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Mathematics Question on Slope of a line

State TT for true and FF 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)(3, -4) and (2,6)(-2, 6) is perpendicular to the line joining the points (3,6)(-3,6) and (9,18)(9,-18). (iii) The angle between the lines y=(23)(x+5)y=\left(2-\sqrt{3}\right)\left(x+5\right) and y=(2+3)(x7)y=\left(2+\sqrt{3}\right)\left(x-7\right) is 4545^{\circ}. (iv) The points A(2,1)A(-2, 1), B(0,5)B(0, 5) and C(1,2)C(-1, 2) are collinear.

A

a

B

b

C

c

D

d

Answer

c

Explanation

Solution

(i) \because In equilateral triangle, tan60=3=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)A(3, -4), B(2,6)B(-2,6), P(3,6)P(-3,6) and Q(9,18)Q(9, -18). Slope of AB=6+423=2AB=\frac{6+4}{-2-3}=-2, Slope of PQ=1869+3=2PQ=\frac{-18-6}{9+3}=-2 Since slope of AB=AB = slope of PQPQ Therefore, line ABAB is parallel to line PQPQ. (iii) Given equation of lines are y=(23)(x+5)(i)y=\left(2-\sqrt{3}\right)\left(x+5\right)\quad\ldots\left(i\right) and y=(2+3)(x7)(ii)y=\left(2+\sqrt{3}\right)\left(x-7\right)\quad\ldots\left(ii\right) \therefore Slope of (i),m1=(23)\left(i\right), m_{1}=\left(2-\sqrt{3}\right) Slope of (ii),m2=(2+3)\left(ii\right), m_{2}=\left(2+\sqrt{3}\right) If θ\theta be the angle between the lines (i)\left(i\right) and (ii)\left(ii\right), then tanθ=m1m21+m1m2tan\,\theta=\left|\frac{m_{1}-m_{2}}{1+m_{1}\,m_{2}}\right| tanθ=(23)(2+3)1+(23)(2+3)\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| tanθ=231+43\Rightarrow tan\,\theta =\left|\frac{-2\sqrt{3}}{1+4-3}\right| tanθ=3\Rightarrow tan\,\theta =\sqrt{3} tanθ=tan(π/3)\Rightarrow tan\,\theta =tan\left(\pi/3\right) θ=π/3=60\Rightarrow \theta=\pi/3=60^{\circ} For obtuse angle, π(π/3)=2π/3=120\pi-\left(\pi/3\right)=2\pi/3=120^{\circ} Hence, the angle between the lines are 6060^{\circ} or 120120^{\circ}. (iv) We have, A(2,1)A(-2,1), B(0,5)B(0,5) and C(1,2)C(-1, 2) Slope of AB=510+2=2AB=\frac{5-1}{0+2}=2, Slope of BC=2510=3BC=\frac{2-5}{-1-0}=3, Slope of AC=211+2=1AC=\frac{2-1}{-1+2}=1 Since, the slopes are different. Therefore, AA, BB and CC are not collinear.