Coordinates of Points PDF

# 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 * $2x^2 + 8y^2 = 8$