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The distance of a point (3, 2) from a line 3x + 4y = 7 is 
1. \(\frac{2}{\sqrt5}\)units 
2. 2 units 
3. 5 units
4. √5  units
5. None of these

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Correct Answer - Option 2 : 2 units 

Concept:

Perpendicular Distance of a Point from a Line:

Let us consider a line Ax + By + C = 0 and a point whose coordinate is (x1, y1)

\(\rm d=|\frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}}|\)

 

Calculation:

Given: equation of line is 3x + 4y = 7 and a point is (3, 2)

 

We know the distance of a line from is given by, \(\rm d=|\frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}}|\)

So, distance of a point (3, 2) from a line 3x + 4y - 7 = 0 is given by,

\(\rm d = |\frac{3(3)+4(2)-7}{\sqrt{3^2+4^2}}|\)

\(\rm = |\frac{9+8-7}{\sqrt{9+16}}|\)

\(\rm = |\frac{10}{\sqrt{25}}|\)

= 10/5

= 2 units 

Hence, option (2) is correct.

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