Midpoint of a Line Segment

Gustavo Felisberto Valente March 30, 2020

1. Determine the coordinates of the midpoint of each line segment.

(a) (1, 5) and (7, 3) (b) (−4, −3) and (5, 2) (c) (−3.2, 4.1) and (5.6, −2.3) 2 3 4 3 (d) (− 5 , − 4 ) and ( 5 , 4 )

(b)

1 3. Find the slope of each median shown. (b) Draw the median from vertex J. Then, ﬁnd an equation in slope y-intercept form for this median. (c) Draw the right bisector of KL. Then, ﬁnd an equation in slope y-intercept form for this right bisector.

9. Write an expression for the coordinates of the midpoint of the line segment with endpoints A(2a, 3b) and B(4a, 5b). Explain your rea- soning. (a) 10. (a) Draw 4ABC with vertices A(−8, 0), B(0, 0), and C(0, −8). (b) Construct the midpoints of AB, BC, and AC and label them D, E, and F , respec- tively. (c) Join the midpoints to form 4DEF . (d) Show that the length of line segment DE is one half the length of line segment AC. (e) Show that the length of line segment DF (b) is one half the length of line segment BC. 4. The endpoints of the diameter of a circle are (f) Show that the length of line segment EF A(−5, −3) and B(3, 7). Find the coordinates is one half the length of line segment of the centre of this circle. AB. 5. The vertices of 4P QR are P (−3, 5), Q(5, 7), and R(3, −3).

(a) Find an equation in slope y-intercept form for the median from vertex P . (b) Find an equation in slope y-intercept form for the median from vertex Q. (c) Find an equation in slope y-intercept form for the median from vertex R.

6. One endpoint of a diameter of a circle centred at the origin is (−5, 2). Find the coordinates of the other endpoint of this diameter.

7. Determine an equation for the right bisector of the line segment with endpoints D(−3, 5) and M(7, −9).

8. (a) Draw 4JKL with vertices J(−6, 4), K(−4, −5), and L(6, 1).

2 Answers (d) From the graph, DE is 4 units long and AC is 8 units long, so DE is one half the 1 1. (a) ( 2 , 2) length of AC. (b) (1, 1) (e) From the graph, DF is 4 units long and (c) (−1, −1) BC is 8 units long, so DF is one half the length of BC. 2. (a) (4, 4) 1 1 (f) By the Pythagorean theorem, (b) ( 2 , − 2 ) (c) (1.2, 0.9) EF 2 = DF 2 + DE2 = 42 + 42 = 32 1 (d) ( 5 , 0) √ √ √ √ √ EF = 32 = 16 × 2 = 16× 2 = 4 2 3. (a) m = − 6 5 2 2 2 2 2 (b) m = 9 AB = AC + BC = 8 + 8 = 128 11 √ √ √ √ √ AB = 128 = 64 × 2 = 64× 2 = 8 2 4. (−1, 2) Line segment AB is twice the length of 5. (a) y = − 3 x + 26 7 7 line segment EF . 6 (b) y = 5 x + 1 9 21 (c) y = − 2 x + 2 6. (5, −2)

5 24 7. y = 7 x − 7

9. (3a, 4b); Use the midpoint formula with x1 = 2a, x2 = 4a, y1 = 3b, and y2 = 5b. 10. (a) (b)

(c)