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

Question

Question

Answer 1

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.