Question

A line through (2,3) makes an angle $\frac{3\pi}{4}$ with the negative direction of $x$ -axis. The length of the line segment cut off between (2,3) and the line $x+y-7=0$.

• A. $\sqrt{2}$
• B. $2\sqrt{2}$
• C. $2\sqrt{3}$
• D. $3\sqrt{2}$

Answer 1

Answer: (A)

$\sqrt{2}$

Answer 2

Let the line passing through $P(2,3)$ be $L_1$. The angle it makes with the negative x-axis is $\frac{3\pi}{4}$. Let $\theta$ be the angle $L_1$ makes with the positive x-axis. The angle of the negative x-axis is $\pi$. The angle between $L_1$ and the negative x-axis is $|\theta - \pi|$. So, $|\theta - \pi| = \frac{3\pi}{4}$. This gives two possibilities:

1. $\theta - \pi = \frac{3\pi}{4} \implies \theta = \pi + \frac{3\pi}{4} = \frac{7\pi}{4}$. The slope is $m = \tan(\frac{7\pi}{4}) = -1$.
2. $\theta - \pi = -\frac{3\pi}{4} \implies \theta = \pi - \frac{3\pi}{4} = \frac{\pi}{4}$. The slope is $m = \tan(\frac{\pi}{4}) = 1$.

The second line is $L_2: x+y-7=0$. Its slope is $m_2 = -1$.

If the slope of $L_1$ were $-1$, then $L_1$ would be parallel to $L_2$. The equation of $L_1$ would be $y-3 = -1(x-2) \implies x+y-5=0$. This line is parallel to $x+y-7=0$, and no finite line segment would be cut off. Therefore, the slope of $L_1$ must be $1$, and $\theta = \frac{\pi}{4}$.

We use the parametric form of the line $L_1$ passing through $(x_0, y_0) = (2,3)$ with an angle $\theta = \frac{\pi}{4}$ and distance $r$: $x = x_0 + r \cos\theta = 2 + r \cos(\frac{\pi}{4}) = 2 + r \frac{1}{\sqrt{2}}$ $y = y_0 + r \sin\theta = 3 + r \sin(\frac{\pi}{4}) = 3 + r \frac{1}{\sqrt{2}}$

Substitute these into the equation of $L_2: x+y-7=0$: $(2 + r \frac{1}{\sqrt{2}}) + (3 + r \frac{1}{\sqrt{2}}) - 7 = 0$ $5 + \frac{2r}{\sqrt{2}} - 7 = 0$ $5 + r\sqrt{2} - 7 = 0$ $r\sqrt{2} - 2 = 0$ $r\sqrt{2} = 2$ $r = \frac{2}{\sqrt{2}} = \sqrt{2}$

The value of $r$ represents the length of the line segment cut off between the point $(2,3)$ and the line $x+y-7=0$. The length of the segment is $\sqrt{2}$.