JEE Mathematics Straight Lines MCQs Set 32

Answer: (b) 2

Explanation:
1. The dividing point is \(\left(\dfrac{5-k}{1+k},\dfrac{7+k}{1+k}\right)\).
2. \(\dfrac{12}{1+k}=4\) gives \(k=2\).

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) 1

Explanation:
1. The line through \((h,0)\) and \((0,k)\) is \(\dfrac xh+\dfrac yk=1\).
2. The point \((a,b)\) lies on it, so \(\dfrac ah+\dfrac bk=1\).

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) 9

Explanation:
1. The second slope is \(\dfrac{6-4}{0+1}=2\).
2. \(\dfrac{7-y}{2-3}=2\) gives \(y=9\).

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) 7

Explanation:
1. \(d=\dfrac{|7-5|}{\sqrt{9+16}}=\dfrac25\), so \(a+b=7\).

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) Assertion is false, Reason is true

Explanation:
1. The line through (3, -2) and (3, 4) is vertical, so its slope is undefined, not 0.
2. The reason is a correct statement.

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) Both are true and Reason explains Assertion

Explanation:
1. \( -3/1 = -6/2 = 9/-3 = -3 \), so the lines coincide by the condition in the reason.

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) Assertion is true, Reason is false

Explanation:
1. Parallel lines have equal slopes, so \( a1/a2 = b1/b2 \) is true.
2. Perpendicular lines satisfy \( a1a2 + b1b2 = 0 \), so the reason is false.

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) Both are true but Reason does not explain Assertion

Explanation:
1. Both are standard results: parallel lines keep the same coefficients of \( x \) and \( y \); perpendicular lines swap them with a sign change.
2. Neither statement proves the other.

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) Both are true and Reason explains Assertion

Explanation:
1. Rewrite the second line as \( 3x - 4y - 7.5 = 0 \).
2. \( d = |9 + 7.5| / 5 = 16.5 / 5 = 33/10 \), using the formula in the reason.

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) Both are true but Reason does not explain Assertion

Explanation:
1. Both statements are true, but they are separate facts about two different families of lines.

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) \(\dfrac3{10}\)

Explanation:
1. Rewrite the second as \(3x+4y=\dfrac{15}2\).
2. \(d=\dfrac{|9-7.5|}5=\dfrac3{10}\).

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) \(\dfrac43\)

Explanation:
1. Slope \(\dfrac13\): \(y-2=\dfrac13(x-2)\).
2. Put \(x=0\): \(y=2-\dfrac23=\dfrac43\).

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) \((a,b)\)

Explanation:
1. The first two lines meet at \((-35,22)\).
2. Concurrency: \(-35a+22b=1\), i.e. \(35a-22b+1=0\), so the line \(35x-22y+1=0\) passes through \((a,b)\).

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) \((-1,-14)\)

Explanation:
1. \(\dfrac{5\cdot4-13+6}{26}=\dfrac12\).
2. Image \(=(4,-13)-2\cdot\dfrac12(5,1)=(-1,-14)\).

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) \((1,-2)\)

Explanation:
1. \(a+c=2b\), so \(a-2b+c=0\): \((1,-2)\) satisfies the equation for every such \(a,b,c\).

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) \(x+7y=31\)

Explanation:
1. The point \((-4,5)\) is not on \(7x-y+8=0\) (\(-28-5+8\ne0\)); the second diagonal through \((-4,5)\) is perpendicular to it.
2. Slope \(-\dfrac17\): \(y-5=-\dfrac17(x+4)\) gives \(x+7y=31\).

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) \(\left(\dfrac{13}5,0\right)\)

Explanation:
1. The reflected ray appears to come from the image \((1,-2)\); the line from \((1,-2)\) to \((5,3)\) has slope \(\dfrac54\).
2. It meets the x-axis at \(x=1+\dfrac85=\dfrac{13}5\).

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: (d) \(\dfrac1{a^2}+\dfrac1{b^2}=\dfrac1{p^2}+\dfrac1{q^2}\)

Explanation:
1. The distance of the line from the origin is unchanged by rotation.
2. For intercepts \(a,b\) it equals \(\dfrac{ab}{\sqrt{a^2+b^2}}\), i.e. \(\dfrac1{d^2}=\dfrac1{a^2}+\dfrac1{b^2}\); the same holds for \(p,q\).

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) \(3x+y+7=0,\ x-3y-31=0\)

Explanation:
1. The third side makes equal angles with the equal sides: \(\left|\dfrac{m-7}{1+7m}\right|=\left|\dfrac{m+1}{1-m}\right|\) gives \(m=-3\) or \(m=\dfrac13\).
2. Through \((1,-10)\): \(3x+y+7=0\) and \(x-3y-31=0\).

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) exactly one value of p

Explanation:
1. Both lines are perpendicular to a common line only if they are parallel: \(p(p^2+1)=-(p^2+1)\).
2. Since \(p^2+1\ne0\), \(p=-1\): exactly one value.

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) \(x-y-1=0,\ x+y+1=0\)

Explanation:
1. Slopes \(\pm1\) and \(c=-1\): \(y=x-1\) and \(y=-x-1\).

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: (a) \(\text{below the x-axis at a distance }\dfrac32\)

Explanation:
1. Multiply the first by \(a\) and the second by \(b\) and add: \((a^2+b^2)x=0\), so \(x=0\).
2. Then \(2by+3b=0\) gives \(y=-\dfrac32\): below the x-axis.

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) \((4-\sqrt5)x-(3-2\sqrt5)y+(2-4\sqrt5)=0\)

Explanation:
1. Bisectors: \(\dfrac{x-2y+4}{\sqrt5}=\pm\dfrac{4x-3y+2}5\).
2. The angle between the lines is about \(26.6^\circ\); the bisector with slope about \(-1.2\) makes \(76.7^\circ\) with each line, so it bisects the obtuse angle.
3. Multiplying out that sign choice gives the equation in the first option.

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) below both the lines

Explanation:
1. The lines are parallel: \(2x+3y=4\) and \(2x+3y=-\dfrac83\).
2. At the point, \(2x+3y=-6\), which is less than both values, so it lies below both lines.

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) \(18\sqrt2\)

Explanation:
1. Distance from \(O\) to the line is \(\dfrac{15}5=3\).
2. \(AB=2\sqrt{81-9}=12\sqrt2\), so the area \(=\dfrac12\cdot12\sqrt2\cdot3=18\sqrt2\).

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.