The equation of the pair of lines perpendicular to the lines given equation is
.
Solution:
Given equation is
.
Let
and
be the slopes of the given lines.
Sum of the roots = ![-\frac{\text {coefficient of } x y}{\text {coefficient of } y^{2}}](https://tex.z-dn.net/?f=-%5Cfrac%7B%5Ctext%20%7Bcoefficient%20of%20%7D%20x%20y%7D%7B%5Ctext%20%7Bcoefficient%20of%20%7D%20y%5E%7B2%7D%7D)
– – – – – (1)
Product of the roots = ![-\frac{\text {coefficient of } x^2}{\text {coefficient of } y^{2}}](https://tex.z-dn.net/?f=-%5Cfrac%7B%5Ctext%20%7Bcoefficient%20of%20%7D%20x%5E2%7D%7B%5Ctext%20%7Bcoefficient%20of%20%7D%20y%5E%7B2%7D%7D)
– – – – – (2)
The required lines are perpendicular to these lines.
Slopes of the required lines are ![$-\frac{1}{m_{1}} \text { and }-\frac{1}{m_{2}}](https://tex.z-dn.net/?f=%24-%5Cfrac%7B1%7D%7Bm_%7B1%7D%7D%20%5Ctext%20%7B%20and%20%7D-%5Cfrac%7B1%7D%7Bm_%7B2%7D%7D)
Required lines also passes through the origin,
therefore their y-intercepts are 0.
Hence their equations are:
![$y=-\frac{1}{m_{1}}x \text { and }y=-\frac{1}{m_{2}}x](https://tex.z-dn.net/?f=%24y%3D-%5Cfrac%7B1%7D%7Bm_%7B1%7D%7Dx%20%5Ctext%20%7B%20and%20%7Dy%3D-%5Cfrac%7B1%7D%7Bm_%7B2%7D%7Dx)
Do cross multiplication, we get
![m_1y=-x \ \text{and} \ m_2y=-x](https://tex.z-dn.net/?f=m_1y%3D-x%20%5C%20%20%5Ctext%7Band%7D%20%5C%20%20m_2y%3D-x)
Add x on both sides of the equation, we get
![x+m_1y=0 \ \text{and} \ x+m_2y=0](https://tex.z-dn.net/?f=x%2Bm_1y%3D0%20%5C%20%20%5Ctext%7Band%7D%20%5C%20%20x%2Bm_2y%3D0)
Therefore, the joint equation of the line is
![\left(x+m_{1} y\right)\left(x+m_{2} y\right)=0](https://tex.z-dn.net/?f=%5Cleft%28x%2Bm_%7B1%7D%20y%5Cright%29%5Cleft%28x%2Bm_%7B2%7D%20y%5Cright%29%3D0)
![x^2+m_2xy+m_1xy+m_1m_2y^2=0](https://tex.z-dn.net/?f=x%5E2%2Bm_2xy%2Bm_1xy%2Bm_1m_2y%5E2%3D0)
![x^{2}+\left(m_{1}+m_{2}\right) x y+m_{1} m_{2} y^{2}=0](https://tex.z-dn.net/?f=x%5E%7B2%7D%2B%5Cleft%28m_%7B1%7D%2Bm_%7B2%7D%5Cright%29%20x%20y%2Bm_%7B1%7D%20m_%7B2%7D%20y%5E%7B2%7D%3D0)
Substitute (1) and (2), we get
![$x^{2}+\left(\frac{-2 h}{b}\right) x y+\left(\frac{a}{b}\right) y^{2}=0](https://tex.z-dn.net/?f=%24x%5E%7B2%7D%2B%5Cleft%28%5Cfrac%7B-2%20h%7D%7Bb%7D%5Cright%29%20x%20y%2B%5Cleft%28%5Cfrac%7Ba%7D%7Bb%7D%5Cright%29%20y%5E%7B2%7D%3D0)
To make the denominator same, multiply and divide first term by b.
![$ \frac{b}{b} x^{2}+\left(\frac{-2 h}{b}\right) x y+\left(\frac{a}{b}\right) y^{2}=0](https://tex.z-dn.net/?f=%24%20%5Cfrac%7Bb%7D%7Bb%7D%20x%5E%7B2%7D%2B%5Cleft%28%5Cfrac%7B-2%20h%7D%7Bb%7D%5Cright%29%20x%20y%2B%5Cleft%28%5Cfrac%7Ba%7D%7Bb%7D%5Cright%29%20y%5E%7B2%7D%3D0)
![$\frac{bx^2-2hxy+ay^2}{b} = 0](https://tex.z-dn.net/?f=%24%5Cfrac%7Bbx%5E2-2hxy%2Bay%5E2%7D%7Bb%7D%20%3D%200)
Do cross multiplication, we get
![b x^{2}-2 h x y+a y^{2}=0](https://tex.z-dn.net/?f=b%20x%5E%7B2%7D-2%20h%20x%20y%2Ba%20y%5E%7B2%7D%3D0)
Hence equation of the pair of lines perpendicular to the lines given equation is
.