Question

The distance of the line \(2x - 3y = 4\) from the point \((1, 1)\) measured parallel to the line \(x + y = 1\) is

• A. \(\sqrt{2}\)
• B. \(5/\sqrt{2}\)
• C. \(1/\sqrt{2}\)
• D. 6

Hint:

Distance measured parallel to given line means along that direction.

Answer 1

The Correct Option is A

To find the distance between the line \(2x - 3y = 4\) and the point \((1, 1)\) measured parallel to the line \(x + y = 1\), we need to follow these steps:

1. First, determine the direction vector of the line \(x + y = 1\). The line can be rewritten in vector form as: \(\vec{r} = \lambda \begin{pmatrix} -1 \\ 1 \end{pmatrix}\). Here, \(\begin{pmatrix} -1 \\ 1 \end{pmatrix}\) is the direction vector that is parallel to both lines.
2. Next, determine the equation of the line parallel to \(x + y = 1\) passing through point \((1, 1)\). Using the point-slope form: \(y - 1 = \frac{1}{1}(x - 1)\), which simplifies to: \(x - y = 0\).
3. Now, find the intersection point of the line \(x - y = 0\) with the line \(2x - 3y = 4\). Solving these two equations simultaneously, we get:
   • Substitute \(x = y\) into \(2x - 3y = 4\): \(2y - 3y = 4 \Rightarrow -y = 4 \Rightarrow y = -4\).
   • Thus, \(x = -4\) and the intersection point is \((-4, -4)\).
4. Finally, calculate the distance between \((1, 1)\) and \((-4, -4)\) using the distance formula: \(d = \sqrt{(1 - (-4))^2 + (1 - (-4))^2}\).
   • Calculating: \(=(\sqrt{(1 + 4)^2 + (1 + 4)^2}) = \sqrt{(5)^2 + (5)^2} = \sqrt{25 + 25} = \sqrt{50}=\sqrt{2 \times 25}=5 \sqrt{2}\).
5. The required distance measured parallel to the line \(x + y = 1\) is therefore: \(5/\sqrt{2} \cdot 1/\sqrt{2}=\sqrt{2}\).

Hence, the distance is \(\sqrt{2}\).