Question

The point on the line \( x + y = 4 \) that lie at a unit distance from the line \( 4x + 3y = 10 \) is

• A. \( (2, 2) \)
• B. \( (3, -1) \)
• C. \( (5, -1) \)
• D. \( (-7, 11) \)

Hint:

When solving geometric problems, remember to use the distance formula to find the perpendicular distance between a point and a line. This is often used in problems involving lines and distances.

Answer 1

The Correct Option is D

Step 1: Equation of the line and distance formula.
We are given two equations. The first one is the line:
\[ x + y = 4 \]
The second equation represents a line: \[ 4x + 3y = 10 \]
We are required to find the point on the first line that lies at a unit distance from the second line.
The formula for the perpendicular distance between a point \( (x_1, y_1) \) and a line \( Ax + By + C = 0 \) is given by:
\[ \text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \] We need to find a point \( (x, y) \) on the first line \( x + y = 4 \) such that the distance from the line \( 4x + 3y - 10 = 0 \) is 1.

Step 2: Set up the distance equation.
The equation of the second line is \( 4x + 3y = 10 \), so \( A = 4, B = 3, C = -10 \). We substitute \( x + y = 4 \) in terms of \( y \) as \( y = 4 - x \).
Substitute \( (x, y) = (x, 4 - x) \) into the distance formula and set the distance to 1:
\[ \frac{|4x + 3(4 - x) - 10|}{\sqrt{4^2 + 3^2}} = 1 \]
Simplify the expression: \[ \frac{|4x + 12 - 3x - 10|}{5} = 1 \] \[ \frac{|x + 2|}{5} = 1 \] \[ |x + 2| = 5 \]

Step 3: Solve the equation.
Now, solve \( |x + 2| = 5 \), which gives two cases:
- Case 1: \( x + 2 = 5 \) \[ x = 3 \]
- Case 2: \( x + 2 = -5 \) \[ x = -7 \]

Step 4: Substitute the value of \( x \) back into the equation of the line.
From the first line \( x + y = 4 \), substitute the values of \( x \) we obtained.
- For \( x = 3 \): \[ y = 4 - 3 = 1 \]
So the point is \( (3, 1) \).
- For \( x = -7 \): \[ y = 4 - (-7) = 4 + 7 = 11 \] So the point is \( (-7, 11) \).

Step 5: Verify the distances.
Now, we check which point satisfies the condition of being a unit distance from the line.
- For \( (3, 1) \), we calculate the distance from the line \( 4x + 3y = 10 \):
\[ \text{Distance} = \frac{|4(3) + 3(1) - 10|}{\sqrt{4^2 + 3^2}} = \frac{|12 + 3 - 10|}{5} = \frac{5}{5} = 1 \] So, the point \( (3, 1) \) satisfies the condition
- For \( (-7, 11) \), we calculate the distance:
\[ \text{Distance} = \frac{|4(-7) + 3(11) - 10|}{\sqrt{4^2 + 3^2}} = \frac{|-28 + 33 - 10|}{5} = \frac{5}{5} = 1 \] Thus, \( (-7, 11) \) also satisfies the condition.

Step 6: Conclusion.
Since \( (-7, 11) \) is the correct point and matches the options, the correct answer is \( \boxed{(-7, 11)} \), which corresponds to option (D).