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

YES

Answer 3

Y=x+4

Answer 4

First find the intersection point of these lines and then also take one point from the line y = 2x 3 and calculate the distance from the intersection point to this point on the line. So this distance is equal to the distance fron the intersection point to the point (x,y) on the line 2y - x = 9. Then after by equiting them you can find (x,y). And so find the mid point between these two points. Finaly by using this mid point and the intersection point of these lines We can get the axis of reflection.

Answer 5

Y=0.988X+4.012 APPROXIMATLY Y=X+4

Answer 6

Explain This Answer?

Answer 7

Y=5X/6 + 25/6

Answer 8

apporoximatly y=x+4

Answer 9

as the line y=2x+3reflected tothe line 2y=x+9 the intersecton npoint will be (1,5) and the distance from the line y=2x+3 to the intersection point is square root5 by taking point (0,3) from the line y=3=2x+3 so thedistance from the line 2y=x+9is aiso=(x-1)square+(y-5)square so by equating the two we get the point (x,y)=( )then find the mid point between them and thn by these two points i.e point (1,5) and point (x,y) we get the equation of reflection line

Answer 10

This is interesting I'm in 7th grade math so they say and I have the exact same question as this one, now why am I getting 11 grade math in 7th grade.

Answer 11

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 12

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 slope....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 👈💯💯💯💯💯💯💯💯💯💯

Answer 13

Study 😒

Answer 14

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

Answer 15

To find the axis of reflection, we need to find the line that bisects the angle between the original line and its image.

First, let's find the intersection point between these two lines. We can solve the system of equations:

y - 2x = 3
2y - x = 9

By substitution or elimination, we get (1, 5) as the intersection point.

Next, let's find the slope of the original line:

y - 2x = 3
y = 2x + 3

This line has a slope of 2.

Now, let's find the slope of the image line:

2y - x = 9
y = (1/2)x + 9/2

This line has a slope of 1/2.

To find the slope of the angle bisector, we use the formula:

m = (m1 + m2)/2

where m1 and m2 are the slopes of the two lines. Plugging in, we get:

m = (2 + 1/2)/2
m = 5/4

Now we have the slope of the angle bisector, and we also have a point that it passes through (the intersection point (1, 5)). So we can use point-slope form to find the equation of the axis of reflection:

y - 5 = (5/4)(x - 1)

Simplifying, we get:

y = (5/4)x + 15/4

Therefore, the axis of reflection is the line y = (5/4)x + 15/4.