COORDINATE GEOMETRY

Class IX mein humne padha tha ki ek plane par kisi point ki position locate karne ke liye humein coordinate axes ke pair ki zaroorat hoti hai.

• Cartesian Coordinate System:
• Do mutually perpendicular lines, x-axis (horizontal) aur y-axis (vertical), jo ek point par intersect karti hain, jise origin (O) kehte hain.
• x-coordinate (Abscissa): Kisi point ki y-axis se distance. Isko \(x\) se denote karte hain.
• y-coordinate (Ordinate): Kisi point ki x-axis se distance. Isko \(y\) se denote karte hain.
• Ek point ke coordinates ko ordered pair \((x, y))\) se represent karte hain.

• Points on Axes:
• x-axis par koi bhi point \((x, 0))\) form ka hota hai. (y-coordinate zero).
• y-axis par koi bhi point \((0, y))\) form ka hota hai. (x-coordinate zero).
• Origin ke coordinates \((0, 0))\) hote hain.

• Quadrants: Coordinate axes plane ko four parts mein divide karte hain, jinhe quadrants kehte hain.
• Quadrant I: \((+, +))\) (x positive, y positive)
• Quadrant II: \((-, +))\) (x negative, y positive)
• Quadrant III: \((-,-))\) (x negative, y negative)
• Quadrant IV: \((+,-))\) (x positive, y negative)

Example: Point \((3, -2))\) fourth quadrant mein hai, jahan x-coordinate 3 aur y-coordinate -2 hai.

Distance Formula ka use karke hum do points ke beech ki distance find karte hain, chahe woh kisi bhi quadrant mein hon ya axes par hon. Yeh Pythagoras Theorem par based hai.

• Derivation ka Idea:
• Do points \(P(x_1, y_1))\) aur \(Q(x_2, y_2))\) ko consider karo.
• In points se x-axis par perpendiculars draw karo.
• Ek right-angled triangle form karo jismein PQ hypotenuse ho.
• Triangle ke legs ki length \((x_2 - x_1))\) aur \((y_2 - y_1))\) hogi.
• Pythagoras Theorem apply karo: \(PQ^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2\).

• Distance Formula:
• Points \(P(x_1, y_1))\) aur \(Q(x_2, y_2))\) ke beech ki distance \(PQ\) is given by:

$$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

• Ya phir \(PQ = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\) bhi use kar sakte hain, kyunki square karne par sign ka farak nahi padta.

• Origin se Distance:
• Kisi point \(P(x, y))\) ki origin \((0, 0))\) se distance:

$$OP = \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2}$$

• Special Cases:
• Agar do points x-axis par hain, e.g., \(A(x_1, 0))\) aur \(B(x_2, 0))\), toh distance \(|x_2 - x_1|\) hogi.
• Agar do points y-axis par hain, e.g., \(C(0, y_1))\) aur \(D(0, y_2))\), toh distance \(|y_2 - y_1|\) hogi.

• Applications:
• Collinear Points: Teen points \(A, B, C\) collinear honge agar \(AB + BC = AC\) (ya koi similar combination) ho.
• Types of Triangles: Sides ki length calculate karke triangle ka type determine karna (e.g., equilateral, isosceles, right-angled).
• Types of Quadrilaterals: Sides aur diagonals ki length calculate karke quadrilateral ka type determine karna (e.g., square, rectangle, rhombus, parallelogram).

Example: Points \((2, 3))\) aur \((4, 1))\) ke beech ki distance: \(x_1 = 2, y_1 = 3, x_2 = 4, y_2 = 1\) \(PQ = \sqrt{(4 - 2)^2 + (1 - 3)^2} = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}\) units.

Section Formula ka use karke hum us point ke coordinates find karte hain jo ek line segment ko internally ek given ratio mein divide karta hai.

• Derivation ka Idea:
• Do points \(A(x_1, y_1))\) aur \(B(x_2, y_2))\) ko consider karo.
• Point \(P(x, y))\) line segment AB ko ratio \(m_1 : m_2\) mein divide karta hai.
• Similar triangles ke concept ka use karte hain (AA similarity criterion).
• Perpendiculars draw karte hain x-axis par aur parallel lines bhi draw karte hain.
• \(\triangle PAQ \sim \triangle BPC\) se ratios equate karte hain.

• Section Formula:
• Point \(P(x, y))\) jo line segment joining \(A(x_1, y_1))\) aur \(B(x_2, y_2))\) ko internally ratio \(m_1 : m_2\) mein divide karta hai, uske coordinates hain:

$$P(x, y) = \left( \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \right)$$

• Special Case: Ratio \(k : 1\):
• Agar ratio \(k : 1\) ho (jahan \(m_1 = k\) aur \(m_2 = 1\)), toh formula ban jaata hai:

$$P(x, y) = \left( \frac{kx_2 + x_1}{k + 1}, \frac{ky_2 + y_1}{k + 1} \right)$$

• Yeh form tab useful hota hai jab humein ratio find karna ho, ya jab ek point kisi line par ho aur uski position pata ho.

Example: Points \(A(4, -3))\) aur \(B(8, 5))\) ko \(3 : 1\) ke ratio mein divide karne wale point ke coordinates find karo. \(x_1 = 4, y_1 = -3, x_2 = 8, y_2 = 5\) \(m_1 = 3, m_2 = 1\) \(x = \frac{3(8) + 1(4)}{3 + 1} = \frac{24 + 4}{4} = \frac{28}{4} = 7\) \(y = \frac{3(5) + 1(-3)}{3 + 1} = \frac{15 - 3}{4} = \frac{12}{4} = 3\) So, point ke coordinates \((7, 3))\) hain.

Mid-point Formula Section Formula ka ek special case hai. Jab ek point line segment ko equal parts (1:1 ratio) mein divide karta hai, toh woh mid-point hota hai.

• Derivation:
• Section Formula mein \(m_1 = 1\) aur \(m_2 = 1\) substitute karo.
• \(x = \frac{1 \cdot x_2 + 1 \cdot x_1}{1 + 1} = \frac{x_1 + x_2}{2}\)
• \(y = \frac{1 \cdot y_2 + 1 \cdot y_1}{1 + 1} = \frac{y_1 + y_2}{2}\)

• Mid-Point Formula:
• Points \(A(x_1, y_1))\) aur \(B(x_2, y_2))\) ke mid-point \(M(x, y))\) ke coordinates hain:

$$M(x, y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$$

Example: Points \(P(2, -3))\) aur \(Q(-6, 7))\) ke mid-point ke coordinates find karo. \(x_1 = 2, y_1 = -3, x_2 = -6, y_2 = 7\) \(x = \frac{2 + (-6)}{2} = \frac{-4}{2} = -2\) \(y = \frac{-3 + 7}{2} = \frac{4}{2} = 2\) So, mid-point ke coordinates \((-2, 2))\) hain.

Teen ya teen se zyada points collinear tab hote hain jab woh ek hi straight line par lie karte hain.

• Distance Formula Method (Most Common for Boards):
• Teen points \(A, B, C\) collinear honge agar sum of any two distances is equal to the third distance.
• Calculate \(AB, BC, AC\) using distance formula.
• Check if \(AB + BC = AC\) or \(AC + CB = AB\) or \(BA + AC = BC\).
• Agar condition satisfy hoti hai, toh points collinear hain.

• Area of Triangle Method (Chapter 7 NCERT mein nahi hai, but useful):
• Agar teen points collinear hain, toh unse banne wale triangle ka area zero hoga.
• Area of Triangle formula: \(\frac{1}{2} [x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]\).
• Agar Area = 0, toh points collinear hain.

Example: Points \(A(1, 5), B(2, 3), C(-2, -11))\) collinear hain ya nahi?

1. \(AB = \sqrt{(2-1)^2 + (3-5)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1+4} = \sqrt{5}\)
2. \(BC = \sqrt{(-2-2)^2 + (-11-3)^2} = \sqrt{(-4)^2 + (-14)^2} = \sqrt{16+196} = \sqrt{212}\)
3. \(AC = \sqrt{(-2-1)^2 + (-11-5)^2} = \sqrt{(-3)^2 + (-16)^2} = \sqrt{9+256} = \sqrt{265}\)

Clearly, \(\sqrt{5} + \sqrt{212} \neq \sqrt{265}\). So, points collinear nahi hain.

Distance Formula ka use karke hum geometric figures ke properties ko verify kar sakte hain.

• Triangles: Points ke vertices ke beech ki distances calculate karke type determine karte hain.
• Equilateral Triangle: All three sides equal (\(AB = BC = CA\)).
• Isosceles Triangle: Any two sides equal (e.g., \(AB = BC\)).
• Scalene Triangle: All three sides different.
• Right-angled Triangle: Pythagoras Theorem satisfy karta hai (\(a^2 + b^2 = c^2\)), jahan \(c\) sabse badi side (hypotenuse) hai.

• Quadrilaterals: Four vertices ke beech ki distances (sides aur diagonals) calculate karke type determine karte hain.
• Square:
• All four sides equal (\(AB = BC = CD = DA\)).
• Diagonals equal (\(AC = BD\)).
• Rhombus:
• All four sides equal (\(AB = BC = CD = DA\)).
• Diagonals unequal (\(AC \neq BD\)).
• Rectangle:
• Opposite sides equal (\(AB = CD, BC = DA\)).
• Diagonals equal (\(AC = BD\)).
• Parallelogram:
• Opposite sides equal (\(AB = CD, BC = DA\)).
• Diagonals unequal (\(AC \neq BD\)).

Important Note: Quadrilateral ke type ko confirm karne ke liye sides ke saath-saath diagonals ki length check karna bhi zaroori hai. Sirf sides se ambiguity ho sakti hai (e.g., square aur rhombus, rectangle aur parallelogram).

Example: Check whether \((5, -2), (6, 4), (7, -2))\) are vertices of an isosceles triangle. Let \(A=(5,-2), B=(6,4), C=(7,-2)\)

1. \(AB = \sqrt{(6-5)^2 + (4-(-2))^2} = \sqrt{1^2 + 6^2} = \sqrt{1+36} = \sqrt{37}\)
2. \(BC = \sqrt{(7-6)^2 + (-2-4)^2} = \sqrt{1^2 + (-6)^2} = \sqrt{1+36} = \sqrt{37}\)
3. \(AC = \sqrt{(7-5)^2 + (-2-(-2))^2} = \sqrt{2^2 + 0^2} = \sqrt{4} = 2\)

Since \(AB = BC = \sqrt{37}\), two sides are equal. So, it is an isosceles triangle.