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Find the equation of the line joining (–1, 3) and (2, 5). Prove that if the point (x, y) is on this line, so is the point (x + 3, y + 2).

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\(\frac{y-3}{x+1}=\frac{5-3}{2+1}=\frac{2}{3}\)

\(\frac{y-3}{x+1}=\frac{2}{3}\)

y - 3 = \(\frac{2}{3}(x+1)\)

y = \(\frac{2}{3}(x+1)+3\) \(\Rightarrow\) 2x - 3y + 11 = 0

If (x + 3, y + 2) is in this line, then

2(x + 3) - 3(y + 2) + 11 = 2x - 3y + 11 = 0

∴ (x + 3, y + 2) lies on this line

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