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Exercise 6.5 · Q16

Q.The image of the point (2,3)(2, 3) in the line y=−xy=-x is

(1) (−3,−2)(-3,-2)
(2) (−3,2)(-3,2)
(3) (−2,−3)(-2,-3)
(4) (3,2)(3,2)
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Reflection of (a,b)(a,b) in the line y=−xy=-x is (−b,−a)(-b,-a); apply it to (2,3)(2,3).

The image of a point in a line is found by ensuring the line is the perpendicular bisector of the segment joining the point and its image.

Step 1. Set up the image point. Let the image of (2,3)(2,3) in the line y=−xy=-x (i.e. x+y=0x+y=0) be (h,k)(h,k).

Step 2. Midpoint lies on the line. The midpoint of (2,3)(2,3) and (h,k)(h,k) is (2+h2,3+k2)\left(\dfrac{2+h}{2},\dfrac{3+k}{2}\right), and it must satisfy x+y=0x+y=0:

2+h2+3+k2=0  ⟹  (2+h)+(3+k)=0  ⟹  h+k=−5(i)\dfrac{2+h}{2}+\dfrac{3+k}{2}=0 \implies (2+h)+(3+k)=0 \implies h+k=-5 \quad (\text{i})

Step 3. The joining segment is perpendicular to the line. The line x+y=0x+y=0 has slope −1-1, so the segment from (2,3)(2,3) to (h,k)(h,k) (slope k−3h−2\dfrac{k-3}{h-2}) must have slope 11 (negative reciprocal of −1-1):

k−3h−2=1  ⟹  k−3=h−2  ⟹  k−h=1(ii)\dfrac{k-3}{h-2}=1 \implies k-3=h-2 \implies k-h=1 \quad (\text{ii}) …

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