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`        Advanced Maths: Short Tricks on Coordinate Geometry pls tell`
2 years ago

```							Important Short Tricks on Coordinate GeometryEquation of line parallel to y-axisX = a For Example: A Student plotted four points on a graph. Find out which point represents the line parallel to y-axis. a) (3,5)b) (0,6) c) (8,0) d) (-2, -4) Solution: Option (C) Equation of line parallel to x-axisY = b For Example: A Student plotted four points on a graph. Find out which point represents the line parallel to x-axis. a) (3,5)b) (0,6) c) (8,0) d) (-2, -4) Solution: Option (B) Equations of linea) Normal equation of line ax + by + c = 0 b) Slope – Intercept Form y = mx + c Where, m = slope of the line & c = intercept on y-axis For Example: What is the slope of the line formed by the equation 5y - 3x - 10 = 0? Solution: 5y - 3x - 10 = 0, 5y = 3x + 10 Y = 3/5 x + 2 Therefore, slope of the line is = 3/5 c) Intercept Form x/A + y/B = 1, Where, A & B are x-intercept & y-intercept respectively For Example: Find the area of the triangle formed the line 4x + 3 y – 12 = 0, x-axis and y-axis? Solution: Area of triangle is = ½ * x-intercept * y-intercept. Equation of line is 4x + 3 y – 12 = 0 4x + 3y = 12, 4x/12 + 3y/12 = 1 x/3 + y/4 = 1 Therefore area of triangle = ½ * 3 * 4 = 6 d) Trigonometric form of equation of line, ax + by + c = 0 x cos θ + y sin θ = p, Where, cos θ = -a/ √(a2 + b2) , sin θ = -b/ √(a2 + b2) & p = c/√(a2 + b2) e) Equation of line passing through point (x1,y1) & has a slope m y - y1 = m (x-x1) Slope of line =
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2 years ago
```							Normal equation of lineax + by + c = 0b) Slope – Intercept Formy = mx + c            Where, m = slope of the line & c = intercept on y-axisFor Example: What is the slope of the line formed by the equation 5y - 3x - 10 = 0?Solution: 5y - 3x - 10 = 0, 5y = 3x + 10Y = 3/5 x + 2Therefore, slope of the line is = 3/5c) Intercept Formx/A + y/B = 1, Where, A & B are x-intercept & y-intercept respectivelyFor Example: Find the area of the triangle formed the line 4x + 3 y – 12 = 0, x-axis and y-axis?Solution: Area of triangle is = ½ * x-intercept * y-intercept.Equation of line is 4x + 3 y – 12 = 04x + 3y = 12,4x/12 + 3y/12 = 1x/3 + y/4 = 1Therefore area of triangle = ½ * 3 * 4 = 6d) Trigonometric form of equation of line, ax + by + c = 0x cos θ + y sin θ = p,Where, cos θ = -a/ √(a2 + b2) ,  sin θ = -b/ √(a2 + b2) & p = c/√(a2 + b2)e) Equation of line passing through point (x1,y1) & has a slope my - y1 = m (x-x1)	Slope of line = y2 - y1/x2 - x1 = - coefficient of x/coefficient of y	Angle between two linesTan θ = ± (m2 – m1)/(1+ m1m2)   where, m1 , m2 = slope of the linesNote:    If lines are parallel, then tan θ = 0If lines are perpendicular, then cot θ = 0For Example: If 7x - 4y = 0 and 3x - 11y + 5 = 0 are equation of two lines. Find the acute angle between the lines?Solution: First we need to find the slope of both the lines.7x - 4y = 0⇒ y = 74xTherefore, the slope of the line 7x - 4y = 0 is 74Similarly, 3x - 11y + 5 = 0⇒ y = 311x + 511Therefore, the slope of the line 3x - 11y + 5 = 0 is = 311Now, let the angle between the given lines 7x - 4y = 0 and 3x - 11y + 5 = 0 is θNow,Tan θ = ± (m2 – m1)/(1+ m1m2) = ±[(7/4)−(3/11)]/[1+(7/4)*(3/11)] = ± 1Since θ is acute, hence we take, tan θ = 1 = tan 45°Therefore, θ = 45°Therefore, the required acute angle between the given lines is 45°.	Equation of two lines parallel to each otherax + by + c1 = 0ax + by + c2 = 0Note:    Here, coefficient of x & y are same.	Equation of two lines perpendicular to each otherax + by + c1 = 0bx - ay + c2 = 0Note:    Here, coefficient of x & y are opposite & in one equation there is negative sign.	Distance between two points (x1, y1), (x2, y2)D = √ (x2 – x1)2 + (y2 – y1)2For Example: Find the distance between (-1, 1) and (3, 4).Solution: D = √ (x2 – x1)2 + (y2 – y1)2= √ (3 – (-1))2 + (4 – 1)2 = √(16 + 9) = √25 = 5	The midpoint of the line formed by (x1, y1), (x2, y2)M = (x1 + x2)/2, (y1 + y2)/2	Area of triangle whose coordinates are (x1, y1), (x2, y2), (x3, y3)½ I x1 (y2 – y3) + x2 (y3 – y1) + x3(y1 – y2) IFor Example: Find area of triangle whose vertices are (1, 1), (2, 3) and (4, 5).Solution: We have (x1, y1) = (1, 1), (x2, y2) = (2, 3) and (x3, y3) = (4, 5)Area of Triangle = ½ I x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) I=1/2 I (1(3−5) +2(5−1) + 4(1−3)) I=1/2 I(−2+8−8) =1/2 (−2) I = I−1I = 1
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2 years ago
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