Math Problem Statement

Consider the point P = (3, 2). In following cases, find the distance from P to the line l.l passes through A(0, 5) and B(8, −1).

Solution

To find the distance from the point P=(3,2)P = (3, 2) to the line passing through points A(0,5)A(0, 5) and B(8,1)B(8, -1), we can use the formula for the distance from a point to a line:

d=Ax1+By1+CA2+B2d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}

Where:

  • AA, BB, and CC are the coefficients of the line equation in standard form Ax+By+C=0Ax + By + C = 0.
  • (x1,y1)(x_1, y_1) are the coordinates of the point PP.

Step 1: Find the equation of the line through A(0,5)A(0, 5) and B(8,1)B(8, -1).

First, calculate the slope mm of the line through A(0,5)A(0, 5) and B(8,1)B(8, -1):

m=y2y1x2x1=1580=68=34m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 5}{8 - 0} = \frac{-6}{8} = -\frac{3}{4}

Now, using the point-slope form of a line equation:

yy1=m(xx1)y - y_1 = m(x - x_1)

Substitute m=34m = -\frac{3}{4} and (x1,y1)=(0,5)(x_1, y_1) = (0, 5):

y5=34(x0)y - 5 = -\frac{3}{4}(x - 0)

y5=34xy - 5 = -\frac{3}{4}x

Multiply through by 4 to eliminate the fraction:

4(y5)=3x4(y - 5) = -3x

4y20=3x4y - 20 = -3x

Rearrange into the standard form:

3x+4y20=03x + 4y - 20 = 0

So, the equation of the line is 3x+4y20=03x + 4y - 20 = 0.

Step 2: Use the distance formula.

Now, apply the distance formula with A=3A = 3, B=4B = 4, and C=20C = -20, and (x1,y1)=(3,2)(x_1, y_1) = (3, 2):

d=3(3)+4(2)2032+42d = \frac{|3(3) + 4(2) - 20|}{\sqrt{3^2 + 4^2}}

d=9+8209+16d = \frac{|9 + 8 - 20|}{\sqrt{9 + 16}}

d=9+8205=17205=35=35d = \frac{|9 + 8 - 20|}{5} = \frac{|17 - 20|}{5} = \frac{|-3|}{5} = \frac{3}{5}

Final Answer:

The distance from point P=(3,2)P = (3, 2) to the line passing through A(0,5)A(0, 5) and B(8,1)B(8, -1) is 35\frac{3}{5} units.


Would you like more details on any part of the solution, or do you have any questions?

Here are some follow-up questions related to the topic:

  1. How would you find the distance between two points instead of from a point to a line?
  2. How can the point-slope form of a line help in different cases of line equations?
  3. What changes in the distance formula if the line equation has different coefficients?
  4. How do you derive the general formula for the distance between a point and a line?
  5. Can the formula be applied in 3D geometry?

Tip: To find the equation of a line given two points, always start by finding the slope and then use the point-slope form.

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Math Problem Analysis

Mathematical Concepts

Analytic Geometry
Point-Line Distance
Line Equation

Formulas

Distance from a point to a line: d = |Ax_1 + By_1 + C| / sqrt(A^2 + B^2)
Slope of a line: m = (y2 - y1) / (x2 - x1)

Theorems

Distance Formula in a Plane

Suitable Grade Level

Grades 9-12