Question

Find the equations to the straight line passing through the foot of the perpendicular from the point (h, k) upon the straight line \[Ax + By + C = 0\] and bisecting the angles between the perpendicular and the given straight line.

Answer 1

Hint: We know the equation of a line passing through the point \[(x,y)\] and having a slope \[m\] is \[Y - y = m(X - x)\] . If a line having slope \[{m_1}\] is perpendicular to a line having a slope \[{m_2}\] then we have \[{m_1} \times {m_2} = - 1\] . We also know that equation of bisector of the angles between the lines \[{A_1}x + {B_1}y + {C_1} = 0\] and \[{A_2}x + {B_2}y + {C_2} = 0\] is \[\dfrac{{{A_1}x + {B_1}y + {C_1}}}{{\sqrt {{a_1}^2 + {b_1}^2} }} = \pm \dfrac{{{A_2}x + {B_2}y + {C_2}}}{{\sqrt {{a_2}^2 + {b_2}^2} }}\] . Using these we can solve the given problem.

Complete step-by-step answer:
Given, \[Ax + By + C = 0{\text{ - - - - - - (1)}}\]
Rearranging the terms in above equation we have,
 \[ \Rightarrow By = - Ax - C\]
Divided by ‘B’ on both sides we get,
 \[ \Rightarrow y = - \dfrac{A}{B}x - \dfrac{C}{B} \]
Comparing this with the equation of straight line \[y = mx + c\] we have slope,
 \[ \Rightarrow m = - \dfrac{A}{B}\] .
Let assume that the slope of the perpendicular line is \[m'\] .
Since both lines are perpendicular we have,
 \[ \Rightarrow m \times m' = - 1\]
Substituting we have,
 \[ \Rightarrow - \dfrac{A}{B} \times m' = - 1\]
Multiply by \[ - \dfrac{B}{A}\] on both sides we have,
 \[ \Rightarrow m' = \dfrac{B}{A}\] .
Now equation of a perpendicular line passing through (h, k) and slope \[\dfrac{B}{A}\] is,
 \[ \Rightarrow y - k = \dfrac{B}{A}(x - h)\]
Multiply by ‘A’ on both sides
 \[ \Rightarrow A(y - k) = B(x - h)\]
Expanding the brackets,
 \[ \Rightarrow Ay - Ak = Bx - Bh\]
Rearranging we have,
 \[ \Rightarrow Bx - Ay - Bh + Ak = 0{\text{ - - - - - - (2)}}\]
Now angel bisector of (1) and (2) is given by,
 \[ \Rightarrow \dfrac{{Bx - Ay - Bh + Ak}}{{\sqrt {{B^2} + {A^2}} }} = \pm \dfrac{{Ax + By + C}}{{\sqrt {{A^2} + {B^2}} }}\]
Denominator on both sides will get cancels,
 \[ \Rightarrow Bx - Ay - Bh + Ak = \pm (Ax + By + C)\]
Rearranging we have,
 \[ \Rightarrow Bx - Bh - Ay + Ak = \pm (Ax + By + C)\]
Taking B and A as common we will have,
 \[ \Rightarrow B(x - h) - A(y - k) = \pm (Ax + By + C)\] , is the required answer.
So, the correct answer is “ \[ B(x - h) - A(y - k) = \pm (Ax + By + C)\]”.

Note: All we used in the above problem is the equation of a line passing through a point with a slope and equation of bisector of the angles between the two lines. Remember them well. Careful in the simplification part, if we miss one term the whole problem will be wrong.