NCERT Exemplar Solutions Class 11 Maths Straight Lines — Objective Type Q22–41

NCERT Exemplar Problems, Class 11 Mathematics — Straight Lines, Objective Type questions 22 to 41. All 20 questions solved.

Twenty multiple-choice questions on lines. Most of them need a single formula: the slope-intercept form (22, 23, 34, 39), perpendicularity (24, 25, 39), intercepts (26, 27, 33), distance of a point from a line (28, 30, 31, 38, 40), and the angle between two lines (29). The last few mix geometry with lines: feet of perpendiculars (32), squares and equilateral triangles (35, 41), and a short transformation question (37).

One answer in the printed key is wrong, Question 41. It is explained where it occurs.

Key insight. Test every option against the facts instead of building a line from scratch when the options are equations. A line “passes through $(3,2)$” and “is perpendicular to $y=x$” are two checks that each take a few seconds, and they eliminate three of the four options.

Question 22

A line cutting off intercept $-3$ from the $y$-axis and the tangent at angle to the $x$-axis is $\frac35$, its equation is

Solution. The slope is $\tan\theta=\frac35$ and the $y$-intercept is $-3$, so $y=\frac35x-3$. Multiply by 5: $5y=3x-15$, that is $5y-3x+15=0$. Option (B) has the slope inverted, and option (C) has the intercept $+3$.

Question 23

Slope of a line which cuts off intercepts of equal lengths on the axes is

Solution. Equal intercepts $a$ give $\frac xa+\frac ya=1$, i.e. $y=-x+a$, whose slope is $-1$. (Equal lengths could also mean opposite signs, $\frac xa-\frac ya=1$ with slope $+1$, but $+1$ is not among the options and the standard reading is equal intercepts.)

Question 24

The equation of the straight line passing through the point $(3,2)$ and perpendicular to the line $y=x$ is

Solution. The line $y=x$ has slope 1, so a perpendicular has slope $-1$: $x+y=k$. Through $(3,2)$, $k=5$. Options (A) and (D) have slope $+1$, so they are parallel to $y=x$, and (C) does not pass through $(3,2)$.

Question 25

The equation of the line passing through the point $(1,2)$ and perpendicular to the line $x+y+1=0$ is

Solution. $x+y+1=0$ has slope $-1$, so the perpendicular has slope $1$: $y-2=x-1$, i.e. $y-x-1=0$. All four options have slope 1, so the constant decides: $(1,2)$ gives $2-1-1=0$ for (B) only.

Question 26

The tangent of angle between the lines whose intercepts on the axes are $a,-b$ and $b,-a$, respectively, is

Solution. The first line has intercepts $a$ and $-b$: $\frac xa-\frac yb=1$, so its slope is $m_1=\frac ba$. The second has intercepts $b$ and $-a$: $\frac xb-\frac ya=1$, so $m_2=\frac ab$. Then

$$\tan\theta=\frac{m_1-m_2}{1+m_1m_2}=\frac{\frac ba-\frac ab}{1+1}=\frac{b^2-a^2}{2ab},$$

where the angle is taken from the second line to the first. Taken the other way the sign flips, and the acute angle has tangent $\frac{|b^2-a^2|}{2ab}$ for $a,b>0$. Of the four options only (C) has this form: (A) is wrong by a factor of 2, and (B) has lost $ab$ from the denominator.

Question 27

If the line $\dfrac xa+\dfrac yb=1$ passes through the points $(2,-3)$ and $(4,-5)$, then $(a,b)$ is

Solution. Substituting both points: $\frac2a-\frac3b=1$ and $\frac4a-\frac5b=1$. Subtracting the first from the second gives $\frac2a-\frac2b=0$, so $a=b$. Then $\frac2a-\frac3a=-\frac1a=1$, so $a=b=-1$. Check: $\frac{2}{-1}+\frac{-3}{-1}=-2+3=1$ and $\frac4{-1}+\frac{-5}{-1}=-4+5=1$. $\checkmark$

Question 28

The distance of the point of intersection of the lines $2x-3y+5=0$ and $3x+4y=0$ from the line $5x-2y=0$ is

Solution. From $3x+4y=0$, $x=-\frac{4y}3$. Substituting into the first, $-\frac{8y}3-3y+5=0$, so $-\frac{17y}3=-5$ and $y=\frac{15}{17}$, $x=-\frac{20}{17}$. The distance from $5x-2y=0$ is

$$\frac{\left|5\left(-\frac{20}{17}\right)-2\cdot\frac{15}{17}\right|}{\sqrt{25+4}}=\frac{\frac{130}{17}}{\sqrt{29}}=\frac{130}{17\sqrt{29}}.$$

Question 29

The equations of the lines which pass through the point $(3,-2)$ and are inclined at $60^\circ$ to the line $\sqrt3x+y=1$ is

Solution. The given line has slope $-\sqrt3$, so its inclination is $120^\circ$. A line at $60^\circ$ to it has inclination $120^\circ-60^\circ=60^\circ$ or $120^\circ+60^\circ=180^\circ$. The inclination $180^\circ$ is a horizontal line, and through $(3,-2)$ that is $y+2=0$. The inclination $60^\circ$ has slope $\sqrt3$:

$$y+2=\sqrt3(x-3)\ \Longrightarrow\ \sqrt3x-y-2-3\sqrt3=0.$$

Both lines are needed, so (C) alone is incomplete.

Question 30

The equations of the lines passing through the point $(1,0)$ and at a distance $\frac{\sqrt3}2$ from the origin, are

Solution. A line through $(1,0)$ is $y=m(x-1)$, i.e. $mx-y-m=0$. Its distance from the origin is $\frac{|m|}{\sqrt{m^2+1}}=\frac{\sqrt3}2$. Squaring, $4m^2=3m^2+3$, so $m^2=3$ and $m=\pm\sqrt3$. The lines are $\sqrt3x-y-\sqrt3=0$ and $-\sqrt3x-y+\sqrt3=0$, that is $\sqrt3x+y-\sqrt3=0$. (The vertical line $x=1$ is at distance 1 from the origin, so it is not a candidate.) Option (B) would pass through $(-1,0)$, not $(1,0)$.

Question 31

The distance between the lines $y=mx+c_1$ and $y=mx+c_2$ is

Solution. The lines are $mx-y+c_1=0$ and $mx-y+c_2=0$. Take the point $(0,c_2)$ on the second; its distance from the first is $\frac{|-c_2+c_1|}{\sqrt{m^2+1}}$. A distance is never negative, so the modulus is essential: options (A) and (C) can be negative.

Question 32

The coordinates of the foot of perpendiculars from the point $(2,3)$ on the line $y=3x+4$ is given by

Solution. Write the line as $3x-y+4=0$. The foot of the perpendicular from $(x_0,y_0)$ is $(x_0,y_0)-\frac{ax_0+by_0+c}{a^2+b^2}(a,b)$. Here $3\cdot2-3+4=7$ and $a^2+b^2=10$, so

$$(x,y)=(2,3)-\frac7{10}(3,-1)=\left(2-\frac{21}{10},\ 3+\frac7{10}\right)=\left(-\frac1{10},\frac{37}{10}\right).$$

Check: $3\left(-\frac1{10}\right)+4=\frac{37}{10}$, so it lies on the line, and the join to $(2,3)$ has direction $\left(\frac{21}{10},-\frac7{10}\right)\propto(3,-1)$, perpendicular to the line’s direction $(1,3)$. Option (A) is the same pair with the coordinates swapped.

Question 33

If the coordinates of the middle point of the portion of a line intercepted between the coordinate axes is $(3,2)$, then the equation of the line will be

Solution. If the intercepts are $a$ and $b$, the midpoint of $(a,0)$ and $(0,b)$ is $\left(\frac a2,\frac b2\right)=(3,2)$, so $a=6$, $b=4$. The line is $\frac x6+\frac y4=1$; multiplying by 12 gives $2x+3y=12$.

Question 34

Equation of the line passing through $(1,2)$ and parallel to the line $y=3x-1$ is

Solution. Parallel lines share the slope 3, and the line passes through $(1,2)$, so $y-2=3(x-1)$. Option (B) uses the point $(-1,-2)$, and (A), (D) have slope 1.

Question 35

Equations of diagonals of the square formed by the lines $x=0$, $y=0$, $x=1$ and $y=1$ are

Solution. The square has vertices $(0,0)$, $(1,0)$, $(1,1)$, $(0,1)$. One diagonal joins $(0,0)$ and $(1,1)$: $y=x$. The other joins $(1,0)$ and $(0,1)$: $x+y=1$.

Question 36

For specifying a straight line, how many geometrical parameters should be known?

Solution. The equation $y=mx+c$ contains exactly two constants, the slope $m$ and the intercept $c$. Equivalently, $\frac xa+\frac yb=1$ has two, and $x\cos\omega+y\sin\omega=p$ has two. In $ax+by+c=0$ only the ratios of $a,b,c$ matter, which again leaves two independent numbers.

Question 37

The point $(4,1)$ undergoes the following two successive transformations: (i) Reflection about the line $y=x$; (ii) Translation through a distance $2$ units along the positive $x$-axis. Then the final coordinates of the point are

Solution. Reflection about $y=x$ swaps the coordinates: $(4,1)\to(1,4)$. Translating $2$ units along the positive $x$-axis adds $2$ to $x$: $(1,4)\to(3,4)$. The order matters; doing the translation first would give $(6,1)\to(1,6)$. Option (C) stops after the reflection.

Question 38

A point equidistant from the lines $4x+3y+10=0$, $5x-12y+26=0$ and $7x+24y-50=0$ is

Solution. The three denominators are $\sqrt{16+9}=5$, $\sqrt{25+144}=13$ and $\sqrt{49+576}=25$. For each option, compute the three distances:

• $(1,-1)$: $\frac{11}5,\ \frac{43}{13},\ \frac{67}{25}$ — unequal;
• $(1,1)$: $\frac{17}5,\ \frac{19}{13},\ \frac{19}{25}$ — unequal;
• $(0,0)$: $\frac{10}5=2,\ \frac{26}{13}=2,\ \frac{50}{25}=2$ — all equal;
• $(0,1)$: $\frac{13}5,\ \frac{14}{13},\ \frac{26}{25}$ — unequal.

The origin is at distance 2 from all three lines.

Question 39

A line passes through $(2,2)$ and is perpendicular to the line $3x+y=3$. Its $y$-intercept is

Solution. The line $3x+y=3$ has slope $-3$, so the perpendicular has slope $\frac13$. Through $(2,2)$: $y-2=\frac13(x-2)$. Putting $x=0$: $y=2-\frac23=\frac43$.

Question 40

The ratio in which the line $3x+4y+2=0$ divides the distance between the lines $3x+4y+5=0$ and $3x+4y-5=0$ is

Solution. All three lines are parallel with the same normal length $\sqrt{9+16}=5$, so distances between them are the differences of the constants divided by 5. The constants are $+5$, $+2$ and $-5$. The distance from the middle line $3x+4y+2=0$ to the first is $\frac{|5-2|}5=\frac35$, and to the second is $\frac{|2-(-5)|}5=\frac75$. The ratio is $3:7$. (The total distance between the outer lines is $\frac{10}5=2$, and $\frac35+\frac75=2$. $\checkmark$)

Question 41

One vertex of the equilateral triangle with centroid at the origin and one side as $x+y-2=0$ is

Solution. In an equilateral triangle the centroid divides each altitude $2:1$ from the vertex. The altitude to the side $x+y-2=0$ passes through the origin, and the origin is at distance $\frac{2}{\sqrt2}=\sqrt2$ from that side, which is one third of the altitude. So the opposite vertex is at distance $2\sqrt2$ from the origin, on the other side of the origin from the line, along the direction perpendicular to the side, i.e. along $y=x$ away from $x+y=2$. That gives $(-2,-2)$ (check: $\sqrt{4+4}=2\sqrt2$).

The hint gives the same: let the vertex be $A(h,k)$ and $D(\alpha,\beta)$ the midpoint of the opposite side. Then $\frac{2\alpha+h}3=0$ and $\frac{2\beta+k}3=0$, with $\alpha+\beta=2$, so $h+k=-2(\alpha+\beta)=-4$. The altitude $AD$ is perpendicular to the side, so $\frac kh=1$. Hence $h=k=-2$. As a final check, the other two vertices then sum to $(2,2)$, and both lie on $x+y=2$, whose coordinates sum to $4$: consistent. If $(2,2)$ were a vertex the other two would sum to $(-2,-2)$ and could not lie on $x+y=2$.

The printed key says (B) $(2,2)$, which is wrong. The point $(2,2)$ lies on $x+y=4$, beyond the given side $x+y=2$ as seen from the origin, so that side would lie between the centroid and the vertex. A centroid lies inside the triangle, between a vertex and the opposite side, never outside it. The correct option is (C).

Common mistakes

• Question 26. Taking the tangent as $\frac{a^2-b^2}{ab}$ by dropping the factor $2$ in $1+m_1m_2=2$. The product of the slopes is $\frac ba\cdot\frac ab=1$, so the denominator is $2$, not $1$.
• Question 28. Using the wrong equation of the second line when finding the intersection, or finding the distance from the intersection to the origin instead of to $5x-2y=0$. Solve the two given lines first, then apply the distance formula with the third.
• Question 29. Stopping at the sloping line and missing $y+2=0$. “Inclined at $60^\circ$” to a line of inclination $120^\circ$ gives two lines, and the second has inclination $180^\circ$, so its slope is $0$; option (C) alone is incomplete.
• Question 32. Choosing (A), which is the correct pair with $x$ and $y$ swapped. The foot must satisfy $y=3x+4$; $\left(\frac{37}{10},-\frac1{10}\right)$ does not.
• Question 38. Comparing the numerators $|4x+3y+10|$, $|5x-12y+26|$, $|7x+24y-50|$ without dividing by $5$, $13$ and $25$. The three lines are not normalised, so equal numerators do not mean equal distances; the origin looks lopsided in the numerators ($10,26,50$) and is only symmetric after the division.
• Question 41. Trusting the printed key’s $(2,2)$ without a sanity check. The vertex must lie on the opposite side of the centroid from the base: the base $x+y=2$ is on the positive side of the origin, so the vertex has negative coordinates.

Practise next

• The other pages of this unit: Short and Long Answer Type Q1–21 and Fill in the Blanks, True or False and Matching Q42–59.
• NCERT textbook practice: Exercise 9.2 for forms of the equation, Exercise 9.3 for distance and intersection questions, and the Miscellaneous Exercise.