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Mathematics Dictionary
Dr. K. G. Shih
Line and two points
Subjects

    Symbol Defintion Example : x^2 = square of x
  • Q01 | - Equation of line
  • Q02 | - Two points (x1,y1) and (x2,y2)
  • Q03 | - Intersection of two lines
  • Q04 | - Angle between two lines
  • Q05 | - Distance of point to a line
  • Q06 | - Distance between two parallel lines
  • Q07 | - Triangle A(0,0), B(6,0) and C(3,4) find centroid of triangle ABC
  • Q08 | - Triangle A(0,0), B(6,0) and C(3,4) find ortho-center of triangle ABC
  • Q09 | - Triangle A(0,0), B(6,0) and C(3,4) find pedal triangle of triangle ABC
  • Q10 | - Triangle A(0,0), B(6,0) and C(3,4) find area of triangle ABC
  • Q11 | - Triangle A(0,0), B(6,0) and C(3,4) find circum-center of triangle ABC
  • Q12 | -
Answers


Q01. Equation of line

Equation of line
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Q02. Two points (x1,y1) and (x2,y2)

Slope
  • Slope = (y2-y1)/(x2-x1)
Distance between two points
  • d = Sqr((x2-x1)^2 + (y2-y1)^2)
Mid point between two points
  • Let (xm,ym) be the mid point
  • xm = (x1 + x2)/2
  • ym = (y2 + y1)/2

Go to Begin

Q03. Intersections of two lines

Point of intersection
  • Line 1 : y = s1*x + b1
  • Line 2 : y = s2*x + b2
  • Intersection at (x, y)
    • At intersection : s1*x + b1 = s2*x + b2
    • x*(s2 - s1) = b1 - b2
    • Intersection
      • x = (b1 - b2)/(s2 - s1)
      • y = s1*(b1 - b2)/(s2 - s1) + b1
If one line is vertical : Slope s = infinite
  • Line 1 : y = s1*x + b
  • Line 2 : x = c
  • Intersection at x = c and y = s1*c + b
If one line is horizontal : slope s = 0
  • Line 1 : y = s1*x + b1
  • Line 2 : y = b2
  • Hence b2 = s1*x + b1
  • Intersection at x = (b2 - b1)/s1 and y = b2
Example : Find itersection point of y = x + 1 and y = 3*x - 1
  • x - y = -1 ....... (1)
  • 3*x - y = 1 ...... (2)
  • Eliminate y : Equation (2) - (1)
  • 2*x = 2 and x = 1
  • Substitute x = 1 into (1) and y = 2

Go to Begin

Q04. Angle between two lines

  • Line 1 : y = s1*x + b1 making angle A1 with x-axis
  • Line 2 : y = s2*x + b2 making angle A2 with x-axis
  • Angle between two lines : A2-A1
    • Tan(A2-A1) = (tan(A2) - tan(A1))/(1 + tan(A1)*tan(A2))
    • Tan(A2-A1) = (s2 - s1)/(1 + s1*s2)
  • if two lines are perpendicular
    • A2 - A1 = 90 degrees and tan(A2-A1) = infinite
    • Hence (1 + s1*s2) = 0
    • Hence s1 = -1/s2 is the condition
Example : Find angle between y = x + 1 and y = 3*x - 1
  • Method 1
    • Let A1 be the angle of y = x + 1 making with x-axis
    • Hence Tan(A1) = 1 and A1 = 45 degrees
    • Let A2 be the angle of y = 3*x - 1 making with x-axis
    • Hence Tan(A2) = 3 and A2 = 71.565 degrees
    • Angle between two lines is (A2 - A1) = 26.565
  • Method 2 : Use formula
    • Let angle between two lines be B
    • tan(B) = (s2 - s1)/(1 + s1*s2)
    • Since s1 = 1 and s2 = 3
    • Hence tan(B) = (3 - 1)/(1 + 1*3) = 0.5
    • B = arctan(0.5) = 26.565

Go to Begin

Q05. Distance of point to a line A*x + B*y + C = 0
  • Let point be P(u1,v1) and line be A*x + B*y + C = 0
  • Make a line passes P(u1,v1) and has intesectiion Q with given line
  • PQ is perpendicular to line A*x + B*y + C = 0
  • Hence slope of line PQ is s2 = -1/s1 and s1 = -A/B
  • Hence s2 = B/A and equation of line PQ is y = s2*x + b2
  • Sine P(u,v) is on line PQ, Hence v1 = s2*u1 + b2 and b2 = v1 - s2*u1
  • Find intersection point Q(u2,v2)
    • Line 1 : y = -Ax/B - C/B = 0
    • Line 2 : y = s2*x + v - s2*u
  • Find distance between P and Q : d = Sqr((u2 - u1)^2 + (v2 - v1)^2)
  • Simplify we have d = (A*u1 + B*v1 + C)/Sqr(A^2 + B^2)

Go to Begin

Q06. Distance between two parallel line
  • Line 1 : y = s1*x + b1 making angle A with x-axis
  • Line 2 : y = s1*x + b2 making angle A with x-axis
  • Construction
    • Draw xy coordinate
    • Draw line 1 which intesect y-axis at P with x = 0 and y = b1
    • Draw line 2 which intesect y-axis at Q with x = 0 and y = b2
    • At point Q draw line QR perpendicular to line 1 and R is on line 1
    • Hence angle PQR = angle A = arctan(s1)
    • PQ = b2 - b1
    • In triangle PQR, QR = PQ*cos(PQR) = (b2 - b1)*aretan(A)

Go to Begin

Q07. Triangle A(0,0), B(6,0) and C(3,4) find the centroid of triangle ABC

1. Use medians AD and BE to find centroid
  • Let D(xd,yd) be mid point of BC
    • Hence xd = (xb + xc)/2 = (6 + 3)/2 = 4.5
    • Hence yd = (yb + yc)/2 = (0 + 4)/2 = 2
  • Let E(xe,ye) be mid point of CA
    • Hence xe = (xc + xa)/2 = (3 + 0)/2 = 1.5
    • Hence ye = (yc + ya)/2 = (4 + 0)/2 = 2
  • Slopes of medians AD and BE
    • Slope AD : s1 = (yd - ya)/(xd - xa) = (2 - 0)/(4.5 - 0) = +0.44444
    • Slope BE : s2 = (ye - yb)/(xe - xb) = (2 - 0)/(1.5 - 6) = -0.44444
  • Equations of medians AD and BE
    • Equation AD : y = s1*x + b1 = 0.44444*x + b1
    • Since A(0,0) on median AD and hence 0 = 0.44444*0 + b1
    • Hence b1 = 0 and y = 0.44444*x
    • Equation BE : y = s2*x + b2 = -0.44444*x + b2
    • Since B(6,0) on median BE and hence 0 = -0.44444*6 + b2
    • Hence b2 = 2.66667 and y = -0.44444*x + 2.66667
  • Solve equations
    • +0.44444*x - y = +0 ........... (1)
    • -0.44444*x - y = -2.66667 ..... (2)
    • Eliminate x : (1) + (2)
    • -2*y = -2.66667 and y = 1.33333
    • From (1) x = y/0.44444 = 1.33333/0.44444 = 3.00000
  • Hence centroid is at (3, 1.33333)
2. Use medians AD and CF to find centroid
  • Median CF where F be mid point of AB
  • xf = (xa + xb)/2 = (0 + 6)/2 = 3
  • yf = (ya + yb)/2 = (0 + 0)/2 = 0
  • Slope of CF : s3 = (yf - yc)/(xf - xc) = (0 - 4)/(3 - 3) = infinite
  • Hence equation CF is x = 3
  • Since equation AD is y = 0.44444*x
  • Hence intersection and AD and CF : x = 3 and y = 0.44444*3 = 1.33333
  • Centroid is at (3, 1.33333)
3. Medians of triangle ABC are concurrent
  • Intersection of AD and BE is (3, 1.33333)
  • Intersection of AD and CF is (3, 1.33333)
  • Hence AD, BE and CF meet at the same point
  • Medians of triangle ABC are concurrent

Go to Begin

Q08. Triangle A(0,0), B(6,0) and C(3,4) find the ortho-center of triangle ABC

1. Use heights AD and BE to find ortho-center
  • Slopes of medians AD and BE
    • Slope BC : m1 = (yb - yc)/(xb - xc) = (0 - 4)/(6 - 3) = -1.33333
    • Slope AD : s1 = -1/m1 = -1/(-1.33333) = 0.75
    • Slope CA : m2 = (yc - ya)/(xc - xa) = (4 - 0)/(3 - 0) = +1.33333
    • Slope BE : s2 = -1/m2 = -1/1.33333 = -0.75
  • Equations of medians AD and BE
    • Equation AD : y = s1*x + b1 = 0.75*x + b1
    • Since A(0,0) on height AD and hence 0 = 0.75*0 + b1
    • Hence b1 = 0 and y = 0.75*x
    • Equation BE : y = s2*x + b2 = -0.75*x + b2
    • Since B(6,0) on height BE and hence 0 = -0.75*6 + b2
    • Hence b2 = 4.5 and y = -0.75*x + 4.5
  • Solve equations
    • +0.75*x - y = +0 ........... (1)
    • -0.75*x - y = -4.5 ......... (2)
    • Eliminate x : (1) + (2)
    • -2*y = -4.5 and y = 2.25
    • From (1) x = y/0.75 = 2.25/0.75 = 3
  • Hence centroid is at (3, 2.25)
2. Use heights AD and CF to find centroid
  • Height CF perpendicular to AB
  • Slope AB = 0 and slope CF = s3 = infinite
  • Hence equation CF is x = 3
  • Since equation AD is y = 0.75*x
  • Hence intersection and AD and CF : x = 3 and y = 0.75*3 = 2.25
  • Orhto-center is at (3, 2.25)
3. Heights of triangle ABC are concurrent
  • Intersection of AD and BE is (3, 2.25)
  • Intersection of AD and CF is (3, 2.25)
  • Hence AD, BE and CF meet at the same point
  • Heights of triangle ABC are concurrent

Go to Begin

Q09. Triangle ABC : A(0,0), B(6,0) and C(3,4) find area of pedal triangle

1. The heights of triangle ABC and pedal triangle
  • AD is height perpendicular to BC
  • BE is height perpendicular to CA
  • CF is height perpendicular to AB
  • Triangle DEF is the pedal trangle of tiriangle ABC
2. Find the heights
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