🇫🇷 Exercise 16: Finding the Intersection of Lines $D_1$ and $D_2$

The intersection of $D_1$ and $D_2$ is a point $(x, y)$ that satisfies the equations for both lines.

a) Cartesian Form

• $D_1$: $x + y = 0$
• $D_2$: $2x + y = 1$

We solve the system of equations: $$\begin{cases} x + y = 0 & (1) \\ 2x + y = 1 & (2) \end{cases}$$ Subtract (1) from (2): $(2x + y) - (x + y) = 1 - 0$, which simplifies to $x = 1$. Substitute $x=1$ into (1): $1 + y = 0$, so $y = -1$. The intersection point is $(1, -1)$.

b) Cartesian Form

• $D_1$: $2x + y = 1$
• $D_2$: $x + 2y = 5$

We solve the system of equations: $$\begin{cases} 2x + y = 1 & (1) \\ x + 2y = 5 & (2) \end{cases}$$ From (1), we have $y = 1 - 2x$. Substitute this into (2): $x + 2(1 - 2x) = 5$ $x + 2 - 4x = 5$ $-3x = 3$ $x = -1$ Substitute $x=-1$ back into $y = 1 - 2x$: $y = 1 - 2(-1) = 1 + 2 = 3$. The intersection point is $(-1, 3)$.

c) Parametric Form

• $D_1$: $\begin{cases} x = 5 - t \\ y = 2t - 1 \end{cases}$
• $D_2$: $\begin{cases} x = 1 + s \\ y = 2 + 3s \end{cases}$

At the intersection, the coordinates must be equal. We set the $x$ and $y$ expressions equal to each other, resulting in a system with variables $t$ and $s$: $$\begin{cases} 5 - t = 1 + s & (1) \\ 2t - 1 = 2 + 3s & (2) \end{cases}$$ From (1), $s = 4 - t$. Substitute this into (2): $2t - 1 = 2 + 3(4 - t)$ $2t - 1 = 2 + 12 - 3t$ $5t = 15$ $t = 3$ Now find $s$: $s = 4 - 3 = 1$. Substitute $t=3$ into $D_1$ (or $s=1$ into $D_2$): $x = 5 - 3 = 2$ $y = 2(3) - 1 = 5$ The intersection point is $(2, 5)$.

d) Parametric and Cartesian Forms

• $D_1$: $\begin{cases} x = 1 + 4t \\ y = 2 + t \end{cases}$
• $D_2$: $x + 2y = 11$

Substitute the parametric expressions for $x$ and $y$ from $D_1$ into the Cartesian equation of $D_2$: $(1 + 4t) + 2(2 + t) = 11$ $1 + 4t + 4 + 2t = 11$ $6t + 5 = 11$ $6t = 6$ $t = 1$ Substitute $t=1$ back into the equations for $D_1$: $x = 1 + 4(1) = 5$ $y = 2 + 1 = 3$ The intersection point is $(5, 3)$.

e) Cartesian and Parametric Forms

• $D_1$: $2x - y = 3$
• $D_2$: $\begin{cases} x = s - 1 \\ y = 3 + 2s \end{cases}$

Substitute the parametric expressions for $x$ and $y$ from $D_2$ into the Cartesian equation of $D_1$: $2(s - 1) - (3 + 2s) = 3$ $2s - 2 - 3 - 2s = 3$ $-5 = 3$ This statement is false. Since there is no value of $s$ that satisfies the equation, the lines are parallel and distinct. The intersection is the empty set $\mathbf{D_1 \cap D_2 = \emptyset}$.

f) Parametric and Cartesian Forms

• $D_1$: $\begin{cases} x = 1 + \lambda \\ y = 2 - \lambda \end{cases}$
• $D_2$: $x + 4y = 3$

Substitute the parametric expressions for $x$ and $y$ from $D_1$ into the Cartesian equation of $D_2$: $(1 + \lambda) + 4(2 - \lambda) = 3$ $1 + \lambda + 8 - 4\lambda = 3$ $9 - 3\lambda = 3$ $-3\lambda = -6$ $\lambda = 2$ Substitute $\lambda=2$ back into the equations for $D_1$: $x = 1 + 2 = 3$ $y = 2 - 2 = 0$ The intersection point is $(3, 0)$.

g) Parametric Form

• $D_1$: $\begin{cases} x = 1 + t \\ y = 2 - t \end{cases}$
• $D_2$: $\begin{cases} x = 1 + s \\ y = 2 + s \end{cases}$

Set the coordinates equal to each other: $$\begin{cases} 1 + t = 1 + s & (1) \\ 2 - t = 2 + s & (2) \end{cases}$$ From (1), $t = s$. Substitute $t=s$ into (2): $2 - s = 2 + s$ $-2s = 0$ $s = 0$ Since $t=s$, we also have $t=0$. Substitute $t=0$ into $D_1$ (or $s=0$ into $D_2$): $x = 1 + 0 = 1$ $y = 2 - 0 = 2$ The intersection point is $(1, 2)$.

h) Parametric Form

• $D_1$: $\begin{cases} x = 2s - 1 \\ y = s + 2 \end{cases}$
• $D_2$: $\begin{cases} x = 1 - 2t \\ y = 3 - t \end{cases}$

Set the coordinates equal to each other: $$\begin{cases} 2s - 1 = 1 - 2t & (1) \\ s + 2 = 3 - t & (2) \end{cases}$$ From (2), $t = 1 - s$. Substitute this into (1): $2s - 1 = 1 - 2(1 - s)$ $2s - 1 = 1 - 2 + 2s$ $2s - 1 = -1 + 2s$ $-1 = -1$ This statement is always true. This means any point on $D_1$ is also on $D_2$, and the lines are the same line. The intersection is the line itself: $\mathbf{D_1 \cap D_2 = D_1 \text{ (or } D_2)}$.

Summary of Intersections: