NCERT Exemplar Solutions Class 11 Maths Straight Lines — Short and Long Answer Type Q1–21

NCERT Exemplar Problems, Class 11 Mathematics — Straight Lines, Short and Long Answer Type questions 1 to 21. All 21 questions solved.

These twenty-one questions run on a small toolkit: the intercept form $\frac xa+\frac yb=1$, the point-slope form, the perpendicular distance $\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$ and the angle formula $\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$. Questions 10 and 17 use the section formula on an intercept, and the triangle questions (12 and 13) turn a side and a vertex into an angle or a distance. The proofs (6, 16, 19, 21) are all two-line algebra once the right form of the line is chosen.

Two printed answers need care. In Question 7 the key lists only one of the two lines, and in Question 12 the key prints $x-7y-12=0$, which misses the given vertex; the correct line is $x-7y+12=0$. Both are explained where they occur.

Key insight. Choose the form of the line that matches what you are given. Two intercepts: $\frac xa+\frac yb=1$. A point and an angle: point-slope. A point and a distance: point-slope with $m$ unknown, then use the distance formula. Almost every question here collapses to one equation in one unknown once you do.

Question 1

Find the equation of the straight line which passes through the point $(1,-2)$ and cuts off equal intercepts from axes.

Solution. Equal intercepts $a$ on both axes means the line is $\frac xa+\frac ya=1$, that is $x+y=a$. Putting in the point $(1,-2)$ gives $a=1-2=-1$. So the line is $x+y=-1$.

(The line $y=-2x$ through the origin also has equal intercepts, both zero, but the intercept form cannot describe it, and the book’s answer leaves it out.)

Question 2

Find the equation of the line passing through the point $(5,2)$ and perpendicular to the line joining the points $(2,3)$ and $(3,-1)$.

Solution. The slope of the line joining $(2,3)$ and $(3,-1)$ is $\frac{-1-3}{3-2}=-4$. A perpendicular line has slope $-\frac{1}{-4}=\frac14$, because perpendicular slopes multiply to $-1$. Through $(5,2)$:

$$y-2=\frac14(x-5)\ \Longrightarrow\ 4y-8=x-5\ \Longrightarrow\ x-4y+3=0.$$

Question 3

Find the angle between the lines $y=(2-\sqrt3)(x+5)$ and $y=(2+\sqrt3)(x-7)$.

Solution. The slopes are read straight off: $m_1=2-\sqrt3$ and $m_2=2+\sqrt3$. Their product is $4-3=1$ and their difference is $m_2-m_1=2\sqrt3$. So

$$\tan\theta=\left|\frac{m_2-m_1}{1+m_1m_2}\right|=\frac{2\sqrt3}{1+1}=\sqrt3,$$

giving $\theta=60^\circ$ for the acute angle between the lines. The obtuse angle is $180^\circ-60^\circ=120^\circ$, which is how the key writes it. (Another way to see it: $2-\sqrt3=\tan15^\circ$ and $2+\sqrt3=\tan75^\circ$, and $75^\circ-15^\circ=60^\circ$.)

Question 4

Find the equation of the lines which pass through the point $(3,4)$ and cut off intercepts from the coordinate axes such that their sum is 14.

Solution. Let the intercepts be $a$ and $b$ with $a+b=14$, so $b=14-a$. The line $\frac xa+\frac yb=1$ passes through $(3,4)$, hence $\frac3a+\frac4b=1$, which clears to $3b+4a=ab$. Substituting $b=14-a$:

$$3(14-a)+4a=a(14-a)\ \Longrightarrow\ 42+a=14a-a^2\ \Longrightarrow\ a^2-13a+42=0\ \Longrightarrow\ (a-6)(a-7)=0.$$

For $a=6$ we get $b=8$ and the line $\frac x6+\frac y8=1$, which is $4x+3y=24$. For $a=7$ we get $b=7$ and the line $x+y=7$. Both pass through $(3,4)$: $12+12=24$ and $3+4=7$.

Question 5

Find the points on the line $x+y=4$ which lie at a unit distance from the line $4x+3y=10$.

Solution. A point on $x+y=4$ can be written $(t,\,4-t)$. Its distance from $4x+3y-10=0$ must be 1:

$$\frac{|4t+3(4-t)-10|}{\sqrt{16+9}}=\frac{|t+2|}{5}=1\ \Longrightarrow\ t+2=\pm5\ \Longrightarrow\ t=3\ \text{or}\ t=-7.$$

So the points are $(3,1)$ and $(-7,11)$.

Question 6

Show that the tangent of an angle between the lines $\dfrac xa+\dfrac yb=1$ and $\dfrac xa-\dfrac yb=1$ is $\dfrac{2ab}{a^2-b^2}$.

Solution. Write each line as $y=mx+c$. The first gives $y=-\frac bax+b$, so $m_1=-\frac ba$. The second gives $y=\frac bax-b$, so $m_2=\frac ba$. Then

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

This is the angle measured from the first line to the second; for $a>b$ it is the acute angle, and for the other angle between the lines the tangent changes sign.

Question 7

Find the equation of lines passing through $(1,2)$ and making angle $30^\circ$ with the $y$-axis.

Solution. A line making $30^\circ$ with the $y$-axis makes $90^\circ-30^\circ=60^\circ$ with the $x$-axis, so its inclination is $60^\circ$ or $120^\circ$ and its slope is $\tan60^\circ=\sqrt3$ or $\tan120^\circ=-\sqrt3$. Both are allowed, which is why the question says “lines”. Through $(1,2)$:

$$y-2=\sqrt3(x-1)\quad\text{and}\quad y-2=-\sqrt3(x-1).$$

The printed key lists only the first, $y-\sqrt3x-2+\sqrt3=0$. The second line, $y+\sqrt3x-2-\sqrt3=0$, is just as valid and belongs in a complete answer.

Question 8

Find the equation of the line passing through the point of intersection of $2x+y=5$ and $x+3y+8=0$ and parallel to the line $3x+4y=7$.

Solution. Solve $2x+y=5$ and $x+3y=-8$. From the first, $y=5-2x$; then $x+15-6x=-8$ gives $x=\frac{23}{5}$ and $y=-\frac{21}{5}$. A line parallel to $3x+4y=7$ has the form $3x+4y=k$. Putting in the intersection point,

$$k=3\cdot\frac{23}{5}+4\cdot\left(-\frac{21}{5}\right)=\frac{69-84}{5}=-3.$$

Question 9

For what values of $a$ and $b$ the intercepts cut off on the coordinate axes by the line $ax+by+8=0$ are equal in length but opposite in signs to those cut off by the line $2x-3y+6=0$ on the axes.

Solution. For $2x-3y+6=0$, setting $y=0$ gives $x=-3$, and setting $x=0$ gives $y=2$. So its intercepts are $-3$ and $2$. The line $ax+by+8=0$ must then have intercepts $+3$ on the $x$-axis and $-2$ on the $y$-axis. Its $x$-intercept is $-\frac8a$ and its $y$-intercept is $-\frac8b$, so

$$-\frac8a=3\ \Longrightarrow\ a=-\frac83,\qquad -\frac8b=-2\ \Longrightarrow\ b=4.$$

Question 10

If the intercept of a line between the coordinate axes is divided by the point $(-5,4)$ in the ratio $1:2$, then find the equation of the line.

Solution. Let the line meet the $x$-axis at $A(a,0)$ and the $y$-axis at $B(0,b)$, and let $P(-5,4)$ divide $AB$ so that $AP:PB=1:2$. By the section formula,

$$P=\left(\frac{2\cdot a+1\cdot0}{3},\ \frac{2\cdot0+1\cdot b}{3}\right)=\left(\frac{2a}{3},\frac b3\right)=(-5,4).$$

So $a=-\frac{15}{2}$ and $b=12$. The line is $\frac{x}{-15/2}+\frac y{12}=1$, and multiplying by 60 gives $-8x+5y=60$. Check: at $(-5,4)$, $40+20=60$. $\checkmark$

If the ratio is taken the other way ($BP:PA=1:2$) the line is $2x-5y+30=0$; the key follows the order above.

Question 11

Find the equation of a straight line on which length of perpendicular from the origin is four units and the line makes an angle of $120^\circ$ with the positive direction of $x$-axis.

Solution. The hint suggests the normal form $x\cos\omega+y\sin\omega=p$ with $p=4$. A line with inclination $120^\circ$ has a normal at $120^\circ-90^\circ=30^\circ$, so $\omega=30^\circ$:

$$x\cos30^\circ+y\sin30^\circ=4\ \Longrightarrow\ \frac{\sqrt3}2x+\frac12y=4\ \Longrightarrow\ \sqrt3x+y=8.$$

Check without the normal form: the slope is $-\sqrt3=\tan120^\circ$ and the distance from the origin is $\frac{8}{\sqrt{3+1}}=4$. $\checkmark$ (The parallel line on the other side of the origin, $\sqrt3x+y=-8$, also satisfies both conditions; the book takes the one with the normal in the first quadrant.)

Question 12

Find the equation of one of the sides of an isosceles right angled triangle whose hypotenuse is given by $3x+4y=4$ and the opposite vertex of the hypotenuse is $(2,2)$.

Solution. In an isosceles right triangle each leg makes $45^\circ$ with the hypotenuse, and the two legs meet at the vertex $(2,2)$. The hypotenuse has slope $-\frac34$. For a leg of slope $m$,

$$\tan45^\circ=1=\left|\frac{m+\frac34}{1-\frac34m}\right|\ \Longrightarrow\ m+\tfrac34=\pm\left(1-\tfrac34m\right).$$

The $+$ sign gives $\frac74m=\frac14$, so $m=\frac17$. The $-$ sign gives $\frac14m=-\frac74$, so $m=-7$. Through $(2,2)$:

$$y-2=\tfrac17(x-2)\ \Longrightarrow\ x-7y+12=0,\qquad y-2=-7(x-2)\ \Longrightarrow\ 7x+y-16=0.$$

The two legs are perpendicular (slopes $\frac17$ and $-7$ multiply to $-1$), as they should be.

The printed key gives $x-7y-12=0$, which is wrong: substituting $(2,2)$ gives $2-14-12=-24\neq0$, so it does not pass through the vertex. The sign of the constant should be $+$.

Question 13

If the equation of the base of an equilateral triangle is $x+y=2$ and the vertex is $(2,-1)$, then find the length of the side of the triangle.

Solution. The perpendicular from the vertex to the base is the altitude $p$:

$$p=\frac{|2+(-1)-2|}{\sqrt{1+1}}=\frac1{\sqrt2}.$$

In an equilateral triangle of side $l$, the altitude is $l\sin60^\circ=\frac{\sqrt3}2l$, as the hint says. So

$$l=\frac{2p}{\sqrt3}=\frac{2}{\sqrt2\,\sqrt3}=\frac{\sqrt2}{\sqrt3}=\sqrt{\tfrac23}=\frac{\sqrt6}{3}.$$

Question 14

A variable line passes through a fixed point $P$. The algebraic sum of the perpendiculars drawn from the points $(2,0)$, $(0,2)$ and $(1,1)$ on the line is zero. Find the coordinates of the point $P$.

Solution. Write the line as $ax+by+c=0$ with the signed distance of a point $(x_0,y_0)$ being $\frac{ax_0+by_0+c}{\sqrt{a^2+b^2}}$. Adding the three signed distances,

$$\frac{(2a+c)+(2b+c)+(a+b+c)}{\sqrt{a^2+b^2}}=\frac{3(a+b+c)}{\sqrt{a^2+b^2}}=0\ \Longrightarrow\ a+b+c=0.$$

But $a+b+c=0$ says exactly that $(1,1)$ lies on the line. So every such line passes through $(1,1)$ (the centroid of the three points). Since the lines also all pass through the fixed point $P$ and are otherwise free, $P=(1,1)$.

Question 15

In what direction should a line be drawn through the point $(1,2)$ so that its point of intersection with the line $x+y=4$ is at a distance $\frac{\sqrt6}3$ from the given point.

Solution. Use the distance form. Let the line make angle $\theta$ with the positive $x$-axis. A point on it at distance $r=\frac{\sqrt6}3$ from $(1,2)$ is $(1+r\cos\theta,\ 2+r\sin\theta)$. It lies on $x+y=4$:

$$3+r(\cos\theta+\sin\theta)=4\ \Longrightarrow\ \cos\theta+\sin\theta=\frac1r=\frac{3}{\sqrt6}=\frac{\sqrt6}2.$$

Since $\cos\theta+\sin\theta=\sqrt2\sin(\theta+45^\circ)$, we get $\sin(\theta+45^\circ)=\frac{\sqrt6}{2\sqrt2}=\frac{\sqrt3}2$. So $\theta+45^\circ=60^\circ$ or $120^\circ$, that is $\theta=15^\circ$ or $75^\circ$. (A sympy check of $\sin\theta+\cos\theta=\frac{\sqrt6}2$ on $[0^\circ,180^\circ)$ returns exactly these two.)

Question 16

A straight line moves so that the sum of the reciprocals of its intercepts made on axes is constant. Show that the line passes through a fixed point.

Solution. Let the line be $\frac xa+\frac yb=1$ with $\frac1a+\frac1b=\frac1k$, where $k$ is the constant. Multiply this relation by $k$:

$$\frac ka+\frac kb=1.$$

That is exactly the statement that the point $(x,y)=(k,k)$ satisfies $\frac xa+\frac yb=1$. Since $k$ does not depend on the position of the line, every line of the family passes through $(k,k)$.

Question 17

Find the equation of the line which passes through the point $(-4,3)$ and the portion of the line intercepted between the axes is divided internally in the ratio $5:3$ by this point.

Solution. Let the line meet the $x$-axis at $A(a,0)$ and the $y$-axis at $B(0,b)$, with $AP:PB=5:3$ for $P(-4,3)$. The section formula gives

$$P=\left(\frac{3a+5\cdot0}{8},\ \frac{3\cdot0+5b}{8}\right)=\left(\frac{3a}8,\frac{5b}8\right)=(-4,3),$$

so $a=-\frac{32}3$ and $b=\frac{24}5$. The line is $\frac{x}{-32/3}+\frac{y}{24/5}=1$, and multiplying by 96 gives $-9x+20y=96$. Check: at $(-4,3)$, $36+60=96$. $\checkmark$

If the ratio were read the other way round ($BP:PA=5:3$), the line would be $5x-4y+32=0$; the key follows the order above.

Question 18

Find the equations of the lines through the point of intersection of the lines $x-y+1=0$ and $2x-3y+5=0$ and whose distance from the point $(3,2)$ is $\frac75$.

Solution. Solving $x-y+1=0$ and $2x-3y+5=0$: put $x=y-1$ into the second, $2y-2-3y+5=0$, so $y=3$ and $x=2$. The lines pass through $(2,3)$. A vertical line $x=2$ is at distance 1 from $(3,2)$, which is not $\frac75$, so each line has a slope $m$: $y-3=m(x-2)$, i.e. $mx-y+(3-2m)=0$. Its distance from $(3,2)$:

$$\frac{|3m-2+3-2m|}{\sqrt{m^2+1}}=\frac{|m+1|}{\sqrt{m^2+1}}=\frac75\ \Longrightarrow\ 25(m+1)^2=49(m^2+1).$$

This simplifies to $24m^2-50m+24=0$, i.e. $12m^2-25m+12=0$, so $m=\frac34$ or $m=\frac43$. The lines are $y-3=\frac34(x-2)$, giving $3x-4y+6=0$, and $y-3=\frac43(x-2)$, giving $4x-3y+1=0$. Check the distances: $\frac{|9-8+6|}5=\frac75$ and $\frac{|12-6+1|}5=\frac75$. $\checkmark$

Question 19

If the sum of the distances of a moving point in a plane from the axes is 1, then find the locus of the point.

Solution. A point $(x,y)$ is at distance $|y|$ from the $x$-axis and $|x|$ from the $y$-axis. The condition is $|x|+|y|=1$. Splitting by quadrant removes the modulus signs:

• first quadrant ($x,y\ge0$): $x+y=1$;
• second quadrant ($x<0,\ y\ge0$): $-x+y=1$;
• third quadrant ($x,y<0$): $-x-y=1$, i.e. $x+y=-1$;
• fourth quadrant ($x\ge0,\ y<0$): $x-y=1$.

These four segments join up into a square with vertices $(\pm1,0)$ and $(0,\pm1)$.

Question 20

$P_1$, $P_2$ are points on either of the two lines $y-\sqrt3|x|=2$ at a distance of 5 units from their point of intersection. Find the coordinates of the foot of perpendiculars drawn from $P_1$, $P_2$ on the bisector of the angle between the given lines.

Solution. For $x\ge0$ the line is $y=\sqrt3x+2$, and for $x<0$ it is $y=-\sqrt3x+2$. Both pass through $(0,2)$, the point of intersection, and they are mirror images in the $y$-axis, so the $y$-axis is the bisector of the angle between them (as the hint says).

The line $y=\sqrt3x+2$ has inclination $60^\circ$, so the point $P_1$ at distance 5 from $(0,2)$ along it is

$$P_1=\left(5\cos60^\circ,\ 2+5\sin60^\circ\right)=\left(\tfrac52,\ 2+\tfrac{5\sqrt3}2\right).$$

By symmetry $P_2=\left(-\frac52,\ 2+\frac{5\sqrt3}2\right)$. The foot of the perpendicular from either point to the $y$-axis keeps the $y$-coordinate and has $x=0$. The $y$-coordinate is $2+5\cos30^\circ=2+\frac{5\sqrt3}2$, as the hint says.

Question 21

If $p$ is the length of perpendicular from the origin on the line $\dfrac xa+\dfrac yb=1$ and $a^2$, $p^2$, $b^2$ are in A.P., then show that $a^4+b^4=0$.

Solution. The line is $bx+ay-ab=0$, so its distance from the origin is $p=\frac{ab}{\sqrt{a^2+b^2}}$, and $p^2=\frac{a^2b^2}{a^2+b^2}$. Three numbers are in A.P. when the middle one is the average of the other two: $2p^2=a^2+b^2$. Substituting,

$$\frac{2a^2b^2}{a^2+b^2}=a^2+b^2\ \Longrightarrow\ 2a^2b^2=(a^2+b^2)^2=a^4+2a^2b^2+b^4\ \Longrightarrow\ a^4+b^4=0.$$

Common mistakes

• Question 1. Writing the equal-intercept line as $x+y=1$. The intercept is not fixed at 1; it is whatever value $a$ makes $x+y=a$ pass through the given point, here $a=-1$.
• Question 3. Using $\tan\theta=\frac{m_1-m_2}{1+m_1m_2}$ without the modulus and then not knowing whether $\sqrt3$ means $60^\circ$ or $120^\circ$. The formula with a modulus gives the acute angle; the obtuse angle is its supplement.
• Question 7. Stopping at one line. “At $30^\circ$ with the $y$-axis” gives two inclinations, $60^\circ$ and $120^\circ$, and students often take $30^\circ$ itself as the inclination with the $x$-axis.
• Question 10 and 17. Applying the section formula with the ratio the wrong way round. If $AP:PB=m:n$ the point is $\frac{n\cdot A+m\cdot B}{m+n}$, so the second number multiplies the first point; swapping them gives a different line.
• Question 12. Sign slips in the final rearrangement. After $y-2=\frac17(x-2)$ the constant becomes $+12$; always substitute the vertex back into the final equation, which would have caught the printed key’s $-12$.
• Question 18. Assuming every line is $y-3=m(x-2)$ and never checking the vertical line $x=2$. It fails here (its distance from $(3,2)$ is 1, not $\frac75$), but the check has to be made; the quadratic $12m^2-25m+12=0$ then gives two slopes, and both lines are required.

Practise next

• The other pages of this unit: Objective Type Q22–41 and Fill in the Blanks, True or False and Matching Q42–59.
• NCERT textbook practice: Exercise 9.1 for slopes and angles between lines, Exercise 9.2 for forms of the equation, and Exercise 9.3 for distances and intersections.
• Harder mixed practice: the Miscellaneous Exercise.