JEE Mathematics Straight Lines MCQs Set 31

Answer: (c) \(x+y-2=0\)

Explanation:
1. Distances: \(\dfrac1{\sqrt2}\approx0.71\), \(\dfrac3{\sqrt5}\approx1.34\), \(\dfrac2{\sqrt5}\approx0.89\) and \(\dfrac2{\sqrt2}\approx1.41\).
2. The largest is for x + y - 2 = 0.

Teacher's Note:
1. Convert the geometric condition into an equation in slopes, intercepts or distances before solving.
2. Check each result by substituting a point or a slope back into the original condition.

Answer: (d) \(x+y-6=0\)

Explanation:
1. Equal intercepts: \(x+y=a\). Through \((2,4)\): \(a=6\).

Teacher's Note:
1. Convert the geometric condition into an equation in slopes, intercepts or distances before solving.
2. Check each result by substituting a point or a slope back into the original condition.

Answer: (c) \(\dfrac{b^2-a^2}{2ab}\)

Explanation:
1. The slopes are \(\dfrac ba\) and \(\dfrac ab\).
2. \(\tan\theta=\dfrac{b/a-a/b}{1+1}=\dfrac{b^2-a^2}{2ab}\).

Teacher's Note:
1. Convert the geometric condition into an equation in slopes, intercepts or distances before solving.
2. Check each result by substituting a point or a slope back into the original condition.

Answer: (b) \(2x+3y=12\)

Explanation:
1. Intercepts are \(6\) and \(4\): \(\dfrac x6+\dfrac y4=1\), i.e. \(2x+3y=12\).

Teacher's Note:
1. Convert the geometric condition into an equation in slopes, intercepts or distances before solving.
2. Check each result by substituting a point or a slope back into the original condition.

Answer: (d) \(2x+y=4\)

Explanation:
1. \(\dfrac xa+\dfrac y{2a}=1\); through \((1,2)\): \(\dfrac1a+\dfrac1a=1\) gives \(a=2\).
2. The line is \(\dfrac x2+\dfrac y4=1\), i.e. \(2x+y=4\).

Teacher's Note:
1. Convert the geometric condition into an equation in slopes, intercepts or distances before solving.
2. Check each result by substituting a point or a slope back into the original condition.

Answer: (b) p = p'

Explanation:
1. Take the point \((8,-1)\) on the first line: distance to \(3x+4y=0\) is \(\dfrac{|24-4|}5=4\).
2. Distance to \(5x-12y=0\) is \(\dfrac{|40+12|}{13}=4\), so \(p=p\prime\). (The first line is the bisector of the angle between the other two.)

Teacher's Note:
1. Convert the geometric condition into an equation in slopes, intercepts or distances before solving.
2. Check each result by substituting a point or a slope back into the original condition.

Answer: (c) form a right angled triangle

Explanation:
1. Vertices (solving pairs): \((1,2),\ (\tfrac34,\tfrac{17}8),\ (\tfrac{23}{26},\tfrac{25}{13})\); the lines are not concurrent since \(6+8-13\ne0\).
2. Squared side lengths: \(\dfrac5{64},\dfrac{49}{832},\dfrac1{52}\), and \(\dfrac{49}{832}+\dfrac1{52}=\dfrac5{64}\), so the triangle is right angled by the converse of Pythagoras.

Teacher's Note:
1. Convert the geometric condition into an equation in slopes, intercepts or distances before solving.
2. Check each result by substituting a point or a slope back into the original condition.

Answer: (a) \(aa\prime+bb\prime=0\)

Explanation:
1. Slopes \(-\dfrac ab\) and \(-\dfrac{a\prime}{b\prime}\) multiply to \(-1\), giving \(aa\prime+bb\prime=0\).

Teacher's Note:
1. Convert the geometric condition into an equation in slopes, intercepts or distances before solving.
2. Check each result by substituting a point or a slope back into the original condition.

Answer: (d) \(\sqrt3x+y+5=0\)

Explanation:
1. Slope \(=\tan120^\circ=-\sqrt3\), y-intercept \(-5\).
2. \(y=-\sqrt3x-5\), i.e. \(\sqrt3x+y+5=0\).

Teacher's Note:
1. Convert the geometric condition into an equation in slopes, intercepts or distances before solving.
2. Check each result by substituting a point or a slope back into the original condition.

Answer: (c) \(2x-3y+6=0\)

Explanation:
1. Slope of given line \(-\dfrac32\), so the required slope is \(\dfrac23\).
2. \(y-4=\dfrac23(x-3)\) gives \(2x-3y+6=0\).

Teacher's Note:
1. Convert the geometric condition into an equation in slopes, intercepts or distances before solving.
2. Check each result by substituting a point or a slope back into the original condition.

Answer: (b) \(x+y=2\)

Explanation:
1. Intercepts are 2 and 2, so \(\dfrac x2+\dfrac y2=1\).

Teacher's Note:
1. Convert the geometric condition into an equation in slopes, intercepts or distances before solving.
2. Check each result by substituting a point or a slope back into the original condition.

Answer: (b) \(60^\circ\)

Explanation:
1. \(\tan\theta=\left|\dfrac{(2+\sqrt3)-(2-\sqrt3)}{1+(2^2-3)}\right|=\dfrac{2\sqrt3}2=\sqrt3\).
2. So \(\theta=60^\circ\).

Teacher's Note:
1. Convert the geometric condition into an equation in slopes, intercepts or distances before solving.
2. Check each result by substituting a point or a slope back into the original condition.

Answer: (d) \(\dfrac{\sqrt3}2\)

Explanation:
1. Distance \(=\dfrac{2p}{\sqrt{1+4b^2}}=p\) gives \(1+4b^2=4\).
2. \(b=\dfrac{\sqrt3}2\) (taking \(b>0\)).

Teacher's Note:
1. Convert the geometric condition into an equation in slopes, intercepts or distances before solving.
2. Check each result by substituting a point or a slope back into the original condition.

Answer: (a) \(\dfrac x2-\dfrac y3=1\text{ and }\dfrac x{-2}+\dfrac y1=1\)

Explanation:
1. \(\dfrac4a+\dfrac3b=1\) with \(b=-1-a\): \(a^2=4\), so \(a=2,b=-3\) or \(a=-2,b=1\).
2. These are the two lines listed.

Teacher's Note:
1. Convert the geometric condition into an equation in slopes, intercepts or distances before solving.
2. Check each result by substituting a point or a slope back into the original condition.

Answer: (c) \((5,6)\)

Explanation:
1. Foot \(=(2,3)-\dfrac{2+3-11}{2}(1,1)=(2+3,3+3)=(5,6)\).

Teacher's Note:
1. Convert the geometric condition into an equation in slopes, intercepts or distances before solving.
2. Check each result by substituting a point or a slope back into the original condition.

Answer: (d) \(2x+y-5=0\)

Explanation:
1. The intersection is \((2,1)\).
2. Of the options only \(2x+y-5=0\) passes through \((2,1)\) and has distance \(\dfrac5{\sqrt5}=\sqrt5\) from the origin.

Teacher's Note:
1. Convert the geometric condition into an equation in slopes, intercepts or distances before solving.
2. Check each result by substituting a point or a slope back into the original condition.

Answer: (a) \(\dfrac54\)

Explanation:
1. Slope \(\dfrac34\): \(y-3=\dfrac34(x-3)\). Intercepts: \(x=-1\) and \(y=\dfrac34\).
2. Length \(=\sqrt{1+\dfrac9{16}}=\dfrac54\).

Teacher's Note:
1. Convert the geometric condition into an equation in slopes, intercepts or distances before solving.
2. Check each result by substituting a point or a slope back into the original condition.

Answer: (c) \(x-y+5=0\)

Explanation:
1. \(\dfrac xa-\dfrac ya=1\) gives \(x-y=a\); through \((-3,2)\): \(a=-5\).
2. So \(x-y+5=0\).

Teacher's Note:
1. Convert the geometric condition into an equation in slopes, intercepts or distances before solving.
2. Check each result by substituting a point or a slope back into the original condition.

Answer: (d) \(2x-y-4=0\)

Explanation:
1. The diagonals bisect each other at \((3,2)\).
2. Line of gradient 2 through \((3,2)\): \(y-2=2(x-3)\), i.e. \(2x-y-4=0\).

Teacher's Note:
1. Convert the geometric condition into an equation in slopes, intercepts or distances before solving.
2. Check each result by substituting a point or a slope back into the original condition.

Answer: (a) II and III only

Explanation:
1. \(A^2=2^2+3^2=13\), \(B^2=1+9=10\), \(C^2=1+4=5\).
2. I: \(13\ne15\) false. II: \(10=2\cdot5\) true. III: \(2\cdot13\cdot5=130=13\cdot10\) true.

Teacher's Note:
1. Convert the geometric condition into an equation in slopes, intercepts or distances before solving.
2. Check each result by substituting a point or a slope back into the original condition.

Answer: (a) All of I, II and III

Explanation:
1. Slope = 3/4. Put y = 0: x = -10/3. Put x = 0: y = 10/4 = 5/2.
2. All three are true.

Teacher's Note:
1. Convert the geometric condition into an equation in slopes, intercepts or distances before solving.
2. Check each result by substituting a point or a slope back into the original condition.

Answer: (a) Both I and II

Explanation:
1. Divide by 2: (sqrt3/2)x + (1/2)y = 4, which is x cos 30 + y sin 30 = 4.
2. So p = 4 and omega = 30 degrees.

Teacher's Note:
1. Convert the geometric condition into an equation in slopes, intercepts or distances before solving.
2. Check each result by substituting a point or a slope back into the original condition.

Answer: (a) 1

Explanation:
1. Slope of the last two points is \(\dfrac{5-1}{4-2}=2\).
2. \(\dfrac{1-(-1)}{2-x}=2\) gives \(x=1\).

Teacher's Note:
1. Convert the geometric condition into an equation in slopes, intercepts or distances before solving.
2. Check each result by substituting a point or a slope back into the original condition.

Answer: (d) 5

Explanation:
1. The line is \(12x-5y+82=0\).
2. \(d=\dfrac{|-12-5+82|}{13}=\dfrac{65}{13}=5\).

Teacher's Note:
1. Convert the geometric condition into an equation in slopes, intercepts or distances before solving.
2. Check each result by substituting a point or a slope back into the original condition.

Answer: (c) 3

Explanation:
1. The slope of the perpendicular \(OP\) is \(-2\), so \(m=\dfrac12\).
2. Substituting the foot: \(2=-\dfrac12+c\), so \(c=\dfrac52\) and \(m+c=3\).

Teacher's Note:
1. Convert the geometric condition into an equation in slopes, intercepts or distances before solving.
2. Check each result by substituting a point or a slope back into the original condition.