Question Statement

Show that the lines $x+y+1=0$, $x-y+1=0$ and the $x$-axis are concurrent. Also prove that the $x$-axis bisects the angle between the lines $x+y+1=0$ and $x-y+1=0$.

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Background and Explanation

Three lines are concurrent if they all intersect at a single common point. For lines $a_{1}x+b_{1}y+c_1=0$, $a_{2}x+b_{2}y+c_2=0$, and $a_{3}x+b_{3}y+c_3=0$, concurrency can be verified by showing that the determinant of their coefficients equals zero. To prove a line bisects the angle between two other lines, we show that it makes equal angles with both lines.

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Solution

Part (a): Proving Concurrency

First, we identify the equations of the three lines:

1. Line 1: $x+y+1=0$
2. Line 2: $x-y+1=0$
3. Line 3 (x-axis): $y=0$, which can be written as $0x+1y+0=0$

To prove these three lines are concurrent, we evaluate the determinant of their coefficients:

$\Delta =\left|\begin{array}{ccc}1& 1& 1\\ 1& -1& 1\\ 0& 1& 0\end{array}\right|$

Expanding along the first row:

$\begin{array}{rl}\Delta & =1\left|\begin{array}{cc}-1& 1\\ 1& 0\end{array}\right|-1\left|\begin{array}{cc}1& 1\\ 0& 0\end{array}\right|+1\left|\begin{array}{cc}1& -1\\ 0& 1\end{array}\right|\\ & =1((-1)(0)-(1)(1))-1((1)(0)-(1)(0))+1((1)(1)-(-1)(0))\\ & =1(0-1)-1(0-0)+1(1-0)\\ & =-1-0+1\\ & =0\end{array}$

Since $\Delta =0$, the three lines are concurrent (they all meet at a single point).

Part (b): Proving the $x$-axis Bisects the Angle

To show that the $x$-axis bisects the angle between the two lines, we calculate the angle that the $x$-axis makes with each line. If these angles are equal, the $x$-axis is the angle bisector.

Finding the angle between the $x$-axis and line $x-y+1=0$:

Let $\alpha$ be the angle from the $x$-axis to the line $x-y+1=0$.

• Slope of $x$-axis: $m=0$
• Slope of line $x-y+1=0$: Rewriting as $y=x+1$, we get $m_1=1$

Using the formula for the tangent of the angle between two lines:

\begin{align*} \tan \alpha &= \frac{m_1 - m}{1 + m_1 m} \\ &= \frac{1 - 0}{1 + (1)(0)} \\ &= \frac{1}{1} \\ &= 1 \end{align*}

Finding the angle between line $x+y+1=0$ and the $x$-axis:

Let $\beta$ be the angle from the line $x+y+1=0$ to the $x$-axis.

• Slope of $x$-axis: $m=0$
• Slope of line $x+y+1=0$: Rewriting as $y=-x-1$, we get $m_2=-1$

Using the angle formula:

\begin{align*} \tan \beta &= \frac{m - m_2}{1 + m m_2} \\ &= \frac{0 - (-1)}{1 + (0)(-1)} \\ &= \frac{1}{1} \\ &= 1 \end{align*}

Conclusion:

From equations (1) and (2): $\tan \alpha =\tan \beta =1$

Therefore: $\alpha =\beta$

Since the $x$-axis makes equal angles with both lines $x+y+1=0$ and $x-y+1=0$, it bisects the angle between them.

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Key Formulas or Methods Used

• Concurrency Condition: Three lines $a_{1}x+b_{1}y+c_1=0$, $a_{2}x+b_{2}y+c_2=0$, $a_{3}x+b_{3}y+c_3=0$ are concurrent if: $\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|=0$

• Angle Between Two Lines: If $\theta$ is the angle between lines with slopes $m_1$ and $m_2$: $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

• Slope from Standard Form: For line $ax+by+c=0$, the slope is $m=-\frac{a}{b}$

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Summary of Steps

1. Identify the three lines: Write the $x$-axis as $y=0$ (or $0x+1y+0=0$) along with the given lines $x+y+1=0$ and $x-y+1=0$.
2. Set up the determinant: Create a $3\times 3$ matrix using the coefficients of $x$, $y$, and the constant term from each line.
3. Evaluate the determinant: Expand and simplify to show it equals zero, proving concurrency.
4. Calculate slopes: Find slopes of both given lines ($m_1=1$ for $x-y+1=0$ and $m_2=-1$ for $x+y+1=0$).
5. Find angles with x-axis: Use $\tan \theta =\frac{m_1-m_2}{1+m_1{m}_2}$ to calculate the angle from the $x$-axis to the first line ($\alpha$) and from the second line to the $x$-axis ($\beta$).
6. Compare angles: Show that $\tan \alpha =\tan \beta =1$, therefore $\alpha =\beta$, proving the $x$-axis is the angle bisector.