Locus of the mid-points of the portion of the line $x \sin \theta+y \cos \theta=p$ intercepted between the axes is $\ldots \ldots \ldots .$.
Given equation of the line is
$$x \sin \theta+y \cos \theta=p\quad\text{... (i)}$$
Let the mid-point of $A B$ is $p(h, k)$.
So, the mid-point of $A B$ are $$\left(\frac{a}{2}, \frac{b}{2}\right)$$
Since, the point $(a, 0)$ lies on the line $(\mathrm{i})$, then
$$\begin{aligned} a \sin \theta+0 & =p \\ a \sin \theta & =p \Rightarrow a=\frac{p}{\sin \theta} \end{aligned}$$
and the point $(0, b)$ also lies on the line, then
$$\begin{aligned} & 0+b \cos \theta=p \\ & \Rightarrow \quad b \cos \theta=p \Rightarrow b=\frac{p}{\cos \theta} \\ & \text { Now, mid-point of } A B=\left(\frac{a}{2}, \frac{b}{2}\right) \text { or }\left(\frac{p}{2 \sin \theta}, \frac{p}{2 \cos \theta}\right) \\ & \because \quad \frac{p}{2 \sin \theta}=h \Rightarrow \sin \theta=\frac{p}{2 h} \\ & \text { and } \quad \frac{p}{2 \cos \theta}=k \Rightarrow \cos \theta=\frac{p}{2 k} \\ & \therefore \quad \sin ^2 \theta+\cos ^2 \theta=\frac{p^2}{4 h^2}+\frac{p^2}{4 k^2} \\ & \Rightarrow \quad 1=\frac{p^2}{4}\left(\frac{1}{h^2}+\frac{1}{k^2}\right) \end{aligned}$$
Locus of the mid-point is
$$4=p^2\left(\frac{1}{x^2}+\frac{1}{y^2}\right)$$
$$\Rightarrow \quad 4 x^2 y^2=p^2\left(x^2+y^2\right)$$
If the vertices of a triangle have integral coordinates, then the triangle cannot be equilateral.
The points $A(-2,1), B(0,5)$ and $C(-1,2)$ are collinear.
Equation of the line passing through the point $\left(a \cos ^3 \theta, a \sin ^3 \theta\right)$ and perpendicular to the line $x \sec \theta+y \operatorname{cosec} \theta=a$ is $x \cos \theta-y \sin \theta=a \sin 2 \theta$.
The straight line $5 x+4 y=0$ passes through the point of intersection of the straight lines $x+2 y-10=0$ and $2 x+y+5=0$.