Use a scalar projection to show that the distance from a point P₁(x₁, y₁) to the line ax + by + c = 0 is as follows. lax₁ + by₂ + cl √a² + b² First note that n = (a, b) is perpendicular to the line. If Q₁ = (a₁, b₁) and Q₂ = (a₂, b₂) lie on the line, then from the equation of the line we have a(a₂) + b(b₂) = a(a) + b(b₂) = c n-Q₁Q₂--Select- = 0. Let P₂= (x2, Y₂) be a point on the line. Then the distance from P₁ to the line comp n. (x₂-x₁ Y₂Y₁) In a(xi) - a(x₁) + b(y:) - b(y₁) √a² + b² the absolute value of the vector projection of P₁P₂ onto n. Since P₂ is on the line, ax₂ + by₂-c |a(x₂) + b(x₂) + c √² + b² ind the distance from the point (-4, 2) to the line 3x - 4y + 3 = 0. ✓. Therefore and we see that

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A:

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