CHAPTER
2
SOLVED PROBLEMS REVIEW COORDINATE GEOMETRY
For the review sessions, I will try to post some of the solved homework since I find that at this age both
taking notes and proofs are still a burgeoning skill.
2.1
Use the slopes, distances, line equations to verify your guesses
It is not enough to guess your figure, or to count squares, etc. on the quad paper. You really
have to write the equations of the lines, of their slopes to check your statements.
The lengths of the sides (distances between the vertex points) establish if a triangle ABC/trapezoid
ABCD is isosceles.
p
dAB = (xB − xA )2 + (yB − yA )2
If a point P is on a line y = mx + n , it verifies the equation of that line : yP = mxP + n
If a point P is at the intersection of two lines it verifies both equations.
The slopes of all the sides can be used to check if two lines are parallel or perpendicular.
Given a quadrilateral with:
• exactly one pair of equal slopes → one pair of parallel lines → trapezoid
• two pairs of equal slopes → two pairs of parallel lines → parallelogram
Given a parallelogram with:
• one pair of adjacent sides with slopes’ product = −1 (i.e. negative reciprocals) → the adjacent sides
are perpendicular → a rectangle
• the slopes of the diagonals are negative reciprocals (i.e. their product = −1) → the diagonals are
perpendicular → a rhombus
• both the slopes of the diagonals, and one pair of adjacent sides are negative reciprocals → a square
1
Finding the length of an altitude/perpendicular AH from a point A to a line BC
• Finding the equation h of a height/altitude from one point A to a line BC :
– Find the slope of BC
mBC =
yB − yA
xB − xA
– Perpendicular lines have negative reciprocal slopes (i.e. their product = −1), thus the slope of
1
the perpendicular h is − mBC
– The point A is on the perpendicular line h so it verifies its slope-intercept equation: yA =
1
− mBC
xA + n and we can find the y-intercept n.
• Finding the coordinates of the point H of intersection between the height/altitude AH and the line BC
from the system of equations:
yH = mBC xH + nBC , H on the BC line
1
xH + n
, H on the h line
yH = − mBC
• Given the coordinates of A(xA , yA ) and H(xH , yH ) the length of the altitude AH is given by the
distance between A and H
p
dAH = (xH − xA )2 + (yH − yA )2
Exercise 5. Consider the triangle ∆ABC with the vertices A(−2, −1), B(2, 0), C(2, 1).
1. Find the coordinates of the midpoint of BC
2. Find the length of the median
The coordinates of the midpoint of BC with B(xB , yB ) and C(xC , yC ) has coordinates
(2 + 2) 1
1
xB + xC yB + yC
,
=
,
= 2,
2
2
2
2
2
the length of the median AM is given by the distance between A(−2, −1), and M 2, 12 :
s
√
2 r
p
1
9
73
2
2
2
dAM = (xM − xA ) + (yM − yA ) = (2 − (−2)) +
− (−1) = 16 + =
2
4
2
Exercise 6. Consider the triangle ∆ABC with the vertices A(−3, 0), B(0, 2), C(3, 0).
1. Which type of triangle is the triangle ∆ABC. Find its area.
2. If we add the points E(3, 2) and F (−3, 2), are the points F, B and E collinear? Why?
3. Which type of figure is ACFE? Prove it.
1. It is an isosceles triangle : dAB = dBC since the Pythagorean Theorem gives us:
dAB =
p
p
p
√
√
(xB − xA )2 + (yB − yA )2 = 32 + 22 = 13, dBC = (xB − xC )2 + (yB − yC )2 = 13
Since BO is on Y axis and AC is on the x-axis BO ⊥ AC. In this case we cannot check their
parallelism using slopes. Why not ? Recall that the slopes of vertical lines ( i.e. with x constant)
are not defined since we cannot divide by zero.
p
In this case we might ”see” the length, but in a general case obtain: dOB = (xB )2 + (yB )2 = 2.
A simple computation gives Area of ∆ABC = BO×AC
2
2
2. The points E(3, 2) and F (−3, 2), B(0, 2) are collinear. They are all situated on the line y = 2
with slope 0. In a general case: find the equation of the line that goes, for example, through E
and B and see if the third point, F verifies its equation.
3. ACFE is a rectangle. First it is a parallelogram (i.e. a convex polygon with two pairs of parallel
lines) We can prove EF k AC either by observing that both are lines of constant y, which makes
them both parallel with axis OX, or by using the fact that parallel lines have equal slopes.
mAC =
yC − yA
= 0 = mEF
xC − xA
To prove AE k CF we cannot use the slopes. Why not? Because both lines are vertical, with an
equation with constant x, and their slopes are not defined (i.e. we cannot divide by zero in the
slope definition). So, we observe AE has the equation x=3, FC has the equation x=-3, thus they
are both parallel with the axis OY. By transitivity they are parallel to each other. Secondly,
ACFE has a right angle, which makes it a rectangle. Again, we cannot compute the slopes to
help us but we use the fact that the axes are perpendicular, and so are our lines.
Exercise 9. Consider the points A(0,-2), B(9,1), C(4,6), D(1,5) are the vertices of a polygon as shown on
the diagram.
1. Show that the polygon ABCD is a trapezoid. Is it isosceles? [Hint: A trapezoid is a convex polygon
with ONY one pair of parallel sides. Parallel lines have equal slopes.]
2. Find the length and midpoint of each of the parallel sides.
3. Find the height of the polygon ABCD and its area.
Exercise 2.1: Isosceles Trapezoid
1. We can prove AB k CD using the fact that parallel lines have equal slopes.
mAB =
yB − yA
1 − (−2)
3
1
yD − yC
5−6
−1
1
=
= = , mCD =
=
=
=
xB − xA
9−0
9
3
xD − xC
1−4
−3
3
The lines AC and BD are not parallel since the slopes mBD 6= mAC . Thus, our figure is only a
trapezoid. To prove that ABCD is an isosceles trapezoid we have to compare the lengths of its
non-parallel sides: AD and BC.
3
√
√
(1 − 0)2 + (5 − (−2))2 = 1 + 49 = 50
p
p
√
√
= (xC − xB )2 + (yC − yB )2 = (4 − 9)2 + (6 − 1)2 = 25 + 25 = 50
dAD =
dBC
p
(xD − xA )2 + (yD − yA )2 =
p
Indeed they are equal so ABCD is an isosceles trapezoid.
2. Finding a midpoint
of a segment
with known endpoints with endpoints (x1 , y1 ) and (x2 , y2 ) has
x1 + x2 y1 + y2
coordinates
,
. For AB with A(0, −2), B(9, 1) the midpoint M has coordinates
2
2 y
+y
x1 +x2
−2+1
, 1 2 2 = 0+9
= (4.5, −0.5). For CD with C(4, 6), D(1, 5) the midpoint N has
2
2 , 2
x1 +x2 y1 +y2
4+1 6+5
coordinates
,
=
= (2.5, 5.5).
2
2
2 , 2
The length of the parallel sides:
p
p
√
√
dAB = (xB − xA )2 + (yB − yA )2 = (9 − 0)2 + (1 − (−2))2 = 90 = 3 10
p
p
√
√
dCD = (xD − xC )2 + (yD − yC )2 = (1 − 4)2 + (5 − 6)2 = 9 + 1 = 10
3. The area of a trapezoid The area of a trapezoid is given by the formula
Area trapezoid =
(b + B)h
2
where b, B are the lengths of the two parallel sides of the trapezoid, also called bases, and h is
length of the altitude (height) which is the perpendicular distance between the two bases. We
have the bases lengths. What we miss is the altitude and its length. The power of coordinate
geometry is that it can transform a ”guess” into a proof very easily We saw in class
the lengthy classical geometrical proof of the fact that in an isosceles trapezoid the height
can be given by connecting the midpoints of the bases.
In coordinate geometry we just have to check our ”guess” using the slopes of the lines involved,
and base use the previous work of a mathematician who proved for us that if two lines AB,
and MN have their slopes as negative reciprocals, i.e. : mAB · mCD = −1, then the lines are
perpendicular on each other.
Checking:
mAB =
yB − yA
3
1
yM − yN
−6
1 − (−2)
−0.5 − 5.5
= = , mM N =
=
= −3
=
=
xB − xA
9−0
9
3
xM − xN
4.5 − 2.5
2
Indeed, mAB · mCD = 13 · (−3) = −1 We found the altitude and now we can compute its length
using our distance formula.
p
p
√
√
√
dM N = (xM − xN )2 + (yM − yN )2 = (2)2 + (−6)2 = 4 + 36 = 40 = 2 10
Now,
√
√
√
(b + B)h
(dAB + dCD ) · dM N
(3 10 + 10) · 10
Area trapezoid =
=
=
= 20units
2
2
2
What if ? Now, as a good mathematician we should ask ourselves: Was this just a lucky case, or is there
something more general to discover? Would coordinate geometry be proven to be so powerful, if we missed
these observations? Moreover, what if we did not have an isosceles trapezoid to start with?
The answer is yes. By mixing algebraic concepts with geometric concepts, coordinate geometry will
provide you with simpler solutions. However, the more observations you make, the more elegant will be
your solutions. Now, let us solve the height problem in case we missed both facts: that we have isosceles
trapezoid and the line connecting the midpoints of the bases is perpendicular on the bases.
4
(a) Exercise 11: Right Triangle 90,60,30
(b) Exercise 12: Right Triangle 90,45,45
Special Right Triangles
How to find the altitude(height) of a general trapezoid, knowing the coordinates of its vertices.
The definition of a height/altitude tells us that it is the perpendicular distance between the two bases. If
any perpendicular is good, we can pick a convenient one, for example one drawn from a vertex we know. Let
us pick D and consider the altitude (height) DH ⊥ AB. If AB is not a vertical line we can use the slopes to
obtain the equation of the line DH
mAB · mDH = −1 ⇒ mDH = −
1
1
= − 1 = −3
mAB
3
Now in its slope intercept form the DH equation is : y = mDH · x + n ⇒ y = −3x + n
(If AB vertical line: x=constant , DH would have been horizontal one : y=n)
Now n =? In order to be able to determine n we have chosen the particular case of the height from ”D”.
We know that the coordinates of D verify the equation of the line DH.
yD = −3xD + n ⇒ 5 = −3 + n ⇒ n = 8
Finally, DH equation in its slope intercept form is : y = −3x + 8. In order to find the length of the
altitude(height) we need to find the coordinates of ”H”. We know that H is the intersection point of the
lines AB and DH. Thus, it verifies both equations. The equation of AB: y = mA B · x + nAB ⇒ y = x3 + nAB
Similarly, since point A verifies AB equation nAB = −2 and the equation of AB becomes y = x3 − 2
Point H verifies: the equation of AB: yD = 13 · xD − 2 and the equation of DH: yD = −3 · xD + 8 t⇒
1
1
10
3 · xH − 2 = −3 · xH + 8 ⇒ 3 · xH + 3 · x = −2 + 8 ⇒ 3 · xH = 10 ⇒ xH = 3 ⇒ yH = −1 ⇒ H(3, −1). Now :
dDH =
p
(xD − xH )2 + (yD − yH )2 =
p
√
√
√
(1 − 3)2 + (5 − (−1))2 = 4 + 36 = 40 = 2 10
Exercise 11. Consider a right triangle with angles of 30◦ , 60◦ . If the size of the leg opposed to 30◦ -angle is
a, find the length of the other leg, and of the hypotenuse.
Draw a ∆ADC so that ∆ABC ≡ ∆ADC. Thus, their angles are equal: BD is a line (m∠ABC =
90◦ = m∠ACD), and m∠ADB = 60◦ and m∠DAB = 30◦ ⇒ ∆AB is a triangle with all angles of
60◦ . So, triangle ∆ABD is equilateral and hence, ⇒ BD ≡ BC ≡ CD = 2a.
√
Using the Pythagorean theorem, a2 + b2 = c2 ⇒ a2 + b2 = (2a)2 = 4a2 . Thus, b2 = 3a2 ⇒ b = a 3.
Exercise 12. Consider a right triangle with an angle of 45◦ . If the size of one leg is a, find the length of
the other leg, and of the hypotenuse.
2
2
2
Triangle ABC is isosceles. Hence, ⇒ AB
√ ≡ AC = a. Using the Pythagorean theorem, a + b = c ⇒
2
2
2
2
2
a + a = c . Thus, c = 2a ⇒ c = a 2.
5
2.2
From simple geometrical proofs to construction proofs: Reduce a problem to a known one
In the review package we used some simple geometrical facts. Let us recall their proofs.
1. The convex polygon ABCD is a parallelogram if and only if it has two opposite sides that are congruent
and parallel.
”=⇒”: Suppose ABCD is a parallelogram with ABkCD and ADkBC.
Connect two non-adjacent vertices,


BD

[
∆ABD ≡ ∆DEB ⇐=
mABD

(ASA)
 mDBC
[
say B and D.
≡ BD
(common)
[ alternate interior angles for AB k DC, BD transversal
= mCDB
[ alternate interior angles for BC k AD, BD transversal
= mBDA
Since corresponding parts of congruent triangles are congruent,
=⇒ AB ≡ CD and AD ≡ BC.
”⇐=”: Suppose ABkCD and AB≡CD. Sketch:
[ =
∆ABD ≡ ∆CDB(SAS). Since corresponding parts of congruent triangles are congruent, =⇒ mADB
[
mCBD, which are alternate interior for AD and BC and transversal BD =⇒ ADkBC, by the converse
of the postulate of parallels.
(a) Parallelogram
(b) Isosceles triangle and apex bisector
2. A triangle ∆ABC is an isosceles triangles if and only if it has two congruent base angles.
\]
[Hint: Construct the angle bisector of the angle BAC
6
”=⇒”: Suppose ∆ABC is an isosceles triangle, with AB ≡ AC.
[ (apex
Recall that with a protractor and a straight hedge we can construct the unique bisector of BAC
angle).
[ then ray AP and segment BC intersect in a unique point D and
If P is an interior point of angle BAC,
D is between B and C. (This statement can be proved as well. It is known as the Crossbar Theorem).

AD ≡ AD
(common)

[
[
∆ABD ≡ ∆ACD ⇐=
mBAD = mCAD (construction of bisector AD )

(SAS)
AB ≡ AC
(hypothesis: ∆ABC isosceles)
[ = mABD.
[
Since corresponding parts of congruent triangles are congruent, we obtain the result: mACD
”⇐=”: Similar proof.
3. Given an isosceles trapezoid ABCD where AD and BC are parallel and AB equals CD, the base angles
\ and CDA
\ are congruent.
BAD
[Hint: Assume BC is the shorter of the two parallel lines. Construct a line BE parallel to CD.]
”=⇒”: Suppose ABCD isosceles trapezoid, with AB ≡ CD and BC < AD .
(a) Isosceles trapezoid and parallelogram construction
Idea: construct something to reduce the problem we do not know how to solve to a known one; we
could thus use parallelograms and isosceles triangles. So, let us draw a ray BP parallel to CD. Since
BC is the smaller base, the ray BP is going to intersect AD in a unique point E. Since now BEkCD,
and BCkED, we have BCDE a parallelogram, and thus BE ≡ CD.
From the hypothesis, ABCD is an isosceles trapezoid with AB ≡ CD, so ∆ABE is now isosceles
(because AB ≡ BE).
[ = mBEA.
[ Also, mBEA
[ = 180−mBED
[ since E is on the line AD. Finally, 180−mBED
[ =
Thus, mBAE
[
[ =
mCDE since they are are same-side interior angles for the parallel lines BE and CD. Therefore, mBAE
[ that is mBAD
[ = mCDA,
[ and we are done.
mCDE,
”⇐=”: Similar proof.
7