Straight Line

          
  value of y-intercept (a)  
  value of x-intercept (b)  
  x-coordinate of one point(x1)  
  Find    
  y-coordinate (y1)  
  value of slope (m)  
  Angle θ w.r.t + ve x-axis (in degree)  
  length of segment between the axes √  (a2 + b2 )  
  Area between axes & line  
       
  x/b + y/a =1 x /   +  y/    =1  
  *x/b - y/a =1 x /   -  y/    =1 (y-intercept is -Ve of above)  
  ax + by + c =0  x +   y +   = 0 ...........(1)  
  b2x-aby+a2b=0(if the line is to given line, passing throughpointof y-intercept) x +   y +   = 0  ...........(2)  
  a2y-abx+ab2=0(if the line isto given line passing throughpoint of x-intercept ) x +   y +   = 0  
  y = mx+ c1 y = x  + (y-intercept is -Ve of above)

 

 
  *y = mx - c1 y = x  -  
  y-y1= [(y2-y1)/(x2-x1)](x-x1) y -  = [ ( - )/( - )](x- )  
  x/b  
  y/a  
  x/a  
  y/b  
  x/b + y/a  
  x/b + (b-x)/b +  
  (a-y)/a +y/a +  
  x/a + y/b  
  comment on x/a+y/b  
  distance from (x1,y1) to the line x/a +y/b =1  
    For what value of (x,y) the point shall lie both on x/b + y/a=1 & x/a + y/b =1  
  x=ab/(a+b)   y= ab/(a+b) x= ;    y=  
  x=    y = Put the above values below to test whether they add up to 1  
  x/b + y/a =  
  x/a + y/b =  
  (x/b + y/a )2 =1 x2 /b2 +y2 /a2  +2xy/ab =1  
  the above equn is recast below Ax2 +Bxy+Cy2 +Dx +Ey+F=0.......(1)  
  x2 /b2 +y2 /a2  
       
  2xy/ab ; xy:; ab:  
     
  A:   B: C: D: E: F: n :  
  Eccentricity e :  
  comment  
       
 

^ When st.line...(1) is rotated by 90 degree about point of y-intercept, the equation becomes ......(2). Now, the point (x1.y1) on 1st st.line is transported to say (x5, y5) on the st.line ...(2). However, the length of line segment say L1 from y-intercept point remains the same. Below, we find out the value of (x5,y5) with respect to global co-ordinate system with origin at (0,0) and also value (x51,y51) in  local co-ordinate system with y-intercept point as origin ( 0,a) and with  the length of 1st st.line as the x-axis.

 
  For anti-clockwise rotation : y5  
  For anti-clockwise rotation : x5  
  For clockwise rotation : y5a  
  For clockwise rotation : x5a  
  Length from y-intercept point to (x5,y5) ---  
  AB  
  AC  
       
       
  ax+by+c=0 Test (a) (x)+(b)(y) +(c)=  
 

How can we get the value of (x5,y5) &(x5a,y5a) without going through the cumbersome process of solving the quadratic equation ? It will be through rotational & translational matrix.

 
  First origin shifted to y-intercept point   +    =  
                   
  Secondly (x1,y1) transformed w.r.t  new origin M    =  
                  
  angle of rotation of the line (in degree)  
  Rotational matrix(2x2) R    -  
       -  
  RM (anti-clockwise rotation)  
     
  RM (clockwise rotaion)  
     
  Add back new origin (anti-clockwise)  
     
  Add back new origin (clockwise)  
     
       
       
 

* These equations represent mirror lines to their plus counterparts. Both lines are parallel.

* In fact, x/b ± y/a =1 and y = mx c1  are various forms of the equation of a st. line. Which form one chooses is a matter of convention. We have chosen the most used format.

* The equation ax+by+c=0 represents a st.line whose x-intercept is b, y-intercept is a and c represents the negative of the products of the intercepts i.e. c=-ab. Hence ay+bx+c=0 represents a st. line whose intercepts have been exchanged.

*The equation of a straight line which is drawn perpendicular to the line given by equation ax+by+c=0 and passes through the point of interception(0,-c/b) or (0,a) of the line with y-axis is given by b2x-aby-ac=0 or b2x-aby+a2b=0.

*The equation of a straight line which is drawn perpendicular to the line given by equation ax+by+c=0 and passes through the point of interception(-c/a,0) or (b,0) of the line with x-axis is given by a2y-abx-bc=0 or or a2y-abx+ab2=0.

* the perpendicular distance from a point (x1,y1) to a st.line with equation ax + by + c =0    is given by d= (ax1+by1+c) / √  (a2 + b2 )

*In A general equation of the type

Ax2 +Bxy+Cy2 +Dx +Ey+F=0, the eccentricity e is given by

e = √(2 √[ (A-C)2 + B2 ]) /√ (n(A+C) +√[ (A-C)2 + B2 ] )

& n =1 if (AC-B2/4) < 0 and  n = -1 if (AC-B2/4) > 0

Pl. note that D,E,F are not contributing to eccentricity. If B2 =4AC, curve is a parabola. The co-efficient B contributes to eccentricity and solely contributes to the slanting of the curve with respect to the co-ordinate axis. Since  B2 is positive, its effect is to increase the eccentricity e whereas the role of second factor AC depends on overall sign. Taking A'=2A and C'=2C. we can write  B2 = > <A'C' or B= > <√(A'C'). Let us state that M=√(A'C')= geometric mean of A',C'. Hence B may be equal to, less than or greater than the geometric mean. In case of non-equality, one can calculate the standard deviation.

* In Ax2 +Bxy+Cy2 +Dx +Ey+F=0

if B2-4AC < 0 ellipse

if B2-4AC < 0  , A=C, B=0 , circle

if B2-4AC = 0 , parabola

if B2-4AC  >  0 , hyperbola

if B2-4AC >  0, A+C=0, Rectangular hyperbola

^ a1x1 +b1y1+c1=0 ; co-ordinate of y-intercept (0,a1). length from this point to (x1.y1) is (x1-0)2 +(y1-a1)2 = L2

Let the st.line perpendicular to the above line, having a point (x2,y2) which is analogous to (x1,y1) have the equation

a2x2+b2y2+c2=0 .......(II)

 and length from this point to (x2.y2) to (0,a1) is (x2-0)2 +(y2-a1)2 = n1=L2 ......(I)

Squaring & rearranging both sides of .....(II), we get

a22x22-b22y22-c22-2b2c2y2=0 .......(III)

multiplying both sides of (I) by a22  , we get

a22x22+a22y22-2a22a1y2+a22a21-n2=0 .......(IV) where n2=a22 *n1

(IV) - (III) = y22(a22+b22) +2y2(b2c2-a22a1 ) +(c22+a22a21 -n2)=0

Taking A= (a22+b22), B= 2(b2c2-a22a1 ), C=(c22+a22a21 -n2)

 this is a quadratic equn A y22 +By2 +C=0

y2=-B ± √( B2 -4AC) /2A 

x2= (-C2-b2y2)/a2 ;

when the rotation is through the x-intercept point, the equn becomes

(IVa) - (IIIa) = x22(a22+b22) +2x2(a2c2-b22b1 ) +(c22+b22b21 -n'2)=0 where

n'2=b22 *n'1 and n'= (x2-b1)2 +(y2)2 = L2

and

 A= (a22+b22), B= 2(a2c2-b22b1 ), C=(c22+b22b21 -n'2)

 this is a quadratic equn A x22 +Bx2 +C=0

x2=-B ± √(B2-4AC) / 2A 

y2= (-C2-a2x2)/b2 ;