1.The line -6x+8y-k=0 cuts the positive x-axis at A and the
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1.The line -6x+8y-k=0 cuts the positive x-axis at A and the positive y-axis at B.If AB is 16.5 units,find k.
2.Find the equation of the line that is the perpendicular bisector of the segment with end-points (3,6) and(7,2) .
2.Find the equation of the line that is the perpendicular bisector of the segment with end-points (3,6) and(7,2) .
1.when x=0,y=k/8;when y=0,x=-k/6,so A is (-k/6,0),B is (0,k/8).(AB)^2=k^2/36+k^2/64,so AB=5|k|/24=16.5,then we can know that |k|=79.2,since the line -6x+8y-k=0 cuts the positive x-axis at A and the positive y-axis at B,we may get k0,so what the subjecet really is on earth only you know .
2.the slope of the segment with end-points(3,6)and (7,2)is-1,so our line is with slope 1.the middle point of the segment with end-points(3,6)and(7,2)is(5,4),so our line passes point(5,4),and with slope 1,we can easily get it:y-4=1(x-5),that is to say y=x+9.
2.the slope of the segment with end-points(3,6)and (7,2)is-1,so our line is with slope 1.the middle point of the segment with end-points(3,6)and(7,2)is(5,4),so our line passes point(5,4),and with slope 1,we can easily get it:y-4=1(x-5),that is to say y=x+9.
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