Question

State $T$ for true and $F$ for false. (i) If the vertices of a triangle have integral coordinates, then the triangle cannot be equilateral. (ii) Line joining the points $(3, -4)$ and $(-2, 6)$ is perpendicular to the line joining the points $(-3,6)$ and $(9,-18)$. (iii) The angle between the lines $y=\left(2-\sqrt{3}\right)\left(x+5\right)$ and $y=\left(2+\sqrt{3}\right)\left(x-7\right)$ is $45^{\circ}$. (iv) The points $A(-2, 1)$, $B(0, 5)$ and $C(-1, 2)$ are collinear.

• A. a
• B. b
• C. c
• D. d

Answer 1

The Correct Option is C

(i) $\because$ In equilateral triangle, $tan60^{\circ}=\sqrt{3}=$ Slope of the line, so with integral coordinates as vertices, the triangle cannot be equilateral. (ii) Given points are $A(3, -4)$, $B(-2,6)$, $P(-3,6)$ and $Q(9, -18)$. Slope of $AB=\frac{6+4}{-2-3}=-2$, Slope of $PQ=\frac{-18-6}{9+3}=-2$ Since slope of $AB =$ slope of $PQ$ Therefore, line $AB$ is parallel to line $PQ$. (iii) Given equation of lines are $y=\left(2-\sqrt{3}\right)\left(x+5\right)\quad\ldots\left(i\right)$ and $y=\left(2+\sqrt{3}\right)\left(x-7\right)\quad\ldots\left(ii\right)$ $\therefore$ Slope of $\left(i\right), m_{1}=\left(2-\sqrt{3}\right)$ Slope of $\left(ii\right), m_{2}=\left(2+\sqrt{3}\right)$ If $\theta$ be the angle between the lines $\left(i\right)$ and $\left(ii\right)$, then $tan\,\theta=\left|\frac{m_{1}-m_{2}}{1+m_{1}\,m_{2}}\right|$ $\Rightarrow tan\,\theta=\left|\frac{\left(2-\sqrt{3}\right)-\left(2+\sqrt{3}\right)}{1+\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\right|$ $\Rightarrow tan\,\theta =\left|\frac{-2\sqrt{3}}{1+4-3}\right|$ $\Rightarrow tan\,\theta =\sqrt{3}$ $\Rightarrow tan\,\theta =tan\left(\pi/3\right)$ $\Rightarrow \theta=\pi/3=60^{\circ}$ For obtuse angle, $\pi-\left(\pi/3\right)=2\pi/3=120^{\circ}$ Hence, the angle between the lines are $60^{\circ}$ or $120^{\circ}$. (iv) We have, $A(-2,1)$, $B(0,5)$ and $C(-1, 2)$ Slope of $AB=\frac{5-1}{0+2}=2$, Slope of $BC=\frac{2-5}{-1-0}=3$, Slope of $AC=\frac{2-1}{-1+2}=1$ Since, the slopes are different. Therefore, $A$, $B$ and $C$ are not collinear.