Answer :
To determine the distance from the point [tex]\(P (1, 2)\)[/tex] to the line given by the equation [tex]\(x + 2y = 3\)[/tex], we follow these steps:
1. Identify the components:
- The point coordinates are [tex]\( (x_1, y_1) = (1, 2) \)[/tex].
- The line equation can be written in the standard form [tex]\( ax + by + c = 0 \)[/tex]. From the given equation [tex]\( x + 2y = 3 \)[/tex], we can rearrange it as [tex]\( x + 2y - 3 = 0 \)[/tex]. Thus, the coefficients are:
[tex]\[ a = 1, \quad b = 2, \quad c = -3 \][/tex]
2. Apply the distance formula:
The formula for the distance [tex]\(d\)[/tex] from a point [tex]\( (x_1, y_1) \)[/tex] to a line [tex]\( ax + by + c = 0 \)[/tex] is:
[tex]\[ d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \][/tex]
3. Substitute values into the formula:
[tex]\[ a = 1, \quad b = 2, \quad c = -3, \quad x_1 = 1, \quad y_1 = 2 \][/tex]
4. Calculate the numerator:
[tex]\[ ax_1 + by_1 + c = (1)(1) + (2)(2) + (-3) = 1 + 4 - 3 = 2 \][/tex]
The absolute value of the numerator is:
[tex]\[ |2| = 2 \][/tex]
5. Calculate the denominator:
[tex]\[ \sqrt{a^2 + b^2} = \sqrt{(1)^2 + (2)^2} = \sqrt{1 + 4} = \sqrt{5} \approx 2.23606797749979 \][/tex]
6. Compute the distance:
[tex]\[ d = \frac{2}{2.23606797749979} \approx 0.8944271909999159 \][/tex]
Thus, the distance from point [tex]\(P (1, 2)\)[/tex] to the line [tex]\( x + 2y = 3 \)[/tex] is approximately [tex]\( 0.8944271909999159 \)[/tex].
1. Identify the components:
- The point coordinates are [tex]\( (x_1, y_1) = (1, 2) \)[/tex].
- The line equation can be written in the standard form [tex]\( ax + by + c = 0 \)[/tex]. From the given equation [tex]\( x + 2y = 3 \)[/tex], we can rearrange it as [tex]\( x + 2y - 3 = 0 \)[/tex]. Thus, the coefficients are:
[tex]\[ a = 1, \quad b = 2, \quad c = -3 \][/tex]
2. Apply the distance formula:
The formula for the distance [tex]\(d\)[/tex] from a point [tex]\( (x_1, y_1) \)[/tex] to a line [tex]\( ax + by + c = 0 \)[/tex] is:
[tex]\[ d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \][/tex]
3. Substitute values into the formula:
[tex]\[ a = 1, \quad b = 2, \quad c = -3, \quad x_1 = 1, \quad y_1 = 2 \][/tex]
4. Calculate the numerator:
[tex]\[ ax_1 + by_1 + c = (1)(1) + (2)(2) + (-3) = 1 + 4 - 3 = 2 \][/tex]
The absolute value of the numerator is:
[tex]\[ |2| = 2 \][/tex]
5. Calculate the denominator:
[tex]\[ \sqrt{a^2 + b^2} = \sqrt{(1)^2 + (2)^2} = \sqrt{1 + 4} = \sqrt{5} \approx 2.23606797749979 \][/tex]
6. Compute the distance:
[tex]\[ d = \frac{2}{2.23606797749979} \approx 0.8944271909999159 \][/tex]
Thus, the distance from point [tex]\(P (1, 2)\)[/tex] to the line [tex]\( x + 2y = 3 \)[/tex] is approximately [tex]\( 0.8944271909999159 \)[/tex].