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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)**

=========================================

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**

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

=========================================

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 :