Question

Find the equation of the line through the point A$(2,3)$ and making an angle of ${45}^{\circ }$ with the $x$ axis Also determine the length of intercept on it between A and the line $x+y+1=0$

Answer 1

The equation of a line through A and making an angle of ${45}^{\circ }$ with the x axis is

$\frac{x-2}{\cos {45}^{\circ }}=\frac{y-3}{\sin {45}^{\circ }}$ or $\frac{x-2}{\frac{1}{\sqrt{2}}}=\frac{y-3}{\frac{1}{\sqrt{2}}}$ or x - y + 1 = 0

Suppose this line meets the line x + y + 1 = 0 at P such that AP = r Then the co-ordinates of P are given by

$\frac{x-2}{\cos {45}^{\circ }}=\frac{y-3}{\sin {45}^{\circ }}=r$

$\Rightarrow x=2+r\cos {45}^{\circ },y=3+r\sin {45}^{\circ }$

$\Rightarrow x=2+\frac{r}{\sqrt{2}},y=3+\frac{r}{\sqrt{2}}$

Thus the co-ordinates of P are $(2+\frac{r}{\sqrt{2}},3+\frac{r}{\sqrt{2}})$

Since P lies on x + y + 1 = 0 so $2+\frac{r}{\sqrt{2}}+3+\frac{r}{\sqrt{2}}+1=0$

$\Rightarrow \sqrt{2}r=-6\Rightarrow r=-3\sqrt{2}$

$\Rightarrow AP=\left|r\right|=3\sqrt{2}$

Thus the length of the intercept = $3\sqrt{2}$