Question

practice: find the perpendicular distance from the given fixed point to the line in each of the following:

1. s(2,3) line: 4x + 3y − 6 = 0
2. d(−2,−3) line: 4x + 3y + 6 = 0

line: 5x − y − 3 = 0

Explanation:

Problem 1: Perpendicular Distance from \( S(2, 3) \) to Line \( 4x + 3y - 6 = 0 \)

The formula for the perpendicular distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[
d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}
\]

Step 1: Identify \( A, B, C, x_0, y_0 \)

For the line \( 4x + 3y - 6 = 0 \), we have \( A = 4 \), \( B = 3 \), \( C = -6 \). The point \( S(2, 3) \) gives \( x_0 = 2 \), \( y_0 = 3 \).

Step 2: Substitute into the formula

Substitute the values into the distance formula:
\[
d = \frac{|4(2) + 3(3) - 6|}{\sqrt{4^2 + 3^2}}
\]

Step 3: Calculate the numerator and denominator

First, calculate the numerator:
\[

4(2) + 3(3) - 6

=

8 + 9 - 6

=

11

= 11

\]
Next, calculate the denominator:
\[
\sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5
\]

Step 4: Compute the distance

Divide the numerator by the denominator:
\[
d = \frac{11}{5} = 2.2
\]

Problem 2: Perpendicular Distance from \( D(-2, -3) \) to Line \( 4x + 3y + 6 = 0 \)

Using the same distance formula \( d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \)

Step 1: Identify \( A, B, C, x_0, y_0 \)

For the line \( 4x + 3y + 6 = 0 \), \( A = 4 \), \( B = 3 \), \( C = 6 \). The point \( D(-2, -3) \) gives \( x_0 = -2 \), \( y_0 = -3 \).

Step 2: Substitute into the formula

\[
d = \frac{|4(-2) + 3(-3) + 6|}{\sqrt{4^2 + 3^2}}
\]

Step 3: Calculate the numerator and denominator

Numerator:
\[

4(-2) + 3(-3) + 6

=

-8 - 9 + 6

=

-11

= 11

\]
Denominator (same as before):
\[
\sqrt{4^2 + 3^2} = 5
\]

Step 4: Compute the distance

\[
d = \frac{11}{5} = 2.2
\]

Problem 3: Perpendicular Distance from \( (-1) \)? (Assuming a typo, let's assume the point is \( (x_0, y_0) \) and line \( 5x - y - 3 = 0 \))

Let's assume the point is \( (x_0, y_0) \), say \( (x_0, y_0) = (a, b) \) (since the point is not fully visible, but using the formula):

Using \( d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \) for line \( 5x - y - 3 = 0 \) (so \( A = 5 \), \( B = -1 \), \( C = -3 \))

Step 1: Substitute \( A, B, C, x_0, y_0 \)

\[
d = \frac{|5x_0 - y_0 - 3|}{\sqrt{5^2 + (-1)^2}} = \frac{|5x_0 - y_0 - 3|}{\sqrt{26}}
\]

If we assume the point is, for example, \( (1, 2) \) (common test point), then:

Step 2: Substitute \( x_0 = 1 \), \( y_0 = 2 \)

\[
d = \frac{|5(1) - 2 - 3|}{\sqrt{26}} = \frac{|0|}{\sqrt{26}} = 0
\]

But since the point is not clear, we'll leave it in terms of \( x_0, y_0 \):

\[
d = \frac{|5x_0 - y_0 - 3|}{\sqrt{26}}
\]

Answer:

s:

1. Perpendicular distance from \( S(2, 3) \) to \( 4x + 3y - 6 = 0 \) is \( \boldsymbol{\frac{11}{5}} \) (or \( 2.2 \)).
2. Perpendicular distance from \( D(-2, -3) \) to \( 4x + 3y + 6 = 0 \) is \( \boldsymbol{\frac{11}{5}} \) (or \( 2.2 \)).
3. For the line \( 5x - y - 3 = 0 \) and point \( (x_0, y_0) \), the distance is \( \boldsymbol{\frac{|5x_0 - y_0 - 3|}{\sqrt{26}}} \) (substitute the correct point to get a numerical value).