Question

in a reflection the image of the line y-2x=3 is the line 2y-x=9.find the axis of reflection.

Answer 1

The axis of reflection is the angle bisector of the two lines.

One of the properties of the axis of reflection is that any point lying on the line is equi-distant from both the original line and the image.

Distance of a point from a line Ax+By+C=0 is
±(Ax+By+C)/sqrt(A²+B²)

Can you take it from here?

Answer 2

To find axis of reflection of this question is more complex but l will show how to find easily now let see it First find the intersection of the two line y-2x=3)€€€€at x=1 and y=5 ****(1,5) 2y-x=9) #at this point the axis of reflection passes at(1,5) *Second take one point from y-2x=3 let (0,3) then find the equation of circle at (1,5) and get radius is the distance between (0,3) and (1,5)=square root of 5 #?Then find intersection b/n y-2x=3 and circle (x-1)²+(y-5)²=5......(x-1)²+((2x+3)-5)²=5 =(x-1)²+(2x-2)²=(x-1)²+4(x-1)²..==5(x-1)²=5 (X-1)²=1....x²-2x+1=1******x(x-2)=0 X=0 and x=2........*then (0,3) and (2,7) point on line y=2x+1. ?Then another find intersection between circle and the line 2y-x=9 ✍y=(x+9)/2........._(x-1)²+((?(x+9)/2)-5)²=5 then .... X= -1 and x =3 then (-1,4) and (3,6) on reflected line 2y=x+9 Now we take our found point that ( smaller x value )on two line that our point (0,3) and (-1,4) M(0,3)=(-1,4)*********find midpoint That is (-1/2,7/2) ......then finally we found one point of axis of reflection pass at mid point then use straight line equation between (1,5) and (-1/2,7/2) slope is =1 then y-5=1(x-1)........y=x-1+5=x+4 Answer is y=x+4...................????

Answer 3

To find axis of reflection of this question is more complex but l will show how to find easily now let see it First find the intersection of the two line y-2x=3)€€€€at x=1 and y=5 ****(1,5) 2y-x=9) #at this point the axis of reflection passes at(1,5) *Second take one point from y-2x=3 let (0,3) then find the equation of circle at (1,5) and get radius is the distance between (0,3) and (1,5)=square root of 5 #?Then find intersection b/n y-2x=3 and circle (x-1)²+(y-5)²=5......(x-1)²+((2x+3)-5)²=5 =(x-1)²+(2x-2)²=(x-1)²+4(x-1)²..==5(x-1)²=5 (X-1)²=1....x²-2x+1=1******x(x-2)=0 X=0 and x=2........*then (0,3) and (2,7) point on line y=2x+1. ?Then another find intersection between circle and the line 2y-x=9 ✍y=(x+9)/2........._(x-1)²+((?(x+9)/2)-5)²=5 then .... X= -1 and x =3 then (-1,4) and (3,6) on reflected line 2y=x+9 Now we take our found point that ( smaller x value )on two line that our point (0,3) and (-1,4) M(0,3)=(-1,4)*********find midpoint That is (-1/2,7/2) ......then finally we found one point of axis of reflection pass at mid point then use straight line equation between (1,5) and (-1/2,7/2) slope is =1 then y-5=1(x-1)........y=x-1+5=x+4 Answer is y=x+4...................????

Answer 4

Today l will show you very simple techniques than previously l posted let me you solution First find intersection b/n the line and the image reflected line that is (1,5) ?find equation of angle bisector to find axis....then form (y-2x-3)/root of 5=+or-(2y-x-9)/root of 5 For +ve ...y-2x-3=2y-x-9**y=-x+6 ??? For -ve .. y-2x-3=-2y+x+9 ..y=x+4 ???????????