Straight Lines — JEE MCQs with answers

Answer: A — Option A is correct. By Definition 1, $\tan\theta = \sqrt{3}$, and page 153 restricts the inclination to $0^\circ \le \theta \le 180^\circ$. In that range $\tan 60^\circ = \sqrt{3}$, so $\theta = 60^\circ$. Option B is wrong: $\tan 120^\circ = -\sqrt{3}$, which is not the given slope $\sqrt{3}$. Option C is wrong: $300^\circ$ lies outside the range page 153 gives for an inclination, $0^\circ \le \theta \le 180^\circ$ (and in any case $\tan 300^\circ = -\sqrt{3}$). Option D is wrong: $\tan 30^\circ = \frac{1}{\sqrt{3}}$, which is not the given slope $\sqrt{3}$.

Answer: B — Option B is correct: p.151 prints “in his book ‘La Géométry, published in 1637”. $1596$ is the first of the two years in the portrait caption “René Descartes (1596 –1650)” on p.151 — a life-date, not a publication date; the sentence carrying the publication year is a different one, “‘La Géométry, published in 1637”. $1650$ is the second year in that same portrait caption, again a life-date; p.151 attaches no book to it. $1367$ is what transposing the middle two digits produces; it also precedes the caption dates $1596$–$1650$, so it cannot be the year of a book Descartes wrote.

Answer: A — Option A is correct. The page supplies the result and then prints "Students may derive this equation themselves by the same method as in Case I" - it is left as an exercise for the reader, the method being the point-slope form applied at $(d, 0)$. Option B is wrong because it is what assuming every formula in the chapter is derived on the page produces, and no derivation is printed here at all - the page explicitly hands it to the student. Option C is wrong because it is what reversing the order of the chapter produces, and the intercept form appears only in the next section, after Case II. Option D is wrong because it is what confusing "not derived" with "not stated" produces, and the equation is displayed on the page, tagged (4), and used at once in Example 7(ii).

Answer: D — Option D is correct: Expanding gives $lx+ly+p=0$ and $lx+ly-r=0$. Matching each against $Ax+By+C=0$ reads the constant term with its own sign, so $C_1=p$ and $C_2=-r$. Option A is wrong: the coefficients of $x$ and $y$ are each $l$, not $l^2$; expanding $l(x+y)$ gives $lx+ly$. Option B is wrong: both signs are wrong. The first equation's constant term is $+p$, as printed. Option C is wrong: the sign of $C_2$ is wrong: the second equation has $-r$ as its constant term, not $+r$. The resulting numerator would be $|p-r|$ instead of $|p+r|$.

Answer: C — Option C is correct: $\tan\theta=\left|\dfrac{m_2-m_1}{1+m_1m_2}\right|=\sqrt{3}$, so the acute angle is $\theta=60^\circ$. Option A ($150^\circ$) is what you get if you invert the fraction and drop the modulus as well, landing on $90^\circ+\theta$; this is $90^\circ+\theta$ with $\theta=60^\circ$; the two adjacent angles here are $60^\circ$ and $120^\circ$, and $150^\circ$ is neither. Option B ($30^\circ$) is what you get if you invert the fraction, which returns $\cot\theta$ and hence the complement $90^\circ-\theta$; $\tan\theta=\sqrt{3}$, so $\cot\theta=\frac{\sqrt{3}}{3}$ and $\tan^{-1}(\cot\theta)=30^\circ$, which is $90^\circ-\theta$, not the angle asked for. Option D ($120^\circ$) is what you get if you drop the modulus bars that p.157 prints round the displayed result, so the calculator returns the obtuse angle $\varphi$ instead of the acute $\theta$; the ratio without the bars is $- \sqrt{3}$, which is negative, and the angle in $(0^\circ,180^\circ)$ whose tangent is negative is the obtuse one, $\varphi=120^\circ$; the bars are there precisely to return the acute $\theta=60^\circ$ instead.

Answer: C — Option C is correct: the opening sentence of sub-section 9.3 is: We know that every line in a plane contains infinitely many points on it. Option A is wrong because two points determine a line but do not exhaust it; the page says infinitely many points. Option B is wrong because a single point does not determine a line at all, and the page says infinitely many points. Option D is wrong because the page says infinitely many, which is precisely why a condition on its points is needed rather than a list of them.

Answer: C — Option C is correct: If $(x_1,y_1)$ lies on the line, then by definition $Ax_1+By_1+C=0$, so the numerator is $|0|=0$ and $d=0$. For instance $(-1,1)$ satisfies $3x-4y+7=0$ because $-3-4+7=0$, and the formula duly gives $\tfrac{0}{5}=0$. Option A is wrong: this is the distance of the line from the origin, which is what the formula gives at $(x_1,y_1)=(0,0)$. It is non-zero whenever $C\neq 0$, so it cannot be the value at a point of the line unless the line passes through the origin. Option B is wrong: the formula returns 0, not 1. A point on the line has zero perpendicular distance from it, since the foot of the perpendicular is the point itself. Option D is wrong: nothing in the formula is undefined: the denominator $\sqrt{A^2+B^2}$ is non-zero because A and B are not both zero, and the numerator is simply $|0|$.

Answer: C — Option C is correct: p.151 prints “A systematic study of geometry by the use of algebra was first carried out by celebrated French philosopher and mathematician René Descartes, in his book ‘La Géométry, published in 1637.” H. Freudenthal is named on p.151 only as the author of the chapter’s epigraph (“… is a means and even the most powerful means to make children feel the strength of the human spirit … – H. FREUDENTHAL”); the epigraph says nothing about algebra or about a systematic study of geometry. Euclid is not named anywhere in §9.1; the only two people named on p.151 are H. Freudenthal, under the epigraph, and René Descartes. Pierre de Fermat is not named anywhere in §9.1; whatever the wider history, this item asks whom NCERT’s passage credits, and the passage credits René Descartes alone.

Answer: B — Option B is correct: the bars are exactly what makes result (1) return the acute angle. Case I and Case II are about the SIGN of the ratio — Case I says a positive ratio makes $\theta$ acute, Case II says a negative ratio makes $\theta$ obtuse — and a modulus there would destroy the very distinction being drawn. Option A is wrong because Example 2's solution does restate the formula with bars, but the displayed result (1) above it already carries them. Option C is wrong because this is the position exactly reversed: the display carries the bars and the two Case sentences do not. Option D is wrong because Case I and Case II turn on the ratio being positive or negative; bars would make it positive always and the two cases would collapse into one.

Answer: B — Option B is correct: $y-1=-2(x-6)$ gives $y=-2x+13$, so at $x=10$, $y=-7$; at $x=6$ the same line returns $y=1$. Option A ($-9$) is wrong because it is the value at $x=10$ of the line you get by writing $y+y_0=m(x-x_0)$, i.e. adding $y_0$ where NCERT subtracts it -- but the line $y=-2x + 11$ it produces gives $y=-1$ at $x=6$, so it does not pass through $(6,\,1)$. Option C ($-25$) is wrong because it is the value at $x=10$ of the line you get by expanding $m(x-x_0)$ as $mx-x_0$, i.e. never multiplying $x_0$ by $m$ -- but the line $y=-2x - 5$ it produces gives $y=-17$ at $x=6$, so it does not pass through $(6,\,1)$. Option D ($-12$) is wrong because it is the value at $x=10$ of the line you get by interchanging the coordinates of the given point and substituting $(y_0,\,x_0)$ -- but the line $y=-2x + 8$ it produces gives $y=-4$ at $x=6$, so it does not pass through $(6,\,1)$.