# find the reflection of the point A(2,1) over the line 12x-5y-41=0

1
by shaalladin

## Answers

2014-10-01T10:24:50+05:30

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given straight line : L =  a x + b y + c = 0   --- equation 1
a = 12,    b = -5     c = -41
slope = m1 = - a/b = 12/5

We want to find the image of a point P (x1, y1) w.r.t. the above line L. Let the image be Q(x2,y2).  Then the mid-point R of PQ lies on the line L.   Also, PQ is perpendicular to L.   x1 =2 ,  y1 = 1

So slope of PQ = m2 =  -1/m1  = -5/12
So  (y2 - 1) / (-5)  = (x2 - 2) / 12   = k (say)    -- equation 2
y 2 = -5k + 1        x2 = 12 k + 2

Mid point R of PQ = [ (x2+2)/2, (y2+1)/2 ],  = [ (6k+2), (-2.5 k +1) ]

R lies on L. so
12 (6 k + 2) - 5 (- 2.5k+1) - 41 = 0    --- equation 3
84.5 k = 22    or,                   k = 22/84.5

x2 = a k + x1 = 12 * 22/84.5 + 2 = 5.124
y2 = b k + y1 = -5 * 22/84.5 + 1 = - 0.301

Image = (5.124,  - 0.301)

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Alternately, we can rotate the coordinate axes and translate them. So that finding the image is easier.

Let the new Y axis is the given line itself.  X is 0 along new Y axis. So
X = 12 x - 5 y - 41        -- y axis.
the x - axis,    Y =  5 x + 12 y + p       , perpendicular to y axis.

Let new x-axis pass through P(x,y) = (2,1). Its new Y coordinate is 0.
Y = 5 * 2 + 12 * 1 + p = 0      =>  p = -22
New X coordinate of P (2,1) =  X = 12*2-5*1-41 = -22

Reflection of P(X,Y) = (-22,0) in new coordinate system is (22,0) wrt new Y axis or L itself.

The coordinates in the old system are :
X = 22 = 12 x - 5 y - 41       => 12 x - 5y = 63           ---   equation 4
Y = 0 = 5 x + 12 y - 22         =>  5x + 12 y = 22      --  equation 5

Solving equations 4 and 5 , we get :
x = 866/169 =5.124
y = (5 * 5.124 - 22 ) / 5 = - 0.301