-1- of of

Mem. Fae. Educ., Kagawa Univ. II, 57 (2007), 1-15

A Locus of the Orthocenter of a Pedal Triangle

- Instruction of Geometry by Use of a Drawing Game on a Display -

by

.Akiko

MATSUSHIMA,

Kazunori

FUITTA

and Hiroo

FUKAISHI

(Received November 30, 2006)

Abstract

An effective use of a computer makes mathematics classes much more interesting and motivates many students to learn. In this paper we present a computer game for drawing a locus of the orthocenter of the pedal triangle of a given triangle with two vertices fixed when one moves the third vertex along a distinguished curve. The drawing game provides an active teaching material of elementary geometry for students.

§

1. Introduction

In the sequel to [4] - [7] our study aims to develop a drawing game on a display as a teaching material for elementary geometry classes with activity using computers.

In elementary geometry we have five significant notions for a triangle; that is, the center of gravity, the center of an inscribed circle, the center of an escribed circle, the circumcenter and the orthocenter of a triangle.

Which curve is drawn as a locus of such a point of the pedal triangle of a triangle with two vertices fixed on the plane when the third vertex moves under a certain condition?

In this study we limit ourselves to the case of a locus of the orthocenter of a

A. MATSUSHIMA, K. FuJITA and H. FUKAISHI

moves along a distinguished curve. Then our main concern is to find various types of remarkable curves with a simple expression as a locus of the orthocenter of a pedal triangle.

In this paper we present a computer game for students to find a solution to the above problem with fun, and obtain some results from our various experiments. In fact,

"Mathematical Discovery" due to G. Polya [11] indicates the spirit of our research in mathematics education.

For terminology of geometry throughout the paper, consult [2], [3] and [10].

As for the present paper, the first author gave a general idea for drawing a locus of the ortho~enter of a pedal triangle of a given triangle on a display and wrote the programs f~r all of the figures under the direction of the second author. The last author joined the discussions and arranged the results for publishing.

§

2. A program for drawing a locus

Let us consider a triangle 6.ABC given on a plane.

Let A= (xa, Ya), B

=

(xb, Yb), C

=

(xe, yJ. Then the coordinates of the orthocenter H of 6.ABC is given by

where

PI= Xb-Xe, qi= Yb- Ye, r1

=

Xa(Xb-Xe)

+

Ya(yb- Ye), P2=Xa-Xe, q2=Ya-Ye, r2 =xb(Xa-Xe)+yb(ya-Ye).

Let D, E, F be the feet of the perpendiculars from the vertices A, B, C to the opposite sides, respectively. The 6.DEF is called the pedal triangle of 6.ABC.

Denote the orthocenter of 6.DEF by K.

Which curve is drawn as a locus of the orthocenter K of the pedal triangle ,6,.DEF of 6.ABC with the vertices B, C fixed when the third vertex A moves along a distinguished curve CC ?

We divide our operation of a drawing game into the drawing part and the printing part, due to the circumstances of our computer machines.

PART ONE : To draw a locus of the orthocenter of a pedal triangle of a given triangle.

A program of our drawing game for Example 4 in § 3 is written in Visual Basic

A Locus of the 0rthocenter of a Pedal Triangle

Ver'. 6.0 by Microsoft Corporation and consists of the following seven steps (see, List 1).

Step 1. Set the coordinates axes and two fixed point B, C in black and a curve CC in blue.

Step 2. Set the initial position of a vertex A of a triangle LABC and draw each side in black.

Step 3. Plot the orthocenter H of LABC in red.

Step 4. Draw each of the perpendiculars from the vertices of LABC to the opposite sides with a broken line in blue.

Step 5. Draw the sides of the pedal triangle LDEF of LABC in magenta and plot its orthocenter K in indigo-blue.

Step 6. When LABC or LDEF is obtuse, extend each of two sides adjacent at the obtuse angle with a broken line in green.

Step 7. When one moves the vertex A continuously along the curve CC by the mouse, the orthocenter K of LDEF continuously draws a locus

.Z

with a solid line in indigo-blue.

PART Two: To process a bitmap file (bmp) by p0TEX2_{8 ,} to exhibit it on another display, and to print it.

OUTLINE of the PROGRAM. Let B (- 1, 0) , C ( 1, 0) be the fixed vertices of a triangle on the plane and let CC be the circle given by the equation

x2 +

(y- 1)2

=

5. When a point A (u, v) on CC is selected by the mouse, the orthocenter K(x, y) is determined by the formula ( *) with D

=

(xa, ya), E= (xh, Yb), F= (xc, ye).

The program for drawing a locus

.Z

of the point K is given in List 1. In this case we will have the ellipse .5t' : x²

+ (

y

2

)

=

1 on a display (Fig. 4).

3v5

²

5

We keep the same notation for the rest of the paper.

A. MATSUSHIMA, K. FUJITA and H. FuKAISHI

§

3. Curves obtained as a locus of the orthocenter of a pedar triangle

As a locus of the orthocenter of a pedal triangle of a given triangle we have various known curves. In this section we exhibit the cases of circles and ellipses.

Proposition 1. A circle can be obtained as a locus of the orthocenter of the pedal triangle of LABC when the third vertex A moves along a straight line.

Example 1. (Fig. 1) Let B (-1, 0), C (1, 0) and

CC :

^y =x+ ^1. Then we have the circle

!£ :

x²

+

y²

=

1 as a locus of the orthocenter of the pedal triangle LDEF of LABC when the third vertex A moves along

CC .

Proof Let A(u, v) and K(x, y), while B=(-1, 0) = (-a, -/3), C= (1, 0) = (a, -/3) are fixed. Then we have the following :

v+

/3

AC: y

+ /3 =

^{u _ a (x - a),}

and

Hence

£=(

^{- (u- 1)2} _{(u- 1)2 +v}^{+v2 2} ' (u-1)2+v^{-2v(u-1) 2}

^),

F=(

^{(u + 1)}_{(u+ 1)2+v}^{2 - v2 2 '} _{(u + 1)}^{2v(u 2+} ₊ ¹⁾ _v²

^),

smce a

=

1, /3

=

0.

Find the Groebner bases (see [1], [9]) by applying Mathematica by Wolfram Research, Inc,. to the following code, where D = (xa, Ya), E = (xb, Yb), F = (xc, ye) :

A Locus of the Orthocenter of a Pedal Triangle

fO:=v-u-1 f1 :=xa-u f2:=ya

f3:=xb-(v-2-(u-1)-2)/((u-1)-2+v-2) f4:=yb+2*(u-1)*v/((u-1)-2+v-2) f5:=xc-((u+1)-2-v-2)/((u+1)-2+v-2) f6:=yc-2*(u+1)*v/((u+1)-2+v-2) f7:=p1-(xc-xb)

f8:=q1-(yc-yb)

f9:=r1-((xc-xb)*xa+(yc-yb)*ya) f10:=p2-(xa-xc)

f11:=q2-(ya-yc)

f12:=r2-((xa-xc)*xb+(ya-yc)*yb) f13:=(p1*q2-p2*q1)*x-(r1*q2-r2*q1) f14:=(p1*q2-p2*q1)*y-(p1*r2-p2*r1)

GroebnerBasis[{fO,f1,f2,f3,f4,f5,f6,f7,f8,f9,f10,f11,f12,f13,f14}, {xa,ya,'xb,yb,xc,yc,p1,p2,q1,q2,r1,r2,v,u,y,x}]

Factor[%]

Then we obtain the term

(-1 +u)u(-1 +x²+y²⁾ in the list of the generating Groebner bases.

Therefore, we have or

Proposition 2. A certain known curve can be obtained as a locus of the orthocenter of the pedal triangle of .h..ABC when the third vertex A moves along a straight line.

Example 2. (Fig. 2) Let B (-1, O), C (1, 0) and

CC:

^x

=

2. Then we have the - -(x-l)(x-2)²

curve ~ :

y2 = - - - -

as a locus of the orthocenter of the pedal triangle of

X

6ABC when the third vertex A moves along CC .

The curve ~ is given as Example 43 in [12, P. 65; P. 67 for its graph].

Proposition 3. An ellipse can be obtained as a locus of the orthocenter of the pedal triangle of 6ABC when the third vertex A moves along a circle.

A. MATSUSHIMA, K. FuJITA and H. FUKAISHI

Example 3. (Fig. 3) Let B (-/o.75, -0.5), C (✓0.75, 0.5) and

X 2

(y+½f

Then we have the ellipse

.!t?: (

~

)2 ^{+ (}

~

r

^{= 1} as a locus of the orthocenter of the pedal triangle of MBC when the vertex A moves along ~.

Proposition 4. Let B ( - 1, 0), C(l, 0) and ~: x²

+

(y-b )²

=

1

+

b² a circle where b is a real constant.

2 2

5£' : ( ; )²

+ ( (

b

+

1

/c

^{b - 1) ) 2}

⁼

^l

✓b²+ 1 ✓b²+ 1 Then we have the ellipse

as a locus of the orthocenter of the pedal triangle of MBC when the third vertex A moves along ~-

Proof Under the same notation as in Proof of Example 1, find the Groebner bases (see [1], [9]) by applying Mathematica to the following code :

f0:=u-2+(v-b) ... 2-(1+b-2) f1 :=xa-u

f2:=ya

f3:=xb-(v ... 2-(u-1) ... 2)/((u-1) ... 2+v-2) f4:=yb+2*(u-1)*v/((u-1)---2+v-2) f5:=xc-((u+1)-2-v-2)/((u+1)-2+v-2) f6 :=yc-2*(u+1)*v/( (u+1) ... 2+v ... 2) f7:=p1-(xc-xb)

f8:=q1-(yc-yb)

f9:=r1-((xc-xb)*xa+(yc-yb)*ya) f10:=p2-(xa-xc)

f 11: =q2-(ya-yc)

f12:=r2-((xa-xc)*xb+(ya-yc)*yb) f13:=(p1*q2-p2*q1)*x-(r1*q2-r2*q1) f14:=(p1*q2-p2*q1)*y-(p1*r2-p2*r1)

GroebnerBasis[{fO,f1,f2,f3,f4,f5,f6,f7,f8,f9,f10,f11,f12,f13,f14}, {xa,ya,xb,yb,xc,yc,p1,p2,q1,q2,r1,r2,v,u,y,x}]

Factor[%]

A Locus of the Orthocenter of a Pedal Triangle

Then we obtain the term

in the list of the generating Groebner bases.

Therefore, we have

that is,

x2 y2

( 2 )² + ((b+l)(b-1))² =1.

•

✓h²+ 1 ✓h²+ 1

Example 4. (Fig. 4) For b

=

2 in Proposition 4 we have the ellipse

.Z:

2

x ²

+ (

3

~

)' 1 as a locus of the orthocenter of the pedal triangle of MBC when the vertex A moves along ~.

§

4. Questions

Question 1. Can all conics be obtained as a locus of the orthocenter of the pedal triangle of a triangle with two vertices fixed when the third vertex moves along a distinguished curve with a simple expression?

Question 2. Can various known curves such as listed in a text [8; pp. 520-521] be obtained as a locus of the orthocenter of the pedal triangle of a triangle with two vertices fixed when the third vertex moves along a distinguished curve with a simple expression?

A. MATSUSHIMA, K. ^FuJITA and ^{H. FuKAISHI}

References

[ 1 J W. W. Adams and P. Loustaunau : An Introduction to Grabner Bases, Graduation Studies in Math., Vol. 3, Amer. Math. Soc., Providence, 1994.

[ 2 J·H. S. M. Coxeter: Introduction to Geometry, 2nd ed., John Wiley and Sons Inc, New York, 1980. 〔コクセタ^ー （銀林 浩・訳）： 幾何学入門 第2版， 明治図書， 東京，

1982.〕

[ 3 J H. S. M. Coxeter and S. L. Greitzer : Geometry Revised, New Mathematical Library, Number 6, School Mathematics Study Group, Random House, Inc., NewYork, 1967.

〔コ^ークスタ^ー， グレイツァ^ー （寺阪英孝・訳）： 幾何学再入門， SMSG双書河

出書房新社， 東京， 1970.〕

[ 4 ] K. Fujita, A. Matsushima and H. Fukaishi : A Locus of the Orthocenter of a Triangle - Instruction in Geometry by a Moving Locus on a Computer, Mem. Fae. Educ, Kagawa

University II, 55 (2005), 1-13.

[ 5 ] K. Fujita, A. Matsushima and H. Fukaishi : A Triangle with Three Distinguished Collinear Points - Instruction of Geometry by Use of a Drawing Game on a Display, Mem. Fae. Educ., Kagawa University II, 55 (2005), 25-41.

[ 6 ] K. Fujita, A. Matsushima and H. Fukaishi : A Triangle with Distinguished Concyclic Points - Instruction of Geometry by Use of a Drawing Game on a Display, Mem. Fae.

Educ, Kagawa University II, 56 (2005), 1-25.

[ 7 ] H. Fukaishi, K. Fujita and A. Matsushima : Alternative Geometric Proofs of Theorems for Concyclic Points for a Triangle, Mem. Fae. Educ, Kagawa University II, 56 (2006), 51-59.

[ 8 J ^一松 信•他： 新数学事典， 大阪書籍， 大阪， 1979. 〔= S. Hitotumatu et al. : Shin-Sugaku-Jiten, Osaka-shoseki, 1979.〕

[ 9 J ^一松 信： 代数学入門 第三課， 近代科学社， 東京， 1994. 〔= S. Hitotumatu : Introduction to Algebra, The Third Lesson, Kindai-kagaku-sha, Tokyo, 1994.〕

[10] D. Pedoe : . Geometry, Dover Puhl. Inc., New York, 1970.

[11] G. Polya: Mathematical Discovery on Understanding, Learning, and Teaching Problem Solving, Vol. 1 - 2, John Wiley and Sons, Inc., New York, 1962. 〔ポリア（柴垣和 三雄， 金山靖夫・訳） ： 数学の問題の発見的解き方， I - II, みすず書房， 東京，

1964.〕

[12]坂井 忠次：グラフと追跡， 培風館， 東京，1963. 〔= C. Sakai : Gurafu to Tsuiseki, Baifukan, Tokyo, 1963.〕

A Locus of the Orthocenter of a Pedal Triangle

Akiko MATSUSHIMA

Okayama Prefectural Goverment

2-4-6 Uchisange, Okayama, 700-8570, JAPAN

Kazunori FUJITA

Department of Mathematics, Faculty of Education, Kagawa University 1-1 Saiwai-cho, Takamatsu-shi, Kagawa, 760-8522, JAPAN

, E-mail address : fujita@ed.kagawa-u.ac.jp

Hiroo FUKAISHI

Department of Mathematics, Faculty of Education, Kagawa University 1-1 Saiwai-cho, Takamatsu-shi, Kagawa, 760-8522, JAPAN

E-mail address : fukaishi@ed.kagawa-u.ac.jp

Fig. 3

A MATSUSHIMA, K. FuJITA and H. FuKAISHI

Fig. 1 ~ : y = x + 1 .2: x2+y2= 1

~ : x2+y2= 1

(Y++ y

.2: ^x2

+

( ^~ y (+Y

^=l

B

Fig. 2 ~: x = 2

.2: y2 = -

---.. \,H

(x-1) (x-2)²

X

Fig.4 ~ : x 2+ (y-2)2=5 .2: x2+ y2

1 (

3

f ^Y

A Locus of the Orthocenter of a Pedal Triangle

List 1. A program for drawing a locus

Specifying statements of the general variables Dim sx As Double, ex As Double

Dim sy As Double, ey As Double Dim ax As Double, ay As Double Dim bx As Double, by As Double Dim ex As Double, cy As Double Dim xd As Double, yd As Double Dim xe As Double, ye As Double Dim xf As Double, yf As Double Dim k As Integer

Dim ra{4) As Double

Drawing the initial figure Private Sub Form_Activate{)

Forml .AutoRedraw = ^True

Forml .Line {sx, 0)-(ex, 0) Form 1 .Line (0, sy)-(0, ey)

Form 1 .Circle (0, 2), Sqr(S), vbBlue Forml .Line (bx, by)-(cx, cy) Forml .DrawMode

=

vbNotXorPen Forml .Line (ax,'ay)-(bx, by) Forml .Line (ex, cy)-(ax, ay)

Orthocenter ax, ay, bx, by, ex, cy, 12, 2, l Pedal_triangle ax, ay, bx, by, ex, cy, 1 3 End Sub

Setting the form, the coordinate axes and the initial values of the coordinates of each vertex of a triangle Private Sub Form_Load()

sx = ^-4.2

ex= 4.2 wx

=

ex - sx

wy

=

^{wx *} Forml .ScaleHeight / Forml .ScaleWidth sy = -0.3 * wy

ey = 0.7 * wy

Form l .Scale (sx, ey)-(ex, sy) Forml .BackColor = ^vbWhite

Forml .DrawWidth = ¹

ax = 0: ay = 2 + Sqr(S) bx= -1 -

by= 0 ex= 1 cy

=

0 k

=

l End Sub

A. MATSUSHIMA, K. FuJITA and H. FuKAISHI

}

Action corresponding to the left button of mouse

Private Sub Form_MouseDown(Button As Integer, Shift As Integer, X As Single, Y As Single) If k = 2 Then

k=l Exit Sub End If k=2

dl =Sqr((ax-X)A2+(ay-Y)A2) If dl < 0.5 Then

Form 1 .Line (ax, ay)-(bx, by) Form 1 .Line (ex, cy)-(ax, ay) r = Sqr(X A 2 + (Y - 2) A 2)

X = Sqr(S) * X / r: Y = Sqr(S) * (Y - 2) / r + 2 Form 1 .Line (X, Y)-(bx, by)

Forml .Line (ex, cy)-(X, Y)

Orthocenter ax, ay, bx, by, ex, cy, 12, 2, 1 Orthocenter X, Y, bx, by, ex, cy, 1 2, 2, 1 PedaL.triangle ax, ay, bx, by, ex, cy, 1 3 Pedal_triangle X, Y, bx, by, ex, cy, 13 ax= X: ay = Y

End If End Sub

Action corresponding to the movement of mouse

Private Sub Form_MouseMove(Button As Integer, Shift As Integer, X As Single, Y As Single) If k = 1 Then Exit Sub

d 1 = Sqr((ax "'" X) A 2 + (ay - Y) A 2) If dl < 0.5 Then

Form 1 .Line (ax, ay)-(bx, by) Form 1 .Line (ex, cy)-(ax, ay) r = Sqr(X A 2 + (Y - 2) A 2)

X = Sqr(S) * X / r: Y = Sqr(S) * (Y - 2) / r + 2 Form 1 .Line (X, Y)-(bx, by)

Form l .Line (ex, cy)-(X, Y)

,Qrthocenter ax, ay, bx, by, ex, cy, 12, 2, 1 Orthocenter X, Y, bx, by, ex, cy, 1 2, 2, 1 Pedal_triangle ax, ay, bx, by, ex, cy, 1 3 Pedal_triangle X, Y, bx, by, ex, cy, 13 ax= X: ay = Y

End If End Sub

Presenting of the orthocenter of a triangle

Private Sub Orthocenter(jxa, jya, jxb, jyb, jxc, jyc, pcl, eel, beaf) xa

=

jxa: ya = jya

A Locus of the Orthocenter of a Pedal Triangle

xb = jxb: yb = jyb xc = jxt: ye = jye

ux = xb - xa: uy = yb - ya vx = xa - xc: vy = ya - ye wx = xc - xb: wy = ye - yb a = Sqr(wx A 2 + wy A 2) b = Sqr(vx A 2 + vy A 2) e = Sqr(ux A 2 + uy A 2) ah = b A 2 + e A 2 - a A 2 bh = c A 2 + a A 2 - b A 2 ch = a A 2 + b A 2 ~ c A 2

If ah>= 0 And bh >= 0 And ch>= 0 Then pl = xb * ye - xc * yb: ql = -wy * ya - wx *. xa p2 = xc * ya - xa * ye: q2 = -vy * yb - vx * xb p3 = xa * yb - xb * ya: q3 = -uy * ye - ux * xc xd = (p 1 *·wy - q l * wx) / a A 2

yd = (-pl * wx - q l * wy) / a A 2 xe = (p2 * vy - q2 * vx) / b A 2 ye = (-p2 * vx - q2 * vy) / b A 2 xf = (p3 * uy - q3 * ux) / c A 2 yf = (-p3 * ^{UX -} q3 * uy) / C A 2

xh = (q2 * wy - ql * vy) / (wx * vy - wy * vx) yh = (ql * vx - q2 * wx) / (wx * vy- wy * vx) Form l .DrawStyle = 2

Forml .Line (xa, ya)-(xd, yd), vbBlue Forml .Line (xb, yb)-(xe, ye), vbBlue Forml .Line (xe, yc)-(xf, yf), vbBlue Forml .DrawStyle = 0

Form 1 .DrawWidth = 5 If beaf = 2 Then

Forml .DrawMode = vbCopyPen Forml .PSet (xh, yh), QBColor(pcl) Forml .DrawMode = vbNotXorPen Else

Forml .PSet (xh, yh), QBColor(pcl) End If

Form 1 .DrawWidth = l Else

If bh < 0 Then

dm = xa: xa = xb: xb = dm dm = ya: ya = yb: yb = dm End If

If ch< 0 Then

dm = xa: xa = xc: xc = dm dm = ya: ya = ye: ye = d m End If

ux = xb - xa: uy = yb - ya vx = xa - xc: vy = ya - ye wx = xc - xb: wy

=

ye - yb d =WX*vy-wy*VX If Abs(d) > 0.001 Then

a = Sqr(wx A 2 + wy A 2)

A. MATSUSHIMA, K. FUJITA and H. FuKAISHI

b = Sqr(vx A 2 + vy A 2) c = Sqr(ux A 2 + uy A 2)

p 1 = xb * ye - xc * yb: q 1 = -wy * ya - wx * xa q2 = -vy * yb - vx * xb

xd = (p 1 * wy - q 1 * wx) / a A 2 yd = (-pl * wx - ql * wy) / a A 2 xh=(q2*wy-q1 *VV)/d yh = (ql * vx - q2 * ~x) / d H_Prolong xb, yb, xa, ya, eel H_Prolong xc, ye, xa, ya, eel Forml .DrawStyle = 2

Forml .Line (xh, yh)-(xd, yd), vbBlue Forml .Line (xb, yb)-(xh, yh), vbBlue Forml .Line (xc, yc)-(xh, yh), vbBlue Forml .DrawStyle = 0

Forml .DrawWidth = 5 If beaf = 2 Then

Form 1 .DrawMode = vbCopyPen Form 1 .PSet (xh, yh), QBColor(pel) Form 1 .DrawMode = vbNotXorPen Else

Form 1 .PSet (xh, yh), QBColor(pel) End If

Forml.DrawWidth = 1 End If

End If End Sub

Extending of two sides adjacent at the vertex with an obtuse angle Private Sub H_Prolong(xl, yl, x2, y2, eel)

Form 1 . DrawWidth = 1

I = Sqr((xl - x2) A 2 + (yl - y2) A 2) ra(l) = Sqr((xl - sx) A 2 + (yl - sy) A 2) ra(2) = Sqr((xl - ex) A 2 + (yl - sy) A 2) ra(3) = Sqr((x l - ex) A 2 + (yl - ey) A 2) ra(4) = Sqr((xl - sx) A 2 + (yl - ey) A 2) r = ra(l)

For i = 2 To 4

If ra(i) > r Then r

=

ra(i) Next i

p

=

r / I

tx = (1 - p) * xl + p * x2 ty = (1 - p) * yl + p * y2 Forml.DrawStyle = 2

Forml .Line (x2, y2)-(tx, ty), QBColor(eel) Forml .DrawStyle = 0

End Sub

A Locus of the Orthocenter of a Pedal Triangle

Presenting of a pedal triangle

Private Sub PedaL.triangle(xa, ya, xb, yb, xc, ye, col) ux

=

xb - xa: uy = yb - ya

vx

=

xa - xc: vy

=

^{ya - ye}

wx

=

^{xc - xb: wy} = ^{ye - yb}

a = Sq r(wx A 2 + wy A 2) b = ^{Sqr(vx A 2} + vy A 2) c = ^{Sqr(ux A 2} + uy A 2) ah = ^{b A} 2 + c A 2 - a A 2 bh

=

c ^A 2 + a ^A 2 - b A 2 ch

=

^{a A 2} + b A 2 - c A 2

p 1

=

xb * ye - xc * yb: q 1

=

-wy * ya - wx * xa p2

=

^xc * ya - xa * ye: q2 = ^-vy * yb - vx * xb p3

=

xa * .vb - xb * ya: q3

=

-uy * ye - ux * xc xd

=

(p 1 * ^{wy - q} 1 * wx) / a A 2

yd

=

(-p 1 * wx - q 1 * wy) / a A 2 xe

=

(p2 * vy - q 2 * vx) / b A 2 ye

=

(-p2 * vx - q2 * vy) / b A 2 xf

=

(p3 * uy - q 3 * ux) / c A 2 yf

=

(-p 3 * U X - q 3 * uy) / C A 2 Forml .DrawWidth = 5

Forml .PSet (xd, yd), QBColor(col) Form 1 .PSet (xe, ye), QBColor(col) Form 1 .PSet (xf, yf), QBColor(col) Forml .DrawWidth = ²

Form 1 .Line (xd, yd)-(xe, ye), QBColor(col) Form 1 .Line (xe, ye)-(xf, yf), QBColor(col) Forml.Line (xf, yf)-(xd, yd), QBColor(col) Forml .DrawWidth

=

1

Orthocenter xd, yd, xe, ye, xf, yf, 9, 10, 2 End Sub