Question: Find the point on the line $ y = 2x + 1 $ closest to $ (3, 4) $.

["How to Find the Point on the Line ( y = 2x + 1 ) Closest to ( (3, 4) ): A Complete Guide", "Finding the point on a line closest to a given point is a common problem in geometry and analytic mathematics. Whether you're optimizing distance, solving engineering problems, or visualizing data, understanding how to pinpoint the nearest point helps improve both accuracy and insight. In this article, we’ll walk through the step-by-step process of finding the point on the line ( y = 2x + 1 ) that is closest to the point ( (3, 4) ), combining algebra and geometry for a clear, practical solution.", "---", "### Understanding the Problem", "Given:", "- A line: ( y = 2x + 1 )\n- A point: ( P = (3, 4) )\n- Goal: Find the point ( Q = (x, y) ) on the line such that the distance ( PQ ) is minimized.", "Intuitively, the shortest distance from a point to a line is along the perpendicular line segment connecting them. This means we can find the closest point by constructing or identifying the line perpendicular to ( y = 2x + 1 ) passing through ( (3, 4) ), then finding where these two lines intersect.", "---", "### Step 1: Determine the Slope of the Given Line", "The equation of the line is:", "[\ny = 2x + 1\n]", "This is in slope-intercept form ( y = mx + b ), so:", "- Slope of the line: ( m = 2 )", "The slope of any perpendicular line is the negative reciprocal:", "[\nm_{\perp} = -\frac{1}{2}\n]", "---", "### Step 2: Write the Equation of the Perpendicular Line", "The perpendicular line passes through ( (3, 4) ) with slope ( -\frac{1}{2} ). Using point-slope form:", "[\ny - 4 = -\frac{1}{2}(x - 3)\n]", "Simplify to slope-intercept form:", "[\ny - 4 = -\frac{1}{2}x + \frac{3}{2}\n]\n[\ny = -\frac{1}{2}x + \frac{3}{2} + 4\n]\n[\ny = -\frac{1}{2}x + \frac{11}{2}\n]", "---", "### Step 3: Find the Intersection of the Two Lines", "The closest point lies at the intersection of:", "1. Original line:\n[\ny = 2x + 1\n]\n2. Perpendicular line:\n[\ny = -\frac{1}{2}x + \frac{11}{2}\n]", "Set the expressions equal:", "[\n2x + 1 = -\frac{1}{2}x + \frac{11}{2}\n]", "Multiply both sides by 2 to eliminate fractions:", "[\n4x + 2 = -x + 11\n]", "Add ( x ) to both sides:", "[\n5x + 2 = 11\n]", "Subtract 2:", "[\n5x = 9 \quad \Rightarrow \quad x = \frac{9}{5}\n]", "Now substitute ( x = \frac{9}{5} ) into the original line equation to find ( y ):", "[\ny = 2\left(\frac{9}{5}\right) + 1 = \frac{18}{5} + \frac{5}{5} = \frac{23}{5}\n]", "---", "### Step 4: The Closest Point Coordinates", "The point on the line ( y = 2x + 1 ) closest to ( (3, 4) ) is:", "[\n\left( \frac{9}{5}, \frac{23}{5} \right)\n]", "---", "### Why This Works: Geometric Insight", "- The line ( y = 2x + 1 ) defines one line.\n- The shortest distance from a point to a line is along the perpendicular line.\n- Intersecting the original line with its perpendicular gives the exact foot of the perpendicular — the closest point.\n- This method preserves both algebraic accuracy and geometric meaning.", "---", "### Bonus: Using Vector Projections (Advanced Insight)", "For those interested in linear algebra, the closest point can be found using vector projection. Represent the line as a parametric vector from a point on the line in the direction of the line’s slope, project ( \vec{P - A} ) onto the direction vector, and solve for the scalar parameter that minimizes distance. This approach confirms our earlier method efficiently.", "---", "### Summary", "Finding the closest point on a line involves:", "1. Identifying the line slope and computing the perpendicular slope.\n2. Writing the equation of the perpendicular line.\n3. Solving the system where the line and its perpendicular intersect.\n4. The intersection point is the closest point on the original line.", "For the line ( y = 2x + 1 ) and point ( (3, 4) ), the closest point is ( \left( \frac{9}{5}, \frac{23}{5} \right) ).", "---", "Key Search Terms:\n- Find closest point on line ( y = 2x + 1 ) to (3, 4)\n- Shortest distance from point to line\n- Geometry problem: line perpendicular to ( y = 2x + 1 )\n- Closest point on line using coordinates", "Use this solution to efficiently compute closest points in real-world applications — from computer graphics to machine learning algorithms."]