points </span><span>y-y1=(-1/3)((x-x1) y-3= (-1/3) (x-2) </span>y= -(1/3)x +(2/3)+3 = -(1/3)x+11/3 equation of the line AB <span>perpendicular line has slope 3 (negative reciprocal) </span><span>goes through midpoint which is half the distance between the two points ((2+8)/2,</span>(3+1)/2)=(5,2) Point C <span>Its equation is y-2=3(x-5) y=3x-13 </span>equation of the line CD calculation of point D point C (5,2) point D (x,y) distance CD=√10 distance CD^2=(x-5)^2+(y-2)^2=10 10=(x-5)^2+(3x-15)^2
<span>solving the quadratic equation </span>x1=4 y1=-1 x2=6 y2=5 y=3x-13 equation of the line CD