Document Details

PlushEuphemism4494

Uploaded by PlushEuphemism4494

كلية التربية قسم الرياضيات

Tags

coordinate geometry cartesian coordinates polar coordinates mathematics

Summary

This document provides an introduction to coordinate geometry, covering Cartesian and polar coordinates, transformations, as well as examples of coordinate transformations.

Full Transcript

# Chapter (1): Coordinates of Points ## CHAPTER (1) ### COORDINATES OF POINTS *** ### (1) Cartesian Coordinates: Any points P in the plane XOY has two components (x, y); where * (i) x is the distance from y axis * (ii) y is the distance from x axis as in figure (1) ### (2) Polar Coordinates...

# Chapter (1): Coordinates of Points ## CHAPTER (1) ### COORDINATES OF POINTS *** ### (1) Cartesian Coordinates: Any points P in the plane XOY has two components (x, y); where * (i) x is the distance from y axis * (ii) y is the distance from x axis as in figure (1) ### (2) Polar Coordinates: The Polar Coordinates of the point P is (r, θ), where * (i) r is the distance OP * (ii) θ is the angle which makes OP with the +ve direction for x-axis as in figure (1) ### (3) The Relation between Cartesian and Polar Coordinates: From figure(1), we get * x = r Cos θ (1) * y = r Sin θ (2) and $x^2 + y^2 = r^2$ → $r = \sqrt{x^2 + y^2}$ (3) y → $θ = tan^{-1} \left( \frac{y}{x} \right)$ (4) * (1). (2) transform from Polar to Cartesian * (3), (4) transform from Cartesian to Polar ### Ex.(1): Find the Polar Coordinate for the Point P (1, √3) and the Cartesian Coordinate for the point Q (2√2,135°) * For the Point P(1, √3) * $x = 1$ and $y = (\sqrt{3})$ * $r = \sqrt{x^2 + y^2} = \sqrt{(1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$ * $θ = tan^{-1} \left( \frac{y}{x} \right) = tan^{-1}(\sqrt{3}) = 60° = \frac{\pi}{3}$ * The Polar Coordinate for P is ($2, \frac{\pi}{3}$). * Similarly: for the Point Q (2√2,135°), we obtain * $r = 2√2$ * $θ = 135°$ * From (1), (2), we get: * $r Cos θ = 2√2 Cos 135° = -2√2 Cos 45° = \left( -2 \cdot \frac{1}{\sqrt{2}} \right) \left( \frac{1}{\sqrt{2}} \right)$ → $x= -2$ * $y = r Sin θ = 2√2 Sin 135° = \left( 2√2 \right) \left( \frac{1}{\sqrt{2}} \right) = 2$ → $y = 2$ * The Cartesian Coordinate for the Point Q is (-2, 2) ### Ex.(2): Transform the equation $x^3 = y^2 (2-x)$ to the Polar form and the equation $r^2 = a^2 Cos 2θ$ to the Cartesian form * $x^3 = y^2 (2 – x)$ * $(r Cos θ)^3 = (r Sin θ)^2 (2-r Cos θ)$ * $r^3 Cos^3θ = r^2 Sin^2θ (2-r Cos θ)$ * $r Cos^3θ = 2 Sin^2θ -r Sin^2θ Cos θ$ * $r Cos θ (Cos^2θ + Sin^2θ) = 2 Sin^2θ$ * Since $Cos^2θ + Sin^2θ = 1$, hence we get * $r Cos θ = 2 Sin^2θ$ * $r = \frac{2 Sin^2θ}{Cos θ}$ → $r = 2 tan θ Sec θ$ * $r^2 = a^2 Cos 2θ$ * $r^² = a^² (Cos²θ - Sin²θ)$ * $r^4 = a² (x²-y²)$ ### (4) Distance Between Two Points The distance between the two points P1 (x1, y1) and P2 (x2, y2) is given by: $P1P2 = \sqrt{(x2-x1)^2+(y2-y1)^2}$ ### (5) Division of a segment in any ratio: The coordinate of the point C (x,y) which divide the segment AB; where A (x1, y1), B (x2, y2) by the ratio λ1:λ2 is given by * $x = \frac{λ2x1 + λ1x2}{λ1 + λ2}$ * $y = \frac{λ2y1 + λ1y2}{λ1 + λ2}$ ### Special Case: If C is the midpoint (λ1 = λ2 = 1): then * $x =\frac{x1 + x2}{2}$ * $y = \frac{y1 + y2}{2}$ ### Ex.(3): Find the ratio which divides the point (-3, 4); the segment between the two points (+3,-2). (-7, y1) hence find y1 Let the ratio be λ1 : λ2 * $-3 = -7 + 3λ2 \quad → \quad -3-3λ1 = -7 +3λ2$ * $4λ1 = 6λ2 \quad → \quad 2λ1 = 3λ2$ * And $4 = 3y1 + 2(-2) \quad → \quad 4 = 3y1 - 4$ * $20 + 4 = 3y1 \quad 24 = 3y1 \quad → \quad y1 = 8$ ### (6) The area of a triangle The area of a triangle ABC whose vertices are known which are denoted by A(x1, y1), B(x2, y2),C(x3, y3) is given by $Area = \frac{1}{2} \begin{vmatrix} x1 & y1 & 1 \\ x2 & y2 & 1 \\ x3 & y3 & 1 \end{vmatrix}$ And if the area equals zero, then the three points lie on the same straight line. ### Ex.(4'): Find the area of the triangle ABC, where A(3,2), B(-1,4), C(0,3); hence find the length of perpendicular line from the points C on AB and the angle ABC. * The area = $\frac{1}{2} \begin{vmatrix} 3 & 2 & 1 \\ -1 & 4 & 1 \\ 0 & 3 & 1 \end{vmatrix} = 1$ And since the area = $\frac{1}{2}$ (AB) (CN) * Where (CN) the perpendicular line from Con AB * (AB) = √(4)+(-2) = √20 = 2√5 * $1 = \frac{1}{2} (2√5) (CN) → CN = \frac{1}{2√5}$ * Also: * Area = (AB) (BC)sin (ABC) * BC =√1+1=√2 * $1 = \frac{1}{2} (2√5) (√2) Sin (ABC) → Sin (ABC) = \frac{1}{√10}$ * $ABC = Sin^{-1} (\frac{1}{√10})$ ### Ex.(5): Show that the three points A(a, b+c), B(b, a+c), C(c+b) lie on the same straight line Since the determinate = $\begin{vmatrix} a & b+c & 1 \\ b & a+c & 1 \\ c & a+b & 1 \end{vmatrix} = (a+b+c) \begin{vmatrix} b & 1 \\ c & 1 \end{vmatrix} = (a+b+c)=0$ .. The three points lie on the same straight line. ### (7) The Geometric Locus:(الموضع الهندى) It is the relation between x and y, when the point P(x, y) moves in the plane under certain conditions. ### Ex.(6): If the point A moves on the line x = 3 and the point B moves on the line OA where; (OA) (OB)=6 and O is the origin. Find the Geometric Locus of the point B. Let B(x, y) and from the two triangles OBD, OAC, we get $\frac{x}{y} = \frac{3}{b}$ → $b = \frac{3y}{x}$ .. The Coordinate of A is ($3, \frac{3y}{x}$) * $OA = \sqrt{9 + \frac{9y^2}{x^2}} = \sqrt{\frac{x^2 + y^2}{x^2}}$ and $OB = \sqrt{x^2 + y^2}$ And since (OA) (OB) = 6 $3\sqrt{\frac{x^2 + y^2}{x^2}} \cdot \sqrt{x^2 + y^2} = 6$ → $3(x^2 + y^2) = 6x $ → $x^2 + y^2 -2x = 0$ This is the Geometric Locus of the point B. ### Ex.(7): Find the Geometric Locus of the point Q which divides the distance OP from interior by the ratio K:1: where O is the origin and P moves on the curve $x^2 + y^2 = r^2$ Let the point P is (x1, y1) and the point Q is (x, y) * $x = \frac{kx1}{k+1}$ * $y = \frac{ky1}{k+1}$ ..$x = \frac{k}{k+1}x1$ ..$y = \frac{k}{k+1} y1$ P moves on the curve $x^2 + y^2 = r^2$ .. P satisfies the equation of the curve $x^2 + y^2 = r^2$ → $\left( \frac{k+1}{k} \right)^2 x^2 + \left( \frac{k+1}{k} \right)^2 y^2 = r^2$ → $(k+1)^2 (x^2 + y^2) = kr^2$ ### (8) Changing the Axes: **1- Transition the axes without changing its direction:** * $x = X + α$ * $y = Y+ β$ Therefore the equation of any curve f(x, y) = 0 with respect to two axes OX, OY becomes to another equation in the form g(X, Y) = 0 ## Ex.(8): Find the new form of the equation of the curve $x^2 + 3xy - y^2 - x + 5y - 6 = 0$. When shifting axes at the point (-1, 1) without change of direction. * x = X-1, y=Y+1 * The equation of the curve becomes * (X-1) + 3(X-1) (+1)-(+1)-(X-1)+5(+1)-6=0 * $X^2-2x+1+3(XY+X-Y-1)-(Y^2+2Y+1)-X+1+5Y+5-6-0$ → $X^2 -Y^2 + 3XY -3 =0$ **Note that:** When shifting axes; the first degree terms are vanished, therefore, we can eliminate the first degree terms by shifting axes at selecting new origin. ### Ex.(9): Discuss the transformation of axes at the point (a. β) on the equation $F(x, y) = ax^2 + 2hxy + by^2 + 2gx + 2fy + C = 0$ * x = X + α * y = Y + β * The equation becomes * $a(X+a)² + 2h(X+α) (Y+β) + b(Y+β)² + 2g(X+a) +2f(Y+B) + C = 0$ * $aX² + 2hXY + bY² + 2(ax + hβ + g) X + 2(ha + bβ + f)Y + (ax² + 2haβ + bβ² + 2ga + 2fβ + C) = 0$ **And we notice that (In the new equation):** * 1- The coefficient of second degree $x^2, xy, y^2$ not change. * 2- The coefficient X is $\frac{\delta F}{\delta x}$ at $x = a, y = β$ * 3- The coefficient Y is $\frac{\delta F}{\delta y}$ at $x = a, y = β$ * 4- The constant term is F(a. β) and in this case, we have $C' = F(α, β) = aa² + 2haβ + bβ² + 2ga + 2fβ + C$ Therefore, we can determine the point O'(α. β); which transform to it the point 0(0.0) such that the first degree terms are vanished from the two equation * $\frac{\delta F}{\delta x} = 2(aa + hβ + g) = 0$ * $\frac{\delta F}{\delta y} = 2(ha + bβ + f) = 0$ And by solving these two equation; we can get (α. β). ### Ex.(10): Find the new origin O' which transform to it; the axes such that the first degree terms are vanished from the equation. $3x^2-2xy + 4y^2 -3x-10y - 7 = 0$ ; hence find the new form of the curve. * $F(x, y) = 3x^2 -2xy + 4y^2 - 3x -10y -7 = 0$ * $\frac{\delta F}{\delta x} = 6x-2y-3$ * $\frac{\delta F}{\delta y} = -2x+8y-10$ * $\frac{\delta F}{\delta x} = 0$ → $6α - 2β - 3 = 0$ * $\frac{\delta F}{\delta y} = 0$ → $2α + 8β - 10 = 0$ By solving the two equation, we get $α = 1, β = \frac{3}{2}$ .. O'(1, $\frac{3}{2}$) and $C= F(α, β)= ga + bβ + C=-(1)-(2)-7=-16$ ..The new equation of the curve is $3x^2-2xy + 4y²-16=0$ ### 2- Rotating the axes without changing the origin If the axes Ox Oy turning with angle θ at the origin O and become the new axes OX, OY, as in figure (8), then: * $x = X cosθ - Y sinθ$ (1) * $y = X sinθ + Y cosθ$ (2) From (1), (2); we get * $X = x cosθ + y sinθ$ (3) * $ Y = -x sinθ + y cosθ$ (4) And, we can write (1), (2), (3), (4) in simplest table |$ $ |$ X $ | $ Y $ | |---|:---:|:---:| |x|$ cosθ $ | $ -sinθ $ | |y | $sinθ $ | $ cosθ $ | ### Ex.(11): If the axes turning be angle $\frac{\pi}{4}$, find the new form of the equation: $5x^2 + 2xy + 5y^2 = 2$ * $x = X Cos \frac{\pi}{4} - Y Sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} (X-Y)$ * $y = X Sin \frac{\pi}{4} + Y Cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} (X+Y)$ the equation becomes * $\frac{5}{2} (X-Y)^2 + \frac{1}{2} (X-Y)(X+Y) + \frac{5}{2} (X + Y)^2 = 2$ → $3X^2 + 2Y^2 - 1=0$ **Note that:** The coefficient of xy is vanished if the axes rotate with appropriate angle θ ### Ex.(12): Find the angle θ which must the axis ox, oy be rotating about the origin such that the term xy be vanished from equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + C = 0$ * Put * $x = X Cos θ - Y Sin θ$ * $y = X Sin θ + Y Cos θ$ * The equation becomes $A(X Cos θ - Y Sin θ)² + 2h(X Cos θ - Y Sin θ)(X Sin θ + Y Cos θ) + b(X Sin θ + Y Cos θ)² + 2g(X Cos θ - Y Sin θ) + 2f(X Sin θ + Y Cos θ) + C = 0$ * By equating the coefficient of xy by zero, we get * $-2a Sine Cose + 2h (Cos² θ - Sin² θ) + 2b Sin 0 Cos 0 = 0$ * $(a-b) Sin 2θ = 2h Cos 2θ$ * $tan 2θ = \frac{2h}{a-b}$ → θ = $\frac{1}{2} tan^{-1} \left( \frac{2h}{a-b} \right)$ * And Put * $a = a Cos² θ + 2h Sin 0 Cos 0 + b Sin² 0$ * $b = a Sin² 0 - 2h Sin 0 Cos 0 + b Cos² 0$ * $a+b=a+b$ Therefore the addition of Coefficient of $x^2, y^2$ does not change under rotation. ### Ex.(13): Find the value of the angle θ when axes ox, oy be rotating such that the Coefficient of xy be vanished from the equation. $5x^2 + 3xy + 5y^2 = 4$ ; hence find the new form * θ = $\frac{1}{2} tan^{-1} \left( \frac{2h} {a-b} \right)$ * a = b = 5, 2h = 3 * θ = $\frac{1}{2} tan^{-1} \left( \frac{3}{0} \right) = \frac{1}{2} tan^{-1} (∞) = \frac{1}{2} \left ( \frac{\pi}{2} \right) = \frac{\pi}{4}$ * Substituting in equation: * $x = \frac{1}{\sqrt{2}} (X-Y)$ * $y = \frac{1}{\sqrt{2}} (X+Y)$ * $\frac{5}{2} (X-Y)^2 + \frac{3}{2} (X-Y)(X+Y) + \frac{5}{2} (X + Y)^2 = 4$ * $13 X^2 + 7Y^2 = 8$ ### Ex.(14): Transform the equation: $F(x, y) = 5x^2-6xy + 5y^2 + 22x - 26y + 29 = 0$ To Simplest form * 1- Firstly, we transition the axes to a new origin O'(α, β), where * $\frac{\delta F}{\delta x} |_{(\alpha, \beta)} = 0$ → $10α - 6β + 22 = 0$ .....(1) * $\frac{\delta F}{\delta y} |_{(\alpha, \beta)} = 0$ → $-6α + 10β - 26 = 0$ ..... (2) * From (1), (2), we get $α = -1, β = 2$ * And $C= F(-1.2) = ga + bβ + C = 11(-1)+(-13) (8) + 29 = -8$ * And the equation after transition becomes $5X^2-6XY+5Y^2-8=0$ .....(3) * 2- Secondly, we rotate the axes by the angle 0. where * $θ= \frac{1}{2} tan^{-1} \left( \frac{2h} {a-b} \right) = \frac{1}{2} tan^{-1} \left( \frac{-6}{5-5} \right) = \frac{1}{2} tan^{-1} (-∞) = \frac{1}{2} \left ( \frac{\pi}{2} \right) = \frac{\pi}{4}$ * $x = \frac{1}{\sqrt{2}} (x'-y')$ * $y = \frac{1}{\sqrt{2}} (x'+ y')$ * Equation (3) becomes * $2x^2 + 8y^2 = 8$

Use Quizgecko on...
Browser
Browser