Class 11 Maths NCERT Solutions Chapter 9 Miscellaneous Exercise

Comparing with the general form Ax+By+C=0: A=k-3, B=-(4-k^2), C=k^2-7k+6.

(a) Parallel to the x-axis

A line is parallel to the x-axis when the coefficient of x is zero (and the coefficient of y is not):

k-3=0\;\Rightarrow\;k=3

Checking the coefficient of y at k=3: -(4-9)=5\neq0, so this is valid.

(b) Parallel to the y-axis

A line is parallel to the y-axis when the coefficient of y is zero (and the coefficient of x is not):

4-k^2=0\;\Rightarrow\;k=\pm2

Checking the coefficient of x at both values: k=2\Rightarrow-1\neq0; k=-2\Rightarrow-5\neq0, so both are valid.

(c) Passing through the origin

A line passes through the origin when its constant term is zero:

k^2-7k+6=0\;\Rightarrow\;(k-1)(k-6)=0\;\Rightarrow\;k=1\text{ or }k=6

Let the x-intercept and y-intercept be a and b, with a+b=1 and ab=-6. So a and b are roots of:

t^2-t-6=0

(t-3)(t+2)=0\;\Rightarrow\;t=3\text{ or }t=-2

So (a,b)=(3,-2) or (a,b)=(-2,3). Using the intercept form \dfrac{x}{a}+\dfrac{y}{b}=1:

Case 1: a = 3, b = −2

\dfrac{x}{3}+\dfrac{y}{-2}=1\;\Rightarrow\;2x-3y=6

Case 2: a = −2, b = 3

\dfrac{x}{-2}+\dfrac{y}{3}=1\;\Rightarrow\;-3x+2y=6\;\Rightarrow\;3x-2y+6=0

Writing the line in general form (multiplying by 12): 4x+3y-12=0. Let the required point on the y-axis be (0,y):

\dfrac{|4(0)+3y-12|}{\sqrt{4^2+3^2}}=4

|3y-12|=20

This gives two cases:

3y-12=20\;\Rightarrow\;y=\dfrac{32}{3}\qquad\text{or}\qquad3y-12=-20\;\Rightarrow\;y=-\dfrac{8}{3}

Using the two-point form, the equation of the line joining P(\cos\theta,\sin\theta) and Q(\cos\phi,\sin\phi), in general form, is:

(\sin\phi-\sin\theta)x-(\cos\phi-\cos\theta)y+(\cos\phi\sin\theta-\cos\theta\sin\phi)=0

Note that \cos\phi\sin\theta-\cos\theta\sin\phi=\sin(\theta-\phi). The distance from the origin:

d=\dfrac{|\sin(\theta-\phi)|}{\sqrt{(\sin\phi-\sin\theta)^2+(\cos\phi-\cos\theta)^2}}

The denominator is the distance PQ. Using the identity for the distance between two points on the unit circle:

(\sin\phi-\sin\theta)^2+(\cos\phi-\cos\theta)^2=2-2\cos(\phi-\theta)=4\sin^2\left(\dfrac{\phi-\theta}{2}\right)

So the denominator is 2\left|\sin\left(\dfrac{\phi-\theta}{2}\right)\right|. For the numerator, using \sin(\theta-\phi)=2\sin\left(\dfrac{\theta-\phi}{2}\right)\cos\left(\dfrac{\theta-\phi}{2}\right):

d=\dfrac{2\left|\sin\left(\dfrac{\theta-\phi}{2}\right)\right|\left|\cos\left(\dfrac{\theta-\phi}{2}\right)\right|}{2\left|\sin\left(\dfrac{\phi-\theta}{2}\right)\right|}=\left|\cos\left(\dfrac{\theta-\phi}{2}\right)\right|

From 3x+y=0: y=-3x. Substituting in x-7y+5=0:

x-7(-3x)+5=0\;\Rightarrow\;x+21x+5=0\;\Rightarrow\;22x=-5\;\Rightarrow\;x=-\dfrac{5}{22}

A line parallel to the y-axis has the form x=c. Since it passes through the intersection point, c=-\dfrac{5}{22}:

The given line meets the y-axis at (0,6) (its y-intercept). Writing the line in slope form: 3x+2y=12\;\Rightarrow\;y=\dfrac{12-3x}{2}, so its slope is:

m_1=-\dfrac{3}{2}

The required line is perpendicular to this, so its slope is:

m=-\dfrac{1}{m_1}=\dfrac{2}{3}

Using the point-slope form through (0,6):

y-6=\dfrac{2}{3}(x-0)

3(y-6)=2x\;\Rightarrow\;2x-3y+18=0

The lines are y=x, y=-x and x=k. Lines y=x and y=-x intersect at the origin O(0,0). The line x=k meets y=x at A(k,k) and meets y=-x at B(k,-k).

Using the area formula for O(0,0), A(k,k), B(k,-k):

\text{Area}=\dfrac{1}{2}\left|0(k-(-k))+k((-k)-0)+k(0-k)\right|

=\dfrac{1}{2}\left|0-k^2-k^2\right|=\dfrac{1}{2}(2k^2)=k^2

Find the intersection of the first and third lines. Solving 3x+y=2 and 2x-y=3 by adding:

5x=5\;\Rightarrow\;x=1,\quad y=2-3(1)=-1

Point of intersection: (1,-1). For all three lines to be concurrent, this point must satisfy the second line:

p(1)+2(-1)-3=0\;\Rightarrow\;p-2-3=0\;\Rightarrow\;p=5

Let the three lines meet at a common point (x_0,y_0). Then each equation is satisfied by this point:

y_0=m_1x_0+c_1\quad\ldots(1)

y_0=m_2x_0+c_2\quad\ldots(2)

y_0=m_3x_0+c_3\quad\ldots(3)

Subtracting (2) from (1), (3) from (2), and (1) from (3):

0=(m_1-m_2)x_0+(c_1-c_2)\quad\ldots(4)

0=(m_2-m_3)x_0+(c_2-c_3)\quad\ldots(5)

0=(m_3-m_1)x_0+(c_3-c_1)\quad\ldots(6)

Multiplying (4) by c_3... instead, a cleaner route: from (4), (5), (6), express x_0 and eliminate. Multiply (1) by (m_2-m_3), (2) by (m_3-m_1), (3) by (m_1-m_2) and add — the coefficients of x_0 and y_0 vanish (since (m_2-m_3)+(m_3-m_1)+(m_1-m_2)=0 for the y0-terms, and m_1(m_2-m_3)+m_2(m_3-m_1)+m_3(m_1-m_2)=0 for the x0-terms, by direct expansion), leaving:

c_1(m_2-m_3)+c_2(m_3-m_1)+c_3(m_1-m_2)=0

which is the same identity as m_1(c_2-c_3)+m_2(c_3-c_1)+m_3(c_1-c_2)=0 (both are the standard determinant-style concurrency identity, differing only in how the terms are grouped).

Slope of x-2y=3, i.e., y=\dfrac{x-3}{2}: m_1=\dfrac{1}{2}. Let the slope of the required line be m. Using the angle formula with \theta=45^\circ:

\tan45^\circ=\left|\dfrac{m-\dfrac{1}{2}}{1+\dfrac{m}{2}}\right|\;\Rightarrow\;1=\left|\dfrac{m-\dfrac{1}{2}}{1+\dfrac{m}{2}}\right|

Case 1

m-\dfrac{1}{2}=1+\dfrac{m}{2}\;\Rightarrow\;\dfrac{m}{2}=\dfrac{3}{2}\;\Rightarrow\;m=3

Line through (3,2) with slope 3: y-2=3(x-3)\;\Rightarrow\;3x-y-7=0

Case 2

m-\dfrac{1}{2}=-1-\dfrac{m}{2}\;\Rightarrow\;\dfrac{3m}{2}=-\dfrac{1}{2}\;\Rightarrow\;m=-\dfrac{1}{3}

Line through (3,2) with slope -\dfrac{1}{3}: y-2=-\dfrac{1}{3}(x-3)\;\Rightarrow\;x+3y-9=0

Solving 4x+7y-3=0 and 2x-3y+1=0 simultaneously — from the second equation, x=\dfrac{3y-1}{2}. Substituting in the first:

4\left(\dfrac{3y-1}{2}\right)+7y-3=0\;\Rightarrow\;2(3y-1)+7y-3=0\;\Rightarrow\;13y-5=0\;\Rightarrow\;y=\dfrac{5}{13}

x=\dfrac{3\left(\dfrac{5}{13}\right)-1}{2}=\dfrac{\dfrac{2}{13}}{2}=\dfrac{1}{13}

Point of intersection: \left(\dfrac{1}{13},\dfrac{5}{13}\right). A line with equal (non-zero) intercepts has the form x+y=a. Substituting the point:

\dfrac{1}{13}+\dfrac{5}{13}=a\;\Rightarrow\;a=\dfrac{6}{13}

Let the inclination of the line y=mx+c be \beta, so \tan\beta=m. Let the required line through the origin have inclination \alpha, so its slope is \tan\alpha. Since the required line makes angle \theta with the given line, its inclination is:

\alpha=\beta\pm\theta

Taking the tangent of both sides and using the tangent addition/subtraction formula:

\tan\alpha=\tan(\beta\pm\theta)=\dfrac{\tan\beta\pm\tan\theta}{1\mp\tan\beta\tan\theta}=\dfrac{m\pm\tan\theta}{1\mp m\tan\theta}

Since the required line passes through the origin with slope \tan\alpha, its equation is y=(\tan\alpha)x, i.e.:

\dfrac{y}{x}=\tan\alpha=\dfrac{m\pm\tan\theta}{1\mp m\tan\theta}

Let the line x+y=4 divide the segment joining (-1,1) and (5,7) in the ratio k:1 at a point R. By the section formula:

R=\left(\dfrac{5k-1}{k+1},\dfrac{7k+1}{k+1}\right)

Since R lies on x+y=4:

\dfrac{5k-1}{k+1}+\dfrac{7k+1}{k+1}=4

\dfrac{12k}{k+1}=4\;\Rightarrow\;12k=4k+4\;\Rightarrow\;8k=4\;\Rightarrow\;k=\dfrac{1}{2}

To find the distance measured along the line 2x-y=0, first find where this line meets 4x+7y+5=0. From 2x-y=0: y=2x. Substituting:

4x+7(2x)+5=0\;\Rightarrow\;18x=-5\;\Rightarrow\;x=-\dfrac{5}{18},\quad y=-\dfrac{5}{9}

The required distance is the distance between (1,2) and this intersection point \left(-\dfrac{5}{18},-\dfrac{5}{9}\right):

d=\sqrt{\left(1+\dfrac{5}{18}\right)^2+\left(2+\dfrac{5}{9}\right)^2}=\sqrt{\left(\dfrac{23}{18}\right)^2+\left(\dfrac{23}{9}\right)^2}

=23\sqrt{\dfrac{1}{18^2}+\dfrac{1}{9^2}}=23\sqrt{\dfrac{1}{324}+\dfrac{4}{324}}=23\sqrt{\dfrac{5}{324}}=\dfrac{23\sqrt{5}}{18}

Let the required line make angle \theta with the positive x-axis. Any point at distance r from (-1,2) along this direction is (-1+r\cos\theta,\,2+r\sin\theta). At r=3, this point must lie on x+y=4:

(-1+3\cos\theta)+(2+3\sin\theta)=4

1+3(\cos\theta+\sin\theta)=4\;\Rightarrow\;\cos\theta+\sin\theta=1

Using \cos\theta+\sin\theta=\sqrt{2}\sin(\theta+45^\circ):

\sqrt{2}\sin(\theta+45^\circ)=1\;\Rightarrow\;\sin(\theta+45^\circ)=\dfrac{1}{\sqrt{2}}

\theta+45^\circ=45^\circ\text{ or }135^\circ\;\Rightarrow\;\theta=0^\circ\text{ or }90^\circ

Since the legs of the right triangle are parallel to the axes, the third vertex (the right-angle vertex) must share its x-coordinate with one end of the hypotenuse and its y-coordinate with the other. There are two such points: (1,1) and (-4,3).

Configuration 1: right angle at (1, 1)

The vertical leg joins (1,1) and (1,3): equation x=1. The horizontal leg joins (1,1) and (-4,1): equation y=1.

Configuration 2: right angle at (−4, 3)

The vertical leg joins (-4,3) and (-4,1): equation x=-4. The horizontal leg joins (-4,3) and (1,3): equation y=3.

Let the image of P(3,8) be Q(h,k). Since the line acts as a mirror, it is the perpendicular bisector of PQ.

Slope of x+3y=7, i.e., y=\dfrac{7-x}{3}, is -\dfrac{1}{3}. Since PQ is perpendicular to this line, its slope is 3:

\dfrac{k-8}{h-3}=3\;\Rightarrow\;k=3h-1\quad\ldots(1)

The mid-point of PQ, \left(\dfrac{3+h}{2},\dfrac{8+k}{2}\right), lies on the line x+3y=7:

\dfrac{3+h}{2}+3\left(\dfrac{8+k}{2}\right)=7\;\Rightarrow\;(3+h)+3(8+k)=14

h+3k=-13\quad\ldots(2)

Substituting (1) into (2):

h+3(3h-1)=-13\;\Rightarrow\;10h-3=-13\;\Rightarrow\;h=-1

From (1): k=3(-1)-1=-4.

Slope of y=3x+1: m_1=3. Slope of 2y=x+3, i.e., y=\dfrac{x}{2}+\dfrac{3}{2}: m_2=\dfrac{1}{2}. Since both lines are equally inclined to y=mx+4, the angle each makes with it is equal:

\dfrac{m_1-m}{1+m_1m}=\dfrac{m-m_2}{1+mm_2}

\dfrac{3-m}{1+3m}=\dfrac{m-\dfrac{1}{2}}{1+\dfrac{m}{2}}

Cross-multiplying and simplifying:

(3-m)\left(1+\dfrac{m}{2}\right)=(1+3m)\left(m-\dfrac{1}{2}\right)

Multiplying through by 2 to clear fractions:

(3-m)(2+m)=(1+3m)(2m-1)

6+3m-2m-m^2=2m-1+6m^2-3m

6+m-m^2=6m^2-m-1

7m^2-2m-7=0

Using the quadratic formula:

m=\dfrac{2\pm\sqrt{4+196}}{14}=\dfrac{2\pm\sqrt{200}}{14}=\dfrac{2\pm10\sqrt{2}}{14}=\dfrac{1\pm5\sqrt{2}}{7}

The perpendicular distances of P(x,y) from the two lines are:

d_1=\dfrac{|x+y-5|}{\sqrt{2}},\qquad d_2=\dfrac{|3x-2y+7|}{\sqrt{13}}

Given d_1+d_2=10. Since P moves continuously and the sum stays fixed at 10, the expressions inside the modulus signs keep a consistent sign along the path (they cannot flip sign without the sum changing discontinuously). Taking both expressions as positive (one of the four possible sign choices):

\dfrac{x+y-5}{\sqrt{2}}+\dfrac{3x-2y+7}{\sqrt{13}}=10

Multiplying throughout by \sqrt{26} (i.e., \sqrt{2}\times\sqrt{13}):

\sqrt{13}(x+y-5)+\sqrt{2}(3x-2y+7)=10\sqrt{26}

(\sqrt{13}+3\sqrt{2})x+(\sqrt{13}-2\sqrt{2})y+(7\sqrt{2}-5\sqrt{13}-10\sqrt{26})=0

This is an equation of the form Ax+By+C=0, which is linear in x and y — the equation of a straight line. (Choosing the other sign combinations for the two distances gives three other such linear equations, one for each region of the plane.)

Writing 9x+6y-7=0 with the same leading coefficients as 3x+2y+6=0 (dividing by 3):

3x+2y-\dfrac{7}{3}=0

Both lines are now of the form 3x+2y+C=0, with C_1=-\dfrac{7}{3} and C_2=6. The line equidistant from both has C equal to the average of C_1 and C_2:

C=\dfrac{-\dfrac{7}{3}+6}{2}=\dfrac{\dfrac{11}{3}}{2}=\dfrac{11}{6}

So the required line is 3x+2y+\dfrac{11}{6}=0. Multiplying throughout by 6:

18x+12y+11=0

By the law of reflection, the reflected ray appears to originate from the mirror image of (1,2) in the x-axis. This image is (1,-2). So the point A where the ray meets the x-axis lies on the straight line joining (1,-2) and (5,3).

Slope of the line joining (1,-2) and (5,3):

m=\dfrac{3-(-2)}{5-1}=\dfrac{5}{4}

Since A lies on the x-axis, its y-coordinate is 0. Using the point-slope form through (1,-2):

0-(-2)=\dfrac{5}{4}(x-1)

2=\dfrac{5}{4}(x-1)\;\Rightarrow\;x-1=\dfrac{8}{5}\;\Rightarrow\;x=\dfrac{13}{5}

Let c=\sqrt{a^2-b^2}. The line, in general form, is \dfrac{\cos\theta}{a}x+\dfrac{\sin\theta}{b}y-1=0. The perpendicular distances from (c,0) and (-c,0) are:

p_1=\dfrac{\left|\dfrac{c\cos\theta}{a}-1\right|}{\sqrt{\dfrac{\cos^2\theta}{a^2}+\dfrac{\sin^2\theta}{b^2}}},\qquad p_2=\dfrac{\left|\dfrac{-c\cos\theta}{a}-1\right|}{\sqrt{\dfrac{\cos^2\theta}{a^2}+\dfrac{\sin^2\theta}{b^2}}}=\dfrac{\left|\dfrac{c\cos\theta}{a}+1\right|}{\sqrt{\dfrac{\cos^2\theta}{a^2}+\dfrac{\sin^2\theta}{b^2}}}

The product is:

p_1p_2=\dfrac{\left|\dfrac{c^2\cos^2\theta}{a^2}-1\right|}{\dfrac{\cos^2\theta}{a^2}+\dfrac{\sin^2\theta}{b^2}}

Since c^2=a^2-b^2:

\dfrac{c^2\cos^2\theta}{a^2}-1=\dfrac{(a^2-b^2)\cos^2\theta}{a^2}-1=\cos^2\theta-\dfrac{b^2\cos^2\theta}{a^2}-1=-\sin^2\theta-\dfrac{b^2\cos^2\theta}{a^2}

Taking the absolute value (the expression is negative):

\left|\dfrac{c^2\cos^2\theta}{a^2}-1\right|=\sin^2\theta+\dfrac{b^2\cos^2\theta}{a^2}

So:

p_1p_2=\dfrac{\sin^2\theta+\dfrac{b^2\cos^2\theta}{a^2}}{\dfrac{\cos^2\theta}{a^2}+\dfrac{\sin^2\theta}{b^2}}

Multiplying numerator and denominator by a^2b^2:

p_1p_2=\dfrac{a^2b^2\sin^2\theta+b^4\cos^2\theta}{b^2\cos^2\theta+a^2\sin^2\theta}=\dfrac{b^2(a^2\sin^2\theta+b^2\cos^2\theta)}{a^2\sin^2\theta+b^2\cos^2\theta}=b^2

The shortest (least-time) path from a point to a line is along the perpendicular to that line. First, find the junction point by solving 2x-3y+4=0 and 3x+4y-5=0 simultaneously. Multiplying the first by 4 and the second by 3:

8x-12y=-16\qquad9x+12y=15

Adding:

17x=-1\;\Rightarrow\;x=-\dfrac{1}{17}

Substituting back: 2\left(-\dfrac{1}{17}\right)-3y+4=0\;\Rightarrow\;y=\dfrac{22}{17}. Junction point: \left(-\dfrac{1}{17},\dfrac{22}{17}\right).

Slope of the path 6x-7y+8=0, i.e., y=\dfrac{6x+8}{7}: m_1=\dfrac{6}{7}. The perpendicular path has slope:

m=-\dfrac{1}{m_1}=-\dfrac{7}{6}

Using the point-slope form through the junction point:

y-\dfrac{22}{17}=-\dfrac{7}{6}\left(x+\dfrac{1}{17}\right)

Multiplying throughout by 102 (i.e., 6\times17) to clear denominators:

119x+102y-125=0