# The distance of a point (3, 2) from a line 3x + 4y = 7 is

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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.