9 questions · self-marked practice — reveal the answer and mark yourself.
Solution:
Given points are A(-2, 1), B(0, 5) and C(-1, 2) are collinear.
Slope of
$\text{AB}=\frac{5-1}{0+2}=2$Slope of
$\text{BC}=\frac{2-5}{-1-0}=3$Since the slopes are different, A, b and C are not collinear.
Solution:
Equation of any line perpendicular to $\text{x}\sec\theta+\text{y cosec}\ \theta+\text{y cosec} \theta = \text{a}$ is $\text{x cosec}\ \theta - \text{y} \sec \theta = \text{k}\ .....(\text{i})$ If eq. (i) passes through $(\text{a}\sin^3\theta,\text{a}\sin^3\theta)$ then $\text{a}\cos^3\theta.\text{cosec} \theta-\text{a}\sin^3\theta.\sec\theta=\text{k}$ $\Rightarrow \frac{\text{a}\cos^3\theta}{\sin\theta}-\frac{\text{a}\sin^3\theta}{\cos\theta}=\text{k}$ $\therefore$ Required equation is $\text{x cosec} \ \theta-\text{y} \sec\theta=\frac{\text{a}\cos^3\theta}{\sin\theta}-\frac{\text{a}\sin^3\theta}{\cos\theta}$ $\Rightarrow \frac{\text{x}}{\sin\theta}-\frac{\text{y}}{\sin\theta}=\text{a}\Big[\frac{\cos^4\theta-\sin^4\theta}{\sin\theta\cos\theta}\Big]$ $\Rightarrow \frac{\text{x}\cos\theta-\text{y}\sin\theta}{\sin\theta\cos\theta}=\text{a}\Big[\frac{(\cos^2\theta+\sin^2\theta)(\cos^2\theta-\sin^2\theta)}{\sin\theta\cos\theta}\Big]$ $\Rightarrow \text{x}\cos\theta-\text{y}\sin\theta=\text{a}(\cos^2\theta-\sin^2\theta)$ $\Rightarrow \text{x}\cos\theta-\text{y}\sin\theta=\text{a}\cos 2\theta$ Hence, the given statement is false.Solution:
We know that if the vertices of triangle has integral coordinates, then the triangle cannot be equilateral. So, the given statement is True.