NCERT Exemplar Solutions Class 11 Maths Straight Lines — Fill in the Blanks, True or False and Matching Q42–59

NCERT Exemplar Problems, Class 11 Mathematics — Straight Lines, Fill in the Blanks, True or False and Matching questions 42 to 59. All 18 questions solved.

Eighteen short-answer items. The blanks (42 to 47) are one-line results from earlier techniques: a family of lines through a fixed point, equal intercepts, the angle formula, the side-of-a-line test, and two locus problems. The true/false statements (48 to 56) each need a quick check or a counterexample, and the matching questions (57 to 59) reuse trisection points, distance from a line, and the family $L_1+\lambda L_2=0$ through an intersection point.

One printed answer has a slip, Question 46, where the constant should be $172$, not $182$. It is explained where it occurs.

Key insight. For a true-or-false statement, decide quickly whether to prove or to find a counterexample, and then test one concrete case. A single line of substitution settles most of 48 to 56. For a family of lines $L_1+\lambda L_2=0$, a condition such as “passes through a point” or “is parallel to an axis” becomes one linear equation in $\lambda$.

Question 42

If $a,b,c$ are in A.P., then the straight lines $ax+by+c=0$ will always pass through ____.

Solution. $a,b,c$ in A.P. means $2b=a+c$, i.e. $a-2b+c=0$. Compare this with putting $(x,y)=(1,-2)$ in $ax+by+c=0$: it gives $a-2b+c=0$. So every line of the family passes through $(1,-2)$.

Question 43

The line which cuts off equal intercept from the axes and pass through the point $(1,-2)$ is ____.

Solution. The line is $\frac xa+\frac ya=1$, i.e. $x+y=a$. Through $(1,-2)$ we get $a=-1$. (The line $y=-2x$ through the origin also has equal intercepts, both zero, but the book’s answer is the intercept-form line.)

Question 44

Equations of the lines through the point $(3,2)$ and making an angle of $45^\circ$ with the line $x-2y=3$ are ____.

Solution. The line $x-2y=3$ has slope $\frac12$. A line of slope $m$ making $45^\circ$ with it satisfies

$$1=\left|\frac{m-\frac12}{1+\frac m2}\right|\ \Longrightarrow\ m-\tfrac12=\pm\left(1+\tfrac m2\right).$$

The $+$ sign gives $\frac m2=\frac32$, so $m=3$. The $-$ sign gives $\frac{3m}2=-\frac12$, so $m=-\frac13$. Through $(3,2)$:

$$y-2=3(x-3)\Rightarrow 3x-y-7=0,\qquad y-2=-\tfrac13(x-3)\Rightarrow x+3y-9=0.$$

Question 45

The points $(3,4)$ and $(2,-6)$ are situated on the ____ of the line $3x-4y-8=0$.

Solution. Substitute each point into $3x-4y-8$. At $(3,4)$: $9-16-8=-15<0$. At $(2,-6)$: $6+24-8=22>0$. The two values have opposite signs, so the points are on opposite sides of the line.

Question 46

A point moves so that square of its distance from the point $(3,-2)$ is numerically equal to its distance from the line $5x-12y=3$. The equation of its locus is ____.

Solution. Let the point be $(x,y)$. The squared distance from $(3,-2)$ is $(x-3)^2+(y+2)^2$, and the distance from the line is $\frac{|5x-12y-3|}{13}$. Taking the condition with the expression $5x-12y-3$ itself (the branch where it is positive):

$$13\left[(x-3)^2+(y+2)^2\right]=5x-12y-3.$$

Expanding the left side, $13\left(x^2+y^2-6x+4y+13\right)=13x^2+13y^2-78x+52y+169$. Moving $5x-12y-3$ across:

$$13x^2+13y^2-83x+64y+172=0.$$

The printed key has the constant $182$ instead of $172$. The constant is $169+3=172$ (from $13\times13$ and the $-3$ in the line).

For completeness, the modulus has a second branch: where $5x-12y-3<0$ the condition reads $13[(x-3)^2+(y+2)^2]=-(5x-12y-3)$, which gives $13x^2+13y^2-73x+40y+166=0$. The book gives only the first.

Question 47

Locus of the mid-points of the portion of the line $x\sin\theta+y\cos\theta=p$ intercepted between the axes is ____.

Solution. Putting $y=0$ gives the $x$-intercept $\frac p{\sin\theta}$, and $x=0$ gives the $y$-intercept $\frac p{\cos\theta}$. The midpoint $(x,y)$ of the intercepted portion is

$$x=\frac p{2\sin\theta},\qquad y=\frac p{2\cos\theta}.$$

So $\sin\theta=\frac p{2x}$ and $\cos\theta=\frac p{2y}$. Eliminating $\theta$ with $\sin^2\theta+\cos^2\theta=1$:

$$\frac{p^2}{4x^2}+\frac{p^2}{4y^2}=1\ \Longrightarrow\ p^2(x^2+y^2)=4x^2y^2.$$

Question 48

State whether the following is true or false. Justify. If the vertices of a triangle have integral coordinates, then the triangle cannot be equilateral.

Solution. The square of the distance between two points with integer coordinates is an integer, so if the triangle is equilateral with side $s$, then $s^2$ is an integer. The area of an equilateral triangle is $\frac{\sqrt3}4s^2$, which is irrational for any non-zero integer $s^2$. But the area of a triangle with integer vertices is $\frac12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$, which is a rational number. A rational cannot equal an irrational, so no such triangle exists.

Question 49

True or false. Justify. The points $A(-2,1)$, $B(0,5)$, $C(-1,2)$ are collinear.

Solution. The slope of $AB$ is $\frac{5-1}{0+2}=2$ and the slope of $AC$ is $\frac{2-1}{-1+2}=1$. The slopes differ, so the points are not collinear. (Equivalently, the area expression is $\frac12|-2(5-2)+0+(-1)(1-5)|=\frac12|-6+4|=1\ne0$.)

Question 50

True or false. Justify. Equation of the line passing through the point $(a\cos^3\theta,\,a\sin^3\theta)$ and perpendicular to the line $x\sec\theta+y\,\mathrm{cosec}\,\theta=a$ is $x\cos\theta-y\sin\theta=a\sin2\theta$.

Solution. The given line has slope $-\frac{\sec\theta}{\mathrm{cosec}\,\theta}=-\tan\theta$, so a perpendicular has slope $\cot\theta=\frac{\cos\theta}{\sin\theta}$. Through $(a\cos^3\theta,a\sin^3\theta)$:

$$y-a\sin^3\theta=\frac{\cos\theta}{\sin\theta}\left(x-a\cos^3\theta\right)\ \Longrightarrow\ y\sin\theta-a\sin^4\theta=x\cos\theta-a\cos^4\theta.$$

So $x\cos\theta-y\sin\theta=a\left(\cos^4\theta-\sin^4\theta\right)=a(\cos^2\theta-\sin^2\theta)=a\cos2\theta$. The right side is $a\cos2\theta$, not $a\sin2\theta$, so the statement is false. Counterexample: at $\theta=45^\circ$ the point is $\left(\frac a{2\sqrt2},\frac a{2\sqrt2}\right)$ and the true perpendicular is $x\cos45^\circ-y\sin45^\circ=a\cos90^\circ=0$, i.e. $y=x$, which passes through that point. The stated line has right side $a\sin90^\circ=a$, i.e. $x-y=a\sqrt2$, and at the point $x-y=0\neq a\sqrt2$, so it misses the point.

Question 51

True or false. Justify. The straight line $5x+4y=0$ passes through the point of intersection of the straight lines $x+2y-10=0$ and $2x+y+5=0$.

Solution. Solve $x+2y=10$ and $2x+y=-5$. Doubling the second, $4x+2y=-10$; subtracting the first gives $3x=-20$, so $x=-\frac{20}3$ and $y=\frac{10+\frac{20}3}2=\frac{25}3$. Then $5x+4y=-\frac{100}3+\frac{100}3=0$. The point satisfies the line.

Question 52

True or false. Justify. The vertex of an equilateral triangle is $(2,3)$ and the equation of the opposite side is $x+y=2$. Then the other two sides are $y-3=(2\pm\sqrt3)(x-2)$.

Solution. The side $x+y=2$ has slope $-1$, i.e. inclination $135^\circ$. The other two sides pass through $(2,3)$ and make $60^\circ$ with it, so their inclinations are $135^\circ\pm60^\circ=195^\circ$ or $75^\circ$. Their slopes are $\tan195^\circ=\tan15^\circ=2-\sqrt3$ and $\tan75^\circ=2+\sqrt3$. So the sides are $y-3=(2\pm\sqrt3)(x-2)$, exactly as stated.

Question 53

True or false. Justify. The equation of the line joining the point $(3,5)$ to the point of intersection of the lines $4x+y-1=0$ and $7x-3y-35=0$ is equidistant from the points $(0,0)$ and $(8,34)$.

Solution. Solve $4x+y=1$ and $7x-3y=35$: multiplying the first by 3, $12x+3y=3$; adding the second gives $19x=38$, so $x=2$, $y=-7$. The line through $(3,5)$ and $(2,-7)$ has slope $12$: $y-5=12(x-3)$, i.e. $12x-y-31=0$. Distances from the two points:

$$\frac{|0-0-31|}{\sqrt{145}}=\frac{31}{\sqrt{145}},\qquad \frac{|96-34-31|}{\sqrt{145}}=\frac{31}{\sqrt{145}}.$$

They are equal, so the statement is true. (The signed values $-31$ and $+31$ have opposite signs, so the two points lie on opposite sides of the line, and the line passes through their midpoint $(4,17)$: $48-17-31=0$.)

Question 54

True or false. Justify. The line $\frac xa+\frac yb=1$ moves in such a way that $\frac1{a^2}+\frac1{b^2}=\frac1{c^2}$, where $c$ is a constant. The locus of the foot of the perpendicular from the origin on the given line is $x^2+y^2=c^2$.

Solution. The perpendicular distance from the origin to $\frac xa+\frac yb=1$ is $p=\frac{1}{\sqrt{\frac1{a^2}+\frac1{b^2}}}=\sqrt{c^2}=c$. The foot of the perpendicular is therefore always at distance $c$ from the origin, which is the circle $x^2+y^2=c^2$.

Question 55

True or false. Justify. The lines $ax+2y+1=0$, $bx+3y+1=0$ and $cx+4y+1=0$ are concurrent if $a,b,c$ are in G.P.

Solution. Three lines are concurrent when the determinant of their coefficients vanishes:

$$\begin{vmatrix}a&2&1\\b&3&1\\c&4&1\end{vmatrix}=a(3-4)-2(b-c)+1(4b-3c)=-a+2b-c.$$

So they are concurrent when $a-2b+c=0$, i.e. $2b=a+c$: $a,b,c$ in A.P., not G.P. For example $a=b=c=1$ is a G.P. as well as an A.P., but $a=1,b=2,c=4$ is a G.P. with $-1+4-4=-1\ne0$, so the lines are not concurrent.

Question 56

True or false. Justify. Line joining the points $(3,-4)$ and $(-2,6)$ is perpendicular to the line joining the points $(-3,6)$ and $(9,-18)$.

Solution. The slope of the first line is $\frac{6+4}{-2-3}=-2$. The slope of the second is $\frac{-18-6}{9+3}=-2$. The slopes are equal, so the lines are parallel, not perpendicular (perpendicular lines would have slopes multiplying to $-1$, whereas here the product is $4$).

Question 57

Match the questions in Column $C_1$ with their appropriate answers in Column $C_2$.

Solution. Each part is a short calculation; the matches are $(a)\to(iii)$, $(b)\to(i)$ and $(c)\to(ii)$.

Question 57 (a)

The coordinates of the points $P$ and $Q$ on the line $x+5y=13$ which are at a distance of $2$ units from the line $12x-5y+26=0$ are

A point on $x+5y=13$ is $(13-5t,\,t)$. Its distance from $12x-5y+26=0$ is $\frac{|12(13-5t)-5t+26|}{13}=\frac{|182-65t|}{13}=|14-5t|$. Setting this equal to $2$: $14-5t=\pm2$, so $t=\frac{12}5$ or $t=\frac{16}5$. The points are $\left(13-12,\frac{12}5\right)=\left(1,\frac{12}5\right)$ and $\left(13-16,\frac{16}5\right)=\left(-3,\frac{16}5\right)$. This is item (iii).

Question 57 (b)

The coordinates of the point on the line $x+y=4$ which are at a unit distance from the line $4x+3y-10=0$ are

This is Question 5 of this exercise: the points are $(3,1)$ and $(-7,11)$. This is item (i).

Question 57 (c)

The coordinates of the point on the line joining $A(-2,5)$ and $B(3,1)$ such that $AP=PQ=QB$ are

The points $P$ and $Q$ trisect $AB$, so they divide it in the ratios $1:2$ and $2:1$. With $B-A=(5,-4)$:

$$P=A+\tfrac13(B-A)=\left(-2+\tfrac53,\ 5-\tfrac43\right)=\left(-\tfrac13,\tfrac{11}3\right),\qquad Q=A+\tfrac23(B-A)=\left(\tfrac43,\tfrac73\right).$$

This is item (ii).

Question 58

The value of the $\lambda$, if the lines $(2x+3y+4)+\lambda(6x-y+12)=0$ are (a) parallel to $y$-axis, (b) perpendicular to $7x+y-4=0$, (c) passes through $(1,2)$, (d) parallel to $x$-axis, matched with (i) $\lambda=-\frac34$, (ii) $\lambda=-\frac13$, (iii) $\lambda=-\frac{17}{41}$, (iv) $\lambda=3$.

Solution. Collect terms: the family is $(2+6\lambda)x+(3-\lambda)y+(4+12\lambda)=0$. Each condition is one equation in $\lambda$.

Question 58 (a)

Parallel to the $y$-axis means a vertical line, so the coefficient of $y$ vanishes: $3-\lambda=0$, so $\lambda=3$.

Question 58 (b)

Perpendicular to $7x+y-4=0$: the normal vectors are perpendicular, i.e. the dot product of $(2+6\lambda,\,3-\lambda)$ and $(7,1)$ is zero. $7(2+6\lambda)+(3-\lambda)=17+41\lambda=0$, so $\lambda=-\frac{17}{41}$.

Question 58 (c)

Passes through $(1,2)$: $(2+6\lambda)+2(3-\lambda)+(4+12\lambda)=12+16\lambda=0$, so $\lambda=-\frac34$.

Question 58 (d)

Parallel to the $x$-axis means a horizontal line, so the coefficient of $x$ vanishes: $2+6\lambda=0$, so $\lambda=-\frac13$.

Question 59

The equation of the line through the intersection of the lines $2x-3y=0$ and $4x-5y=2$ and (a) through the point $(2,1)$, (b) perpendicular to the line $x+2y+1=0$, (c) parallel to the line $3x-4y+5=0$, (d) equally inclined to the axes, matched with (i) $2x-y=4$, (ii) $x+y-5=0$, (iii) $x-y-1=0$, (iv) $3x-4y-1=0$.

Solution. First find the intersection point. From $2x=3y$, $x=\frac{3y}2$; then $4x-5y=6y-5y=y=2$, so $x=3$. The intersection is $(3,2)$. Each part is a line through $(3,2)$.

Question 59 (a)

Through $(2,1)$ as well: the slope is $\frac{2-1}{3-2}=1$, so $y-1=x-2$, i.e. $x-y-1=0$.

Question 59 (b)

Perpendicular to $x+2y+1=0$ (slope $-\frac12$): slope $2$, so $y-2=2(x-3)$, i.e. $2x-y-4=0$.

Question 59 (c)

Parallel to $3x-4y+5=0$: the line is $3x-4y=k$, and through $(3,2)$, $k=9-8=1$. So $3x-4y-1=0$.

Question 59 (d)

Equally inclined to the axes means slope $\pm1$. Slope $+1$ through $(3,2)$ is $x-y-1=0$, which is already item (iii) for part (a); the remaining item is slope $-1$: $y-2=-(x-3)$, i.e. $x+y-5=0$.

Common mistakes

• Question 42. Looking for a point that depends on $a,b,c$. The A.P. condition $2b=a+c$ is what removes them: it says $a\cdot1+b\cdot(-2)+c=0$, so $(1,-2)$ satisfies every line of the family.
• Question 44. Using the angle formula without the modulus and getting only one slope. Both signs of $\pm1$ give valid lines, $m=3$ and $m=-\frac13$.
• Question 46. Dropping the constant $169$ from $13(x-3)^2+13(y+2)^2$. The expansion of $13(x^2-6x+9+y^2+4y+4)$ has constant $13\times13=169$, and adding $3$ gives $172$, not the printed $182$.
• Question 50. Accepting the statement because it “looks like” a standard result. The perpendicular line through the point has right side $a\cos2\theta$; $\sin2\theta$ is a plausible-looking but wrong replacement.
• Question 55. Concluding “G.P.” because the question says so. The determinant condition $a-2b+c=0$ is the condition for an A.P.; test with $a=1,b=2,c=4$ to see the G.P. claim fail.
• Question 56. Computing both slopes correctly and still calling the lines perpendicular because they are “related”. Equal slopes mean parallel; perpendicular needs $m_1m_2=-1$.

Practise next

• The other pages of this unit: Short and Long Answer Type Q1–21 and Objective Type Q22–41.
• NCERT textbook practice: Exercise 9.1 for collinearity and perpendicular or parallel conditions, Exercise 9.3 for distances, and the Miscellaneous Exercise for families of lines through an intersection point.