Class 11 Maths NCERT Solutions Chapter 9 Ex 9.3

The slope-intercept form of a line is y=mx+c, where m is the slope and c is the y-intercept.

(i) x + 7y = 0

7y=-x\;\Rightarrow\;y=-\dfrac{1}{7}x+0

(ii) 6x + 3y − 5 = 0

3y=5-6x\;\Rightarrow\;y=-2x+\dfrac{5}{3}

(iii) y = 0

This is already of the form y=0\cdot x+0.

The intercept form of a line is \dfrac{x}{a}+\dfrac{y}{b}=1, where a is the x-intercept and b is the y-intercept.

(i) 3x + 2y − 12 = 0

3x+2y=12

Dividing throughout by 12:

\dfrac{x}{4}+\dfrac{y}{6}=1

(ii) 4x − 3y = 6

Dividing throughout by 6:

\dfrac{x}{3/2}+\dfrac{y}{-2}=1

(iii) 3y + 2 = 0

y=-\dfrac{2}{3}

This line is parallel to the x-axis, so it never meets the x-axis and has no x-intercept in the usual sense.

First, write the line in general form Ax+By+C=0:

12x+72=5y-10\;\Rightarrow\;12x-5y+82=0

Here A=12, B=-5, C=82. Using the distance formula for the point (-1,1):

d=\dfrac{|A x_1+B y_1+C|}{\sqrt{A^2+B^2}}=\dfrac{|12(-1)-5(1)+82|}{\sqrt{12^2+(-5)^2}}

=\dfrac{|-12-5+82|}{\sqrt{144+25}}=\dfrac{65}{\sqrt{169}}=\dfrac{65}{13}=5

Writing the line in general form (multiplying by 12):

4x+3y-12=0

Let the required point on the x-axis be (x,0). Using the distance formula:

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

\dfrac{|4x-12|}{5}=4\;\Rightarrow\;|4x-12|=20

This gives two cases:

4x-12=20\;\Rightarrow\;x=8\qquad\text{or}\qquad4x-12=-20\;\Rightarrow\;x=-2

The distance between two parallel lines Ax+By+C_1=0 and Ax+By+C_2=0 is d=\dfrac{|C_1-C_2|}{\sqrt{A^2+B^2}}.

(i) 15x + 8y − 34 = 0 and 15x + 8y + 31 = 0

Here A=15, B=8, C_1=-34, C_2=31:

d=\dfrac{|-34-31|}{\sqrt{15^2+8^2}}=\dfrac{65}{\sqrt{225+64}}=\dfrac{65}{\sqrt{289}}=\dfrac{65}{17}

(ii) l(x + y) + p = 0 and l(x + y) − r = 0

Rewriting: lx+ly+p=0 and lx+ly-r=0. Here A=l, B=l, C_1=p, C_2=-r:

d=\dfrac{|p-(-r)|}{\sqrt{l^2+l^2}}=\dfrac{|p+r|}{\sqrt{2}\,|l|}

Any line parallel to 3x-4y+2=0 has the same A, B coefficients and differs only in the constant, so it can be written as:

3x-4y+k=0

Since this line passes through (-2,3):

3(-2)-4(3)+k=0\;\Rightarrow\;-6-12+k=0\;\Rightarrow\;k=18

Slope of the given line x-7y+5=0, i.e., y=\dfrac{x+5}{7}:

m_1=\dfrac{1}{7}

Since the required line is perpendicular to this line, its slope is the negative reciprocal:

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

The required line has x-intercept 3, so it passes through (3,0). Using the point-slope form:

y-0=-7(x-3)

y=-7x+21

7x+y-21=0

Slope of \sqrt{3}x+y=1, i.e., y=-\sqrt{3}x+1:

m_1=-\sqrt{3}

Slope of x+\sqrt{3}y=1, i.e., y=\dfrac{1-x}{\sqrt{3}}:

m_2=-\dfrac{1}{\sqrt{3}}

Using the formula for the acute angle between two lines:

\tan\theta=\left|\dfrac{m_2-m_1}{1+m_1m_2}\right|=\left|\dfrac{-\dfrac{1}{\sqrt{3}}+\sqrt{3}}{1+(-\sqrt{3})\left(-\dfrac{1}{\sqrt{3}}\right)}\right|

=\left|\dfrac{\dfrac{-1+3}{\sqrt{3}}}{1+1}\right|=\left|\dfrac{2/\sqrt{3}}{2}\right|=\dfrac{1}{\sqrt{3}}

\theta=\tan^{-1}\left(\dfrac{1}{\sqrt{3}}\right)=30^\circ

Slope of the given line 7x-9y-19=0, i.e., y=\dfrac{7x-19}{9}:

m_1=\dfrac{7}{9}

Slope of the line through (h,3) and (4,1):

m_2=\dfrac{1-3}{4-h}=\dfrac{-2}{4-h}

Since the two lines intersect at right angles, m_1m_2=-1:

\dfrac{7}{9}\times\dfrac{-2}{4-h}=-1

\dfrac{-14}{9(4-h)}=-1\;\Rightarrow\;14=9(4-h)

14=36-9h\;\Rightarrow\;9h=22\;\Rightarrow\;h=\dfrac{22}{9}

Any line parallel to Ax+By+C=0 has the same coefficients of x and y, differing only in the constant term, so it can be written as:

Ax+By+k=0

for some constant k. Since this line passes through (x_1,y_1):

Ax_1+By_1+k=0\;\Rightarrow\;k=-(Ax_1+By_1)

Substituting this value of k back into the line:

Ax+By-(Ax_1+By_1)=0

A(x-x_1)+B(y-y_1)=0

Let the slope of the other line be m. Using the formula for the angle between two lines, with m_1=2 and \theta=60^\circ:

\tan60^\circ=\left|\dfrac{m-2}{1+2m}\right|\;\Rightarrow\;\sqrt{3}=\left|\dfrac{m-2}{1+2m}\right|

Case 1

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

m(1-2\sqrt{3})=\sqrt{3}+2\;\Rightarrow\;m=\dfrac{\sqrt{3}+2}{1-2\sqrt{3}}=-\dfrac{8+5\sqrt{3}}{11}

Using the point-slope form through (2,3):

(8+5\sqrt{3})x+11y-49-10\sqrt{3}=0

Case 2

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

m(1+2\sqrt{3})=2-\sqrt{3}\;\Rightarrow\;m=\dfrac{2-\sqrt{3}}{1+2\sqrt{3}}=\dfrac{5\sqrt{3}-8}{11}

Using the point-slope form through (2,3):

(8-5\sqrt{3})x+11y-49+10\sqrt{3}=0

The right bisector (perpendicular bisector) of a segment passes through its mid-point and is perpendicular to it. Mid-point of the segment joining (3,4) and (-1,2):

M=\left(\dfrac{3+(-1)}{2},\dfrac{4+2}{2}\right)=(1,3)

Slope of the segment:

m_1=\dfrac{2-4}{-1-3}=\dfrac{-2}{-4}=\dfrac{1}{2}

Since the right bisector is perpendicular to the segment, its slope is:

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

Using the point-slope form through M(1,3):

y-3=-2(x-1)

y-3=-2x+2

2x+y-5=0

Let the foot of the perpendicular be F(x_0,y_0). Since F lies on the given line:

3x_0-4y_0-16=0\quad\ldots(1)

Slope of the given line 3x-4y-16=0, i.e., y=\dfrac{3x-16}{4}, is \dfrac{3}{4}. Since PF is perpendicular to this line, its slope is -\dfrac{4}{3}:

\dfrac{y_0-3}{x_0-(-1)}=-\dfrac{4}{3}

3(y_0-3)=-4(x_0+1)\;\Rightarrow\;3y_0-9=-4x_0-4

4x_0+3y_0-5=0\quad\ldots(2)

Solving (1) and (2) simultaneously — multiply (1) by 3 and (2) by 4:

9x_0-12y_0=48\qquad16x_0+12y_0=20

Adding these:

25x_0=68\;\Rightarrow\;x_0=\dfrac{68}{25}

Substituting back into (2):

4\left(\dfrac{68}{25}\right)+3y_0=5\;\Rightarrow\;3y_0=5-\dfrac{272}{25}=\dfrac{-147}{25}\;\Rightarrow\;y_0=-\dfrac{49}{25}

Since (-1,2) lies on the line y=mx+c:

2=-m+c\quad\ldots(1)

The perpendicular from the origin meets the line at (-1,2), so the segment joining O(0,0) and (-1,2) is perpendicular to the line. Slope of this segment:

m_{OP}=\dfrac{2-0}{-1-0}=-2

Since the line and this segment are perpendicular, m\times m_{OP}=-1:

m\times(-2)=-1\;\Rightarrow\;m=\dfrac{1}{2}

Substituting in (1):

c=2+m=2+\dfrac{1}{2}=\dfrac{5}{2}

Finding p

The line x\cos\theta-y\sin\theta-k\cos2\theta=0 has A=\cos\theta, B=-\sin\theta, C=-k\cos2\theta. The distance from the origin:

p=\dfrac{|{-k\cos2\theta}|}{\sqrt{\cos^2\theta+\sin^2\theta}}=|k\cos2\theta|

so that p^2=k^2\cos^2 2\theta.

Finding q

The line x\sec\theta+y\csc\theta-k=0 has A=\sec\theta, B=\csc\theta, C=-k. The distance from the origin:

q=\dfrac{|{-k}|}{\sqrt{\sec^2\theta+\csc^2\theta}}

Simplifying the denominator:

\sec^2\theta+\csc^2\theta=\dfrac{1}{\cos^2\theta}+\dfrac{1}{\sin^2\theta}=\dfrac{\sin^2\theta+\cos^2\theta}{\sin^2\theta\cos^2\theta}=\dfrac{1}{\sin^2\theta\cos^2\theta}

so that:

q=|k|\cdot|\sin\theta\cos\theta|\;\Rightarrow\;q^2=k^2\sin^2\theta\cos^2\theta

Combining

Using \sin2\theta=2\sin\theta\cos\theta\;\Rightarrow\;\sin^22\theta=4\sin^2\theta\cos^2\theta:

4q^2=4k^2\sin^2\theta\cos^2\theta=k^2\sin^22\theta

Therefore:

p^2+4q^2=k^2\cos^22\theta+k^2\sin^22\theta=k^2(\cos^22\theta+\sin^22\theta)=k^2

The altitude from A is the line through A perpendicular to the opposite side BC. Slope of BC:

m_{BC}=\dfrac{2-(-1)}{1-4}=\dfrac{3}{-3}=-1

Since the altitude is perpendicular to BC, its slope is:

m=-\dfrac{1}{m_{BC}}=1

Using the point-slope form through A(2,3):

y-3=1(x-2)

x-y+1=0

Length of the altitude

Equation of side BC, using the point-slope form through B(4,-1) with slope -1:

y-(-1)=-1(x-4)\;\Rightarrow\;x+y-3=0

The length of the altitude from A is the perpendicular distance from A(2,3) to this line:

\text{length}=\dfrac{|2+3-3|}{\sqrt{1^2+1^2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}

The line with x-intercept a and y-intercept b is, in intercept form:

\dfrac{x}{a}+\dfrac{y}{b}=1

Multiplying throughout by ab to write it in general form:

bx+ay-ab=0

Using the distance formula for the perpendicular distance from the origin (0,0):

p=\dfrac{|b(0)+a(0)-ab|}{\sqrt{a^2+b^2}}=\dfrac{|ab|}{\sqrt{a^2+b^2}}

Squaring both sides:

p^2=\dfrac{a^2b^2}{a^2+b^2}

Taking the reciprocal:

\dfrac{1}{p^2}=\dfrac{a^2+b^2}{a^2b^2}=\dfrac{a^2}{a^2b^2}+\dfrac{b^2}{a^2b^2}=\dfrac{1}{b^2}+\dfrac{1}{a^2}