Solution: The closest point on a line to a given point is the perpendicular projection. The slope of the line is $ m = -\frac{1}{2} $, so the perpendicular slope is $ 2 $. The equation of the perpendicular line through $ (4, 3) $ is $ y - 3 = 2(x - 4) $. Solving the system:

["The Solution: Closest Point on a Line Using Perpendicular Projection", "When finding the closest point from a given point to a straight line, a powerful geometric principle applies: the shortest distance is achieved when the line segment connecting the point to the line is perpendicular to the original line. This concept is fundamental in geometry, computer graphics, optimization, and data analysis.", "### The Problem Setup", "Let’s solve a classic case: Determine the closest point on a line to a specific point using the perpendicular projection method.", "Suppose the line has slope ( m = -\frac{1}{2} ), and we are given a point ( P(4, 3) ). We want the point ( Q ) on the line such that segment ( PQ ) is perpendicular to the line.", "### Step 1: Determine the perpendicular slope", "Since the slope of the original line is ( m = -\frac{1}{2} ), the slope of any line perpendicular to it is the negative reciprocal, which is:", "[\nm_{\perp} = 2\n]", "### Step 2: Write the equation of the perpendicular line", "This perpendicular line passes through point ( P(4, 3) ), so we use point-slope form:", "[\ny - 3 = 2(x - 4)\n]", "Simplify:", "[\ny = 2x - 8 + 3 = 2x - 5\n]", "So the perpendicular line is:\n[\ny = 2x - 5\n]", "### Step 3: Find the intersection point (closest point on the original line)", "We now solve the system formed by the original line and the perpendicular line to find their intersection point ( Q ), which is the closest point.", "But wait—we don’t yet have the equation of the original line. However, since the slope of the original line is given as ( m = -\frac{1}{2} ), we can express it in point-slope form using point ( P(4, 3) ), though note: this assumes ( P ) lies on the line—which it does not necessarily. But since we are projecting perpendicularly from ( P(4, 3) ) to the line, the shortest point ( Q ) lies at the intersection of ( P )’s perpendicular ray with the line. So we only need the general form of the original line.", "But here’s a key insight: although the full line equation is not given, the method relies only on slopes and a point projected perpendicularly.", "However, for a complete solution, suppose the original line is not fully specified—but since we’re computing the projection, we can express the closest point algebraically.", "Actually, in most projection problems, if the original line is given (e.g., with a specific equation), we plug it in. Here, we are only told the slope—so we proceed with symbolic form.", "But let’s assume for illustration that the original line has equation derived from slope and passes through a reference—yet to keep it general, consider this:", "Since we know the perpendicular has slope 2 and passes through ( (4,3) ), and the projection finds where it meets the original line, we solve the system:", "1. ( y = m_1 x + b_1 ) — but we don’t know ( b_1 ).", "Wait—this reveals a gap without the full line equation. But suppose instead we reframe: the closest point depends only on the direction, not the specific line. However, for a valid problem, we need an explicit line.", "Let’s revise: Let’s assume the original line passes through point ( (0, b) ), but since no intercept is given, instead solve the system parametrically.", "But a better approach: suppose the line has slope ( -\frac{1}{2} ) and pass through a known point—but since it’s not stated, we must recognize: in standard projection problems, often the line is given.", "However, for instructional clarity, let’s complete the problem by assuming the line is given as passing through ( (0, 1) )—a common choice. But without loss of generality, let's suppose the line is:", "[\ny = -\frac{1}{2}x + 5\n]", "—which has slope ( -\frac{1}{2} ) and passes through ( (0, 5) ). This is consistent with the perpendicular slope of 2.", "Now proceed.", "Let line ( L ) be:\n[\ny = -\frac{1}{2}x + 5\n]", "And point ( P(4, 3) ).", "### Step 1: Perpendicular line through ( P )\nAs before:\n[\ny - 3 = 2(x - 4) \Rightarrow y = 2x - 5\n]", "### Step 2: Solve the system", "Set the two equations equal:", "[\n-\frac{1}{2}x + 5 = 2x - 5\n]", "Multiply both sides by 2 to eliminate fractions:", "[\n- x + 10 = 4x - 10\n]", "Bring like terms together:", "[\n10 + 10 = 4x + x \Rightarrow 20 = 5x \Rightarrow x = 4\n]", "Wait—this gives ( x = 4 ), then substitute into either equation:", "( y = 2(4) - 5 = 8 - 5 = 3 )", "So the intersection is ( (4, 3) ) itself.", "But that means ( P(4,3) ) lies on the line ( y = -\frac{1}{2}x + 5 )? Let’s check:", "[\ny = -\frac{1}{2}(4) + 5 = -2 + 5 = 3\n]", "Yes! So ( P(4,3) ) is already on the line. Therefore, the closest point is ( (4, 3) ), and the distance is zero.", "But this is a degenerate case—let’s adjust the example to reflect a nontrivial projection.", "Let’s instead define the line differently with the same slope but different intercept.", "Wait—alternate plan: The principle remains: the closest point is the intersection of the line and the perpendicular from the point. But if the point lies on the line, that point is itself.", "To make a useful, non-degenerate problem, suppose the line is:", "[\ny = -\frac{1}{2}x + 3\n]", "Now check: at ( x = 4 ), ( y = -2 + 3 = 1 <br/>\ne 3 ), so ( P(4,3) ) does not lie on the line.", "Now set:", "Let original line:\n[\ny = -\frac{1}{2}x + 3\n]", "Perpendicular slope: 2\nThrough point ( P(4, 3) ):\n[\ny - 3 = 2(x - 4) \Rightarrow y = 2x - 5\n]", "Now solve:", "[\n-\frac{1}{2}x + 3 = 2x - 5\n]", "Multiply by 2:", "[\n- x + 6 = 4x - 10\n\Rightarrow 6 + 10 = 4x + x \Rightarrow 16 = 5x \Rightarrow x = \frac{16}{5} = 3.2\n]", "Then:", "[\ny = 2\left(\frac{16}{5}\right) - 5 = \frac{32}{5} - \frac{25}{5} = \frac{7}{5} = 1.4\n]", "So the closest point is ( \left( \frac{16}{5}, \frac{7}{5} \right) )", "### Final Answer", "The closest point on the line ( y = -\frac{1}{2}x + 3 ) to the point ( (4, 3) ) is ( \left( \frac{16}{5}, \frac{7}{5} \right) ), found by solving the system of the line and its perpendicular from ( (4, 3) ).", "### Key Takeaway", "The shortest distance from a point to a line occurs along the perpendicular line. The solution involves:", "- Determining the perpendicular line’s"]