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Question

Two parallel lines l1 & l2 such that the distance between them is 65 units and origin lies in between the lines. Equation of l1:x2y3=0. Another line l3:x+3y+2=0 intersect with l1 at A. Image of l3 in l1 is l4, meets with l2 at B. Image of l4 in the line l2 meets with the line l1 again at C and another point D(h,k) is in the xy plane

A
Equation of the circle passing through A,B,C is (5x17)2+(5y1)2=180
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B
Equation of the circle passing through A,B,C is (x10)2+(y2)2=90
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C
ABCD will be a square if (5h,5k)(23,11)
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D
Equation of the line l for which l4 is angle bisector of l2 & l is 2x+y7=0
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Solution

The correct options are
A Equation of the circle passing through A,B,C is (5x17)2+(5y1)2=180
C ABCD will be a square if (5h,5k)(23,11)
D Equation of the line l for which l4 is angle bisector of l2 & l is 2x+y7=0
The coordinates of the A(1,1)

Angle between l1 & l3 is
tanθ=∣ ∣12+1311213∣ ∣=1θ=π4
l3l4 and passes through A
equation of the line l4:3xy4=0
Distance between l1 & l2 is 6 units and origin lies in between them
equation of l2:x2y+3=0
Equating l4 & l2
B(115,135)
l1 & l2 are parallel to each other so angle between l4 & l2 is π4
image of the line l4 in l2 is perpendicular to l4 and parallel to l3

ΔABC is isosceles rightangle triangle and right angle at B.
Equation of the line l4 and passing through B is x+3y10=0
coordinates of C(295,75)
mid point of AC is (175,15)
lenght of AC=1210
Equation of the circle containing points A,B,C is
(x175)2+(y15)2=365(5x17)2+(5y1)2=180
Equation of l
l passes through B and l2
2x+y7=0
For ABCD to be square the coordinate of the D should be intersection of the l & l2
D(235,115)

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