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Exercise 2(a) · Q1

Q.When the origin is shifted to the point (2,3)(2,3) by translation of axes, find the transformed equation of the straight line 2x+3y−5=02x + 3y - 5 = 0.

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Step 1. The origin is shifted to (h,k)=(2,3)(h,k)=(2,3), so the translation formulas are x=x′+h=x′+2x=x'+h=x'+2 and y=y′+k=y′+3y=y'+k=y'+3.

Step 2. Substitute into the given line 2x+3y−5=02x+3y-5=0:

2(x′+2)+3(y′+3)−5=02(x'+2)+3(y'+3)-5=0

Step 3. Expand each bracket:

2x′+4+3y′+9−5=02x'+4+3y'+9-5=0

Step 4. Collect the constant terms 4+9−5=84+9-5=8:

2x′+3y′+8=02x'+3y'+8=0

[!ANSWER] The transformed equation of the line, referred to the new origin (2,3)(2,3), is 2x′+3y′+8=02x'+3y'+8=0.

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