Partition a directed line segment in a given ratio and compute perimeters and areas of polygons from coordinates

Learning goal: Find the point on a directed line segment between two given points that partitions the segment in a given ratio; and use coordinates to compute perimeters of polygons and areas of triangles and rectangles, using the distance formula and decomposition into right triangles and rectangles.

Directed segments. A directed line segment from A to B has a starting point and an ending point; the segment from B to A is the same set of points traveled the other way. To partition the directed segment from A to B in the ratio m : n is to find the point P on it with AP : PB = m : n, where AP is measured from the start. Because the ratio is read from the starting point, the answer changes when the direction changes: the point that partitions A to B in the ratio 2 : 3 is not the point that partitions B to A in the ratio 2 : 3.

Finding the point. If AP : PB = m : n then AP is the fraction m/(m + n) of the whole length AB. The x-coordinate of P is therefore that fraction of the way from A's x-coordinate to B's, and the same for y: P = (x₁ + (m/(m + n))(x₂ − x₁), y₁ + (m/(m + n))(y₂ − y₁)). The reasoning is similar triangles: the horizontal and vertical legs of the slope triangle from A to P are the same fraction of the legs from A to B as AP is of AB, because the small triangle is a dilation of the large one with center A and factor m/(m + n). The midpoint is the special case m = n, fraction 1/2.

A worked model inside the lesson. Find the point P that partitions the directed segment from A(−2, 1) to B(8, 6) in the ratio 2 : 3. The fraction is 2/(2 + 3) = 2/5. The change from A to B is (8 − (−2), 6 − 1) = (10, 5). Two fifths of that is (4, 2), so P = (−2 + 4, 1 + 2) = (2, 3). Check the distances: AP = √(4² + 2²) = √20 = 2√5 and PB = √(6² + 3²) = √45 = 3√5, so AP : PB = 2√5 : 3√5 = 2 : 3. Check that P is on the line: the slope from A to P is 2/4 = 1/2 and from P to B is 3/6 = 1/2, the same. Now reverse the direction: the point that partitions B to A in the ratio 2 : 3 starts at B and moves 2/5 of the way toward A: (8 + (2/5)(−10), 6 + (2/5)(−5)) = (8 − 4, 6 − 2) = (4, 4). Different point, as expected.

A second partition, ratio 3 : 1. The point Q partitioning (−1, −2) to (7, 10) in the ratio 3 : 1 is 3/4 of the way: change (8, 12), three quarters of it (6, 9), so Q = (5, 7). Check: the distance from (−1, −2) to Q is √(36 + 81) = √117 = 3√13 and from Q to (7, 10) is √(4 + 9) = √13, ratio 3 : 1.

Perimeters from coordinates. The perimeter of a polygon is the sum of its side lengths, and each side length comes from the distance formula, d = √((x₂ − x₁)² + (y₂ − y₁)²). For the triangle with vertices (1, 2), (7, 2) and (4, 6): the first side is horizontal with length 7 − 1 = 6; the second is √((4 − 7)² + (6 − 2)²) = √(9 + 16) = 5; the third is √((1 − 4)² + (2 − 6)²) = √(9 + 16) = 5. Perimeter 6 + 5 + 5 = 16. Because two sides are equal, the triangle is isosceles, a fact the coordinates prove rather than suggest. When lengths are irrational, keep them exact until the end: the rectangle A(0, 0), B(6, 3), C(4, 7), D(−2, 4) from the previous lesson has sides 3√5 and 2√5, so its perimeter is 2(3√5) + 2(2√5) = 10√5 ≈ 22.36.

Areas of rectangles and triangles. For a rectangle, multiply two adjacent side lengths: the rectangle above has area 3√5 · 2√5 = 6·5 = 30 square units, the radicals combining to a whole number. For a triangle with a horizontal or vertical side, use ½·base·height with the height read from the coordinates: the isosceles triangle above has base 6 along y = 2 and apex at height 6 − 2 = 4, so its area is ½·6·4 = 12. When no side is horizontal or vertical, enclose the triangle in the smallest axis-aligned rectangle and subtract the right triangles in the corners. For the triangle (0, 0), (6, 3), (4, 7): the enclosing rectangle runs from x = 0 to 6 and y = 0 to 7, area 42. The three corner right triangles have areas ½·6·3 = 9 (below the side to (6, 3)), ½·2·4 = 4 (the corner near (6, 7)) and ½·4·7 = 14 (along the left edge). Triangle area = 42 − 9 − 4 − 14 = 15 square units. A different route confirms it: (0, 0), (6, 3) and (4, 7) are three vertices of the rectangle ABCD above, so the triangle is half of that rectangle, with a right angle at (6, 3) (slopes 1/2 and −2) and legs 3√5 and 2√5; its area is ½·3√5·2√5 = 15. The right-angle route works for any right triangle: the triangle (−2, 1), (2, 4), (5, 0) has a right angle at (2, 4) with legs both 5, so its area is ½·5·5 = 12.5, and its perimeter is 5 + 5 + √50 = 10 + 5√2 ≈ 17.07.

Exact and approximate. √20 is exactly 2√5; writing 4.47 is an approximation. Perimeters that mix rational and irrational lengths, like 10 + 5√2, should be left exact and then rounded once at the end with the precision stated. Areas of rectangles with sides a√k and b√k are exact whole or rational numbers, ab·k, even though the sides are irrational.

The misconception: the partition point is the average, or the ratio can be read from either end. The midpoint formula averages the coordinates, and students often reach for it whatever the ratio. The average is right only for 1 : 1. The general rule uses the fraction m/(m + n), and it is measured from the starting point of the directed segment; reversing the direction swaps the roles of m and n. A second slip in area problems is subtracting only two of the three corner triangles from the enclosing rectangle, or using a slanted side as the base with a height that is not perpendicular to it. The height must be measured perpendicular to the base, which is why enclosing in a rectangle, where every height is horizontal or vertical, is the safer method.

Worked partition: Find the point that partitions the directed segment from C(3, −4) to D(−7, 1) in the ratio 3 : 2. Fraction 3/5. Change from C to D: (−10, 5). Three fifths: (−6, 3). Point: (3 − 6, −4 + 3) = (−3, −1). Check: distance C to (−3, −1) is √(36 + 9) = √45 = 3√5; distance (−3, −1) to D is √(16 + 4) = √20 = 2√5; ratio 3 : 2. Slopes: from C, 3/(−6) = −1/2; from the point to D, 2/(−4) = −1/2, so the point is on the segment.

Worked partition in the other direction: The point that partitions D(−7, 1) to C(3, −4) in the ratio 3 : 2 starts at D: change (10, −5), three fifths (6, −3), point (−1, −2). This is a different point from (−3, −1), because the ratio is measured from the starting point. Both points lie on the same line: slope from D to (−1, −2) is −3/6 = −1/2.

Worked perimeter of a quadrilateral: Vertices (−3, −1), (5, −1), (8, 3), (0, 3). Sides: horizontal, 5 − (−3) = 8; √(3² + 4²) = 5; horizontal, 8 − 0 = 8; √(3² + 4²) = 5. Perimeter 8 + 5 + 8 + 5 = 26. Opposite sides are congruent and the horizontal sides are parallel, and the slopes of the other two sides are both 4/3, so it is a parallelogram; its area is base 8 times the vertical distance between the horizontal sides, 3 − (−1) = 4, giving 32 square units.

Worked triangle area by enclosing rectangle: Vertices (1, 1), (7, 3), (3, 8). Enclosing rectangle from x = 1 to 7 and y = 1 to 8: area 6·7 = 42. Corner triangles: ½·6·2 = 6 (below the side from (1, 1) to (7, 3)); ½·4·5 = 10 (the corner near (7, 8), legs 7 − 3 = 4 and 8 − 3 = 5); ½·2·7 = 7 (the corner near (1, 8), legs 3 − 1 = 2 and 8 − 1 = 7). Area = 42 − 6 − 10 − 7 = 19 square units. Perimeter: √(36 + 4) + √(16 + 25) + √(4 + 49) = √40 + √41 + √53 = 2√10 + √41 + √53 ≈ 20.01 to two decimal places. (Rounding each term first, 6.32 + 6.40 + 7.28 = 20.00, is off in the last digit; round once, at the end.)

Worked rectangle area with irrational sides: The rectangle with vertices (0, 0), (6, 3), (4, 7), (−2, 4) has adjacent sides √45 = 3√5 and √20 = 2√5. Area = 3√5 · 2√5 = 6 · 5 = 30. Check by the enclosing-rectangle method: the box from x = −2 to 6 and y = 0 to 7 has area 56; the four corner triangles have areas ½·6·3 = 9, ½·2·4 = 4, ½·6·3 = 9 and ½·2·4 = 4, total 26; 56 − 26 = 30.

Partition-and-measure record: For the directed segment from (−4, 3) to (8, −3), find the points that partition it in the ratios 1 : 1, 1 : 2, 2 : 1 and 1 : 5, and for the reversed segment in the ratios 1 : 2 and 2 : 1. Plot all the points on one grid, check each with the distance formula, and write one sentence about how the 1 : 2 and 2 : 1 points relate when the direction is reversed. Then compute the perimeter and area of the triangle formed by (−4, 3), (8, −3) and (8, 3).

Independent performance: A park is to be laid out on a coordinate grid measured in meters. Choose four points forming a rectangle that is not aligned with the grid, prove it is a rectangle, and compute its perimeter and area exactly and to the nearest tenth. Place a fountain at the point that divides the path from one vertex to the opposite vertex in the ratio 3 : 5, starting from the vertex you name, and find its coordinates. Then choose a triangular flower bed with no horizontal or vertical side and find its perimeter and its area by the enclosing-rectangle method. Success criteria: every length is computed with the distance formula and left exact before rounding; the area of the rectangle is checked two ways; the fountain point is verified by the distance ratio; and all three corner triangles are shown in the flower-bed calculation with units stated.