The distance between the point (4, 3) and the intersection of y = 2x + 4 and y = 7 - x is

1. Analyze the question

We need to find the distance between a fixed point $(4, 3)$ and the point where two lines, $y = 2x + 4$ and $y = 7 - x$, intersect.

2. Explain the concept

• Intersection of lines: To find the point of intersection, we solve the two equations simultaneously.
• Distance formula: The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

3. Show the solution

Step 1: Find the point of intersection. Set the two equations equal to each other since both are equal to $y$: $2x + 4 = 7 - x$

Add $x$ to both sides: $3x + 4 = 7$

Subtract 4 from both sides: $3x = 3$

Divide by 3: $x = 1$

Now, substitute $x = 1$ into either equation to find $y$: $y = 7 - (1) = 6$ The point of intersection is (1, 6).

Step 2: Calculate the distance between (4, 3) and (1, 6). Using the distance formula where $(x_1, y_1) = (4, 3)$ and $(x_2, y_2) = (1, 6)$: $d = \sqrt{(1 - 4)^2 + (6 - 3)^2}$ $d = \sqrt{(-3)^2 + (3)^2}$ $d = \sqrt{9 + 9}$ $d = \sqrt{18}$

Step 3: Simplify the radical. $\sqrt{18} = \sqrt{9 \times 2}$ $d = 3\sqrt{2}$

4. Identify the answer

Comparing our result $3\sqrt{2}$ to the given options, it matches option B.

5. Tip

When solving for the intersection of two lines, always double-check your coordinates by plugging them back into both original equations. If the values satisfy both, you are guaranteed to have the correct point before calculating the distance!