Question

Find the equation of the straight lines passing through the following pair of points:
(i) (0, 0) and (2, −2)
(ii) (a, b) and (a + c sin α, b + c cos α)
(iii) (0, −a) and (b, 0)
(iv) (a, b) and (a + b, a − b)
(v) (at₁, a/t₁) and (at₂, a/t₂)
(vi) (a cos α, a sin α) and (a cos β, a sin β)

Answer 1

(i) (0, 0) and (2, −2)

$$\begin{gathered} \mathrm{Here},\left(x_1,y_1\right)\equiv \left(0,0\right) \\ \left(x_2,y_2\right)\equiv \left(2,-2\right) \end{gathered}$$

So, the equation of the line passing through the two points (0, 0) and (2, −2) is

$$\begin{gathered} y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right) \\ \Rightarrow y-0=\frac{-2-0}{2-0}\left(x-0\right) \\ \Rightarrow y=-x \end{gathered}$$

(ii) (a, b) and (a + csin α, b + ccos α)

$$\begin{gathered} \mathrm{Here},\left(x_1,y_1\right)\equiv \left(a,b\right) \\ \left(x_2,y_2\right)\equiv \left(a+c\sin \alpha ,b+c\cos \alpha \right) \end{gathered}$$

So, the equation of the line passing through the two given points is

$$\begin{gathered} y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right) \\ \Rightarrow y-b=\frac{b+c\cos \alpha -b}{a+c\sin \alpha -a}\left(x-a\right) \\ \Rightarrow y-b=\cot \alpha \left(x-a\right) \end{gathered}$$

(iii) (0, −a) and (b, 0)

$$\begin{gathered} \mathrm{Here},\left(x_1,y_1\right)\equiv \left(0,-a\right) \\ \left(x_2,y_2\right)\equiv \left(b,0\right) \end{gathered}$$

So, the equation of the line passing through the two points is

$$\begin{gathered} y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right) \\ \Rightarrow y+a=\frac{0+a}{b-0}\left(x-0\right) \\ \Rightarrow ax-by=ab \end{gathered}$$

(iv) (a, b) and (a + b, a − b)

$$\begin{gathered} \mathrm{Here},\left(x_1,y_1\right)\equiv \left(a,b\right) \\ \left(x_2,y_2\right)\equiv \left(a+b,a-b\right) \end{gathered}$$

So, the equation of the line passing through the two points is

$$\begin{gathered} y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right) \\ \Rightarrow y-b=\frac{a-b-b}{a+b-a}\left(x-a\right) \\ \Rightarrow by-b^2=\left(a-2b\right)x-a^2+2ab \\ \Rightarrow \left(a-2b\right)x-by+b^2+2ab-a^2=0 \end{gathered}$$

(v) (at₁, a/t₁) and (at₂, a/t₂)

$$\begin{gathered} \mathrm{Here},\left(x_1,y_1\right)\equiv \left(a{t}_1,\frac{a}{t_1}\right) \\ \left(x_2,y_2\right)\equiv \left(a{t}_2,\frac{a}{t_2}\right) \end{gathered}$$

So, the equation of the line passing through the two points is

$$\begin{gathered} y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right) \\ \Rightarrow y-\frac{a}{t_1}=\frac{{\displaystyle \frac{a}{t_2}}-{\displaystyle \frac{a}{t_1}}}{a{t}_2-a{t}_1}\left(x-a{t}_1\right) \\ \Rightarrow y-\frac{a}{t_1}=\frac{-1}{t_2{t}_1}\left(x-a{t}_1\right) \\ \Rightarrow x+t_1{t}_{2}y=a\left(t_1+t_2\right) \end{gathered}$$

(vi) (acos α, asin α) and (acos β, asin β)

$$\begin{gathered} \mathrm{Here},\left(x_1,y_1\right)\equiv \left(a\cos \alpha ,a\sin \alpha \right) \\ \left(x_2,y_2\right)\equiv \left(a\cos \beta ,a\sin \beta \right) \end{gathered}$$

So, the equation of the line passing through the two points is

$$\begin{gathered} y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right) \\ \Rightarrow y-a\sin \alpha =\frac{a\sin \beta -a\sin \alpha }{a\cos \beta -a\cos \alpha }\left(x-a\cos \alpha \right) \\ \Rightarrow y-a\sin \alpha =\frac{\sin \beta -\sin \alpha }{\cos \beta -\cos \alpha }\left(x-a\cos \alpha \right) \end{gathered}$$

$$\begin{gathered} \Rightarrow y\left(\cos \beta -\cos \alpha \right)-x\left(\sin \beta -\sin \alpha \right)-a\sin \alpha \cos \beta +a\sin \alpha \cos \alpha +a\cos \alpha \sin \beta -a\cos \alpha \sin \alpha =0 \\ \Rightarrow y\left(\cos \beta -\cos \alpha \right)-x\left(\sin \beta -\sin \alpha \right)=a\sin \alpha \cos \beta -a\cos \alpha \sin \beta \\ \Rightarrow 2y\sin \left(\frac{\alpha +\beta }{2}\right)\sin \left(\frac{\alpha -\beta }{2}\right)-2x\sin \left(\frac{\beta -\alpha }{2}\right)\cos \left(\frac{\alpha +\beta }{2}\right)=a\sin \left(\alpha -\beta \right) \\ \Rightarrow 2y\sin \left(\frac{\alpha +\beta }{2}\right)\sin \left(\frac{\alpha -\beta }{2}\right)+2x\sin \left(\frac{\alpha -\beta }{2}\right)\cos \left(\frac{\alpha +\beta }{2}\right)=2a\sin \left(\frac{\alpha -\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right) \\ \Rightarrow x\cos \left(\frac{\alpha +\beta }{2}\right)+y\sin \left(\frac{\alpha +\beta }{2}\right)=a\cos \left(\frac{\alpha -\beta }{2}\right)\left[\mathrm{dividing}\mathrm{by}\sin \left(\frac{\alpha -\beta }{2}\right)\right] \end{gathered}$$