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Question: Find the condition that the diagonals of the parallelogram formed by the lines $ax + by + c = 0; ax...

Find the condition that the diagonals of the parallelogram formed by the lines

ax+by+c=0;ax+by+c′=0;a′x+b′y+c=0ax + by + c = 0; ax + by + c' = 0; a'x + b'y +c=0 & a′x+b′y+c′=0a'x + b'y + c' = 0 are at right angles. Also find the equation to the diagonals of the parallelogram.

Answer

The condition that the diagonals of the parallelogram are at right angles is: a2+b2=(a′)2+(b′)2a^2 + b^2 = (a')^2 + (b')^2

The equations to the diagonals of the parallelogram are:

  1. (a−a′)x+(b−b′)y=0(a-a')x + (b-b')y = 0
  2. (a+a′)x+(b+b′)y+(c+c′)=0(a+a')x + (b+b')y + (c+c') = 0
Explanation

Solution

The four given lines are: L1:ax+by+c=0L_1: ax + by + c = 0 L2:ax+by+c′=0L_2: ax + by + c' = 0 L3:a′x+b′y+c=0L_3: a'x + b'y + c = 0 L4:a′x+b′y+c′=0L_4: a'x + b'y + c' = 0

These lines form a parallelogram because L1∥L2L_1 \parallel L_2 (both have slope −a/b-a/b) and L3∥L4L_3 \parallel L_4 (both have slope −a′/b′-a'/b').

1. Condition for diagonals to be at right angles

Let the vertices of the parallelogram be: A = L1∩L3L_1 \cap L_3 B = L1∩L4L_1 \cap L_4 C = L2∩L4L_2 \cap L_4 D = L2∩L3L_2 \cap L_3

The diagonals are AC and BD.

The slope of the lines L1L_1 and L2L_2 is m1=−a/bm_1 = -a/b. The slope of the lines L3L_3 and L4L_4 is m2=−a′/b′m_2 = -a'/b'.

The vertices are: A=L1∩L3:(c(b′−b)ab′−a′b,c(a−a′)ab′−a′b)A = L_1 \cap L_3: \left(\frac{c(b'-b)}{ab'-a'b}, \frac{c(a-a')}{ab'-a'b}\right) B=L1∩L4:(bc′−b′cab′−a′b,a′c−ac′ab′−a′b)B = L_1 \cap L_4: \left(\frac{bc'-b'c}{ab'-a'b}, \frac{a'c-ac'}{ab'-a'b}\right) C=L2∩L4:(c′(b′−b)ab′−a′b,c′(a−a′)ab′−a′b)C = L_2 \cap L_4: \left(\frac{c'(b'-b)}{ab'-a'b}, \frac{c'(a-a')}{ab'-a'b}\right) D=L2∩L3:(bc−b′c′ab′−a′b,a′c′−acab′−a′b)D = L_2 \cap L_3: \left(\frac{bc-b'c'}{ab'-a'b}, \frac{a'c'-ac}{ab'-a'b}\right)

Slope of diagonal AC (mACm_{AC}): mAC=yC−yAxC−xA=c′(a−a′)ab′−a′b−c(a−a′)ab′−a′bc′(b′−b)ab′−a′b−c(b′−b)ab′−a′b=(c′−c)(a−a′)(c′−c)(b′−b)=a−a′b′−bm_{AC} = \frac{y_C - y_A}{x_C - x_A} = \frac{\frac{c'(a-a')}{ab'-a'b} - \frac{c(a-a')}{ab'-a'b}}{\frac{c'(b'-b)}{ab'-a'b} - \frac{c(b'-b)}{ab'-a'b}} = \frac{(c'-c)(a-a')}{(c'-c)(b'-b)} = \frac{a-a'}{b'-b} (assuming c≠c′c \neq c')

Slope of diagonal BD (mBDm_{BD}): mBD=yD−yBxD−xB=a′c′−acab′−a′b−a′c−ac′ab′−a′bbc−b′c′ab′−a′b−bc′−b′cab′−a′b=a′c′−ac−a′c+ac′bc−b′c′−bc′+b′c=(a+a′)(c′−c)−(b+b′)(c′−c)=−a+a′b+b′m_{BD} = \frac{y_D - y_B}{x_D - x_B} = \frac{\frac{a'c'-ac}{ab'-a'b} - \frac{a'c-ac'}{ab'-a'b}}{\frac{bc-b'c'}{ab'-a'b} - \frac{bc'-b'c}{ab'-a'b}} = \frac{a'c'-ac-a'c+ac'}{bc-b'c'-bc'+b'c} = \frac{(a+a')(c'-c)}{-(b+b')(c'-c)} = -\frac{a+a'}{b+b'} (assuming c≠c′c \neq c')

For the diagonals to be at right angles, mAC⋅mBD=−1m_{AC} \cdot m_{BD} = -1. (a−a′b′−b)⋅(−a+a′b+b′)=−1\left(\frac{a-a'}{b'-b}\right) \cdot \left(-\frac{a+a'}{b+b'}\right) = -1 (a−a′)(a+a′)(b−b′)(b+b′)=−1\frac{(a-a')(a+a')}{(b-b')(b+b')} = -1 a2−(a′)2b2−(b′)2=−1\frac{a^2 - (a')^2}{b^2 - (b')^2} = -1 a2−(a′)2=−(b2−(b′)2)a^2 - (a')^2 = -(b^2 - (b')^2) a2−(a′)2=−b2+(b′)2a^2 - (a')^2 = -b^2 + (b')^2 a2+b2=(a′)2+(b′)2a^2 + b^2 = (a')^2 + (b')^2

This is the condition for the diagonals to be at right angles.

2. Equation to the diagonals of the parallelogram

The two diagonals are given by the equations: Diagonal 1: L1−L2=k(L3−L4)L_1 - L_2 = k(L_3 - L_4) (ax+by+c)−(ax+by+c′)=k((a′x+b′y+c)−(a′x+b′y+c′))(ax+by+c) - (ax+by+c') = k((a'x+b'y+c) - (a'x+b'y+c')) c−c′=k(c−c′)c - c' = k(c - c') This implies k=1k=1. So, one diagonal is L1−L3=L2−L4L_1 - L_3 = L_2 - L_4. (ax+by+c)−(a′x+b′y+c)=(ax+by+c′)−(a′x+b′y+c′)(ax+by+c) - (a'x+b'y+c) = (ax+by+c') - (a'x+b'y+c') (a−a′)x+(b−b′)y=(a−a′)x+(b−b′)y(a-a')x + (b-b')y = (a-a')x + (b-b')y This is an identity, not an equation of a diagonal.

The correct method for finding the equations of the diagonals of a parallelogram formed by the lines L1=0,L2=0,L3=0,L4=0L_1=0, L_2=0, L_3=0, L_4=0 is to use the fact that the diagonals pass through the intersection points of opposite vertices. Let the lines be P1=ax+by+c=0P_1=ax+by+c=0, P2=ax+by+c′=0P_2=ax+by+c'=0, Q1=a′x+b′y+c=0Q_1=a'x+b'y+c=0, Q2=a′x+b′y+c′=0Q_2=a'x+b'y+c'=0. The vertices are (P1,Q1)(P_1, Q_1), (P1,Q2)(P_1, Q_2), (P2,Q2)(P_2, Q_2), (P2,Q1)(P_2, Q_1). The diagonals connect (P1,Q1)(P_1, Q_1) to (P2,Q2)(P_2, Q_2) and (P1,Q2)(P_1, Q_2) to (P2,Q1)(P_2, Q_1).

Equation of diagonal AC (connecting L1∩L3L_1 \cap L_3 and L2∩L4L_2 \cap L_4): This diagonal passes through the intersection of L1=0L_1=0 and L3=0L_3=0, and the intersection of L2=0L_2=0 and L4=0L_4=0. The general equation of a line passing through the intersection of L1=0L_1=0 and L3=0L_3=0 is L1+λL3=0L_1 + \lambda L_3 = 0. ax+by+c+λ(a′x+b′y+c)=0ax+by+c + \lambda(a'x+b'y+c) = 0 This line also passes through the intersection of L2=0L_2=0 and L4=0L_4=0. So, (ax+by+c′)+λ(a′x+b′y+c′)=0(ax+by+c') + \lambda(a'x+b'y+c') = 0. Subtracting the two equations: (ax+by+c)+λ(a′x+b′y+c)−[(ax+by+c′)+λ(a′x+b′y+c′)]=0(ax+by+c) + \lambda(a'x+b'y+c) - [(ax+by+c') + \lambda(a'x+b'y+c')] = 0 (c−c′)+λ(c−c′)=0(c-c') + \lambda(c-c') = 0 (c−c′)(1+λ)=0(c-c')(1+\lambda) = 0. Since c≠c′c \neq c', we must have 1+λ=0⇒λ=−11+\lambda = 0 \Rightarrow \lambda = -1. Substitute λ=−1\lambda = -1 into L1+λL3=0L_1 + \lambda L_3 = 0: (ax+by+c)−(a′x+b′y+c)=0(ax+by+c) - (a'x+b'y+c) = 0 (a−a′)x+(b−b′)y=0(a-a')x + (b-b')y = 0 This is the equation of one diagonal.

Equation of diagonal BD (connecting L1∩L4L_1 \cap L_4 and L2∩L3L_2 \cap L_3): This diagonal passes through the intersection of L1=0L_1=0 and L4=0L_4=0, and the intersection of L2=0L_2=0 and L3=0L_3=0. The general equation of a line passing through the intersection of L1=0L_1=0 and L4=0L_4=0 is L1+μL4=0L_1 + \mu L_4 = 0. ax+by+c+μ(a′x+b′y+c′)=0ax+by+c + \mu(a'x+b'y+c') = 0 This line also passes through the intersection of L2=0L_2=0 and L3=0L_3=0. So, (ax+by+c′)+μ(a′x+b′y+c)=0(ax+by+c') + \mu(a'x+b'y+c) = 0. Subtracting the two equations: (ax+by+c)+μ(a′x+b′y+c′)−[(ax+by+c′)+μ(a′x+b′y+c)]=0(ax+by+c) + \mu(a'x+b'y+c') - [(ax+by+c') + \mu(a'x+b'y+c)] = 0 (c−c′)+μ(c′−c)=0(c-c') + \mu(c'-c) = 0 (c−c′)−μ(c−c′)=0(c-c') - \mu(c-c') = 0 (c−c′)(1−μ)=0(c-c')(1-\mu) = 0. Since c≠c′c \neq c', we must have 1−μ=0⇒μ=11-\mu = 0 \Rightarrow \mu = 1. Substitute μ=1\mu = 1 into L1+μL4=0L_1 + \mu L_4 = 0: (ax+by+c)+(a′x+b′y+c′)=0(ax+by+c) + (a'x+b'y+c') = 0 (a+a′)x+(b+b′)y+(c+c′)=0(a+a')x + (b+b')y + (c+c') = 0 This is the equation of the other diagonal.

The two equations of the diagonals are:

  1. (a−a′)x+(b−b′)y=0(a-a')x + (b-b')y = 0
  2. $(a+a')x + (b+b')y + (c+c') = 0