﻿ Coordinate Geometry Proofs Worksheet | Problems & Solutions
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Coordinate Geometry Proofs Worksheet
• Page 1
1.
Prove that the diagonals of a parallelogram bisect each other.

#### Solution: Let OABC be the parallelogram.

Coordinates of A are (a, 0)

Coordinates of B are (b, c)

OA = CB
OA || CB
OA = a

y-coordinate of C = c
[OA || CB.]

x-coordinate of C = x - coordinate of B - a = b - a.

Coordinates of C are (b - a , c)

Midpoint of OB = (b2, c2)

Midpoint of AC = (a + b - a2, 0 + c2) = (b2, c2)

Midpoint of OB and mid point of AC are same.

So, the diagonals bisect each other.

Correct answer : (0)
2.
Prove that the diagonals of a rectangle are equal and bisect each other.

#### Solution: Let P(a, b) and Q (c, b) and R (c, d) be the three vertices of the rectangle.

The fourth vertex S is (a, d)
[PQ = RS, PQ || RS.]

PR = (a-c)2+(b-b)2 = (a - c)

RS = (a-c)2+(d-d)2 = (a - c)

The diagonals are equal.

Midpoint of PR = (a+c2, b+d2)

Midpoint of QS = (a+c2, b+d2)

The diagonals are equal and bisect each other.

Correct answer : (0)
3.
Prove that the midpoints of the sides of a quadrilateral form a parallelogram.

#### Solution: Let the four vertices of the quadrilateral be A(p, q), B(r, s), C(t, u) and D(v, w).

Midpoint M of AB = (p+r2, s+q2)

Midpoint N of BC = (r+t2, s+u2)

Midpoint O of CD = (t+v2, u+w2)

Midpoint P of AD = (p+v2, q+w2)

MN = (p+r2-r+t2)2+(s+q2-s+u2)2
= (p-t2)2+(q-u2)2

OP = (t+v2-p+v2)2+(u+w2-q+w2)2
= (t-p2)2+(u-q2)2 (p-t2)2+(q-u2)2

MN = OP
[From step6 and step7.]

ON = (t+v2-r+t2)2+(u+w2-s+u2)2
= (v-r2)2+(w-s2)2

PM = (p+r2-p+v2)2+(s+q2-q+w2)2
= (r-v2)2+(s-w2)2 (v-r2)2+(w-s2)2

ON = PM

Opposite sides are congruent

PQRS is a parallelogram.

Correct answer : (0)
4.
Prove that the mid points of the sides of a rectangle form a rhombus.

#### Solution: The rectangle is as shown.

Coordinates of M = (a+c2, b)

Coordinates of N = (c, b+d2)

Coordinates of O = (a+c2, d)

Coordinates of P = (a, b+d2)

MN = (a+c2-c)2+(b -b+d2)2
= (a-c2)2+(b-d2)2

OP = (a+c2-a)2+(d -b+d2)2
= (c-a2)2+(d-b2)2 (a-c2)2+(b-d2)2

MN = OP

ON = (a+c2-c)2+(d -b+d2)2
= (a-c2)2+(d-b2)2

MP = (a+c2-a)2+(b -b+d2)2
= (c-a2)2+(b-d2)2 (a-c2)2+(d-b2)2

ON = PM

MN = OP = ON = PM.

All the sides are equal.

MNOP is a rhombus.

Mid points of teh rectangle form a rhombus.

Correct answer : (0)
5.
Prove that the midpoints of a rhombus form a rectangle.

#### Solution: Let P(a, b) and Q(c, d) be the two vertices of the rhombus.

Coordinates of O are (a, d)
OQ = (c - a)
OP = (d - b)

Coordinates of R are (a, d + d - b) = (a, 2d - b).

Coordinates of S are (a - c -a, d) = ((2a - c), d)

Coordinates of A = (a+c2, b+d2)

Coordinates of B = (a+c2, d +2d-b2)

Coordinates of C = (2a-c+a2, 2d-b+d2) = (3a-c2, 3d-b2)

Coordinates of D = (2a-c+a2, b+d2) = (3a-c2, b+d2)

AB = (b+d2-3d-b2)2 = (b - d)

CD = (3d-b2-b+d2)2 = (b - d)

AD = (a+c2-3a-c2)2 = (c - a)

BC = (a+c2-3a-c2)2 = (c - a)

y-coordinates of A and D are same
[AD || x-axis.]

x-coordinates of A and B are same
[AB || y-axis.]

AD and AB are perpendicular.

ABCD is a rectangle.

Correct answer : (0)
6.
Prove that the diagonals of a kite are perpendicular.

#### Solution: Adjacent sides AB = AD , CB = CD.

AB =(p - r)2+(q - s)2

AD =(p - v)2+(q - w)2

CD =(t - v)2+(u - w)2

BC =(t - r)2+(u - s)2

(p-r)2 + (q-s)2 = (p-v)2 + (q-w)2
[Step1,2, 3, square the sides.]

r² + p² - 2pr + s² + q² - 2qs = v² + p² - 2pv + w² + q² - 2qw

r² + s² - v² - w² = 2pr + 2qs - 2pv - 2qw.
[Simplify.]

(t-v)2 + (u-w)2 = (t-r)2 + (u-s)2

v² - 2tv + w² - 2uw = r² - 2tr + s² - 2us.
[Simplify.]

r² + s² - v² - w² = 2tr + 2us - 2tv - 2uw

pr +qs -pv - qw = tr + us - tv - uw
[Steps 7 and 11.]

p(r -v) + q(s - w) = t(r - v) + u(s - w)

(r - v)(p - t) = - (s - w)(q -u)

(q-u)(s-w)(p-t)(r-v) = - 1

Slope of AC = q-up-t

Slope of BD =s-wr-v

Slope of AC × slope of BD = - 1.

AC ^ BD

Correct answer : (0)
7.
Prove that diagonals of a rhombus are $\perp$ to another.

#### Solution:

The given figure KL = KN.

The length of sides are (a-c)2+(b-d)2 = (a-g)2+(b-h)2

a² + c² - 2ac + b² + d² - 2bd = a² + g² - 2ag + b² + h² - 2bh

= c² + d² - g² - h² = 2ac + 2bd - 2ag - 2bh.
[Simplify.]

And also MN = ML.

(e-g)2+(f-h)2 = (e-c)2+(f-d)2

e² + g² - 2ge + f² + h² - 2fh = e² + c² - 2ec + f² + d² - 2fd

g² + h² - c² - d² = 2ge + 2fh - 2ec - 2fd
[Simplify.]

c² + d² - g² - h² = 2ec + 2fd - 2ge - 2fh

[From step 4 and step 9.]
ac + bd - ag - bh = ec + fd - ge - fh

a(c - g) + b (d - h) = e( c - g) + f(d - h).

(a - e)(c - g) = (d - h)(f - b)

f - ba - e = c - gd - h.

(f - b)-(c - a) = c - gd - h

(f - b)(c - a) × (d - h)(c - g) = - 1

Slope of KM × Slope of LN = - 1.

KM LN.

Correct answer : (0)
8.
Prove that the midsegment of a trapezoid is equal in length to the average of the length of the bases.

#### Solution: Let ABCD be the trapezoid.

Let the coordinates of A(- a, 0), B(a, 0), C(b, c) and D(- b , c).

Let E and F be the midpoints of AD and BC.

Coordinates of E = (-(a + b)2, c2).
[Midpoint form.]

Coordinates of F are (a + b2, c2)
[Midpoint form.]

Length of EF = (a + b2-(-a+b2))2 = a + b

Length of AB = (a-(-a))2+02 = 2a.

Length of CA = (b-(-b))2+02 = 2b.

So, length of EF = 1 / 2[AB + CB].
[From 6, 7 and 8.]

So, the midsegment of a trapizoid is equal in length to the average of the lengths of bases.

Correct answer : (0)
9.
The vertices of a parallelogram are A(3, 1), B(7, 1), C(10, 5) and D(6, 5). If P, Q, R and S are the midpoint of the sides of $\stackrel{‾}{\mathrm{AB}}$, $\stackrel{‾}{\mathrm{BC}}$, $\stackrel{‾}{\mathrm{CD}}$ and $\stackrel{‾}{\mathrm{AD}}$ respectively. Show that $\stackrel{‾}{\mathrm{PQ}}$ || $\stackrel{‾}{\mathrm{RS}}$ and $\stackrel{‾}{\mathrm{RS}}$ || $\stackrel{‾}{\mathrm{QR}}$.

#### Solution: Coordinates of P are (5, 1).

Coordinates of Q are (8.5, 3).

Coordinates of R are (8, 5).

Coordinates of S are (4.5, 3)

Slope of PQ = 3 - 18.5 - 5 = 23.5.

Slope of RS = 5 - 38 - 4.5 = 23.5.

Slope of PQ and RS are equal.
PQ || RS.

Slope of PS = 3 -14.5 - 5 = - 20.5 = - 4.

Slope of QR = 5 -38 - 8.5 = - 20.5 = - 4.

Slope of PS = Slope of QR.

PS || QR

Correct answer : (0)
10.
Prove that the diagonals of the rhombus bisects each other.

#### Solution: Let the vertices of the rhombus be as shown figure.

Let LK = LM = MN = KN = R.

LK² = LM² = MN² = KN² .

(a - c)2 +(b - d)2 = (a - g)2 +(b - h)2 = (g - e)2 +(h - f)2 = (e - c)2 +(f - h)2

Slope of KL = Slope of NM.

b - ha - g = d - fc - e a - gb - h = c - ed - f

(b-h)2(a - c)2 = (d - f)2(g - e)2
[Squaring on both sides.]

(b-h)2+(a-c)2(a - c)2 = (d - f)2+(g-e)2(g - e)2

R2(a - c)2 = R2(g - e)2

a - c = g - e a + e = g + c.

(a - g)2(b - h)2 = (c - e)2(d - f)2

(a-g)2+(b-h)2(b-h)2 = (c-e)2+(d-f)2(d-f)2

R2(b - h)2 = R2(d - f)2

b - h = d - f b + f = d + h

Midpoint of KM is (a+c2 , b+f2)

Midpoint of LN is (g+c / 2, d+h / 2)

Since a + c = g + c and b + f = d + h,

Midpoint of KM and midpoint of LN coinside.

(i.e)., LN and KM meet at the midpoints.

(i.e)., The diagonals bisects each other.

Correct answer : (0)

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