Question 1: Find the distance of the point (4, 5) from the straight line 3x -5y + 7 = 0 .

Answer:

Given equation: 3x -5y + 7 = 0

Comparing with ax + by + c = 0   we get a = 3, \ \ \ \ b = -5 , \ \ \ \ c = 7

Therefore perpendicular distance from point ( 4, 5)   from 3x -5y + 7 = 0

d = \Bigg|  \frac{ax_1+ by_1 + c}{\sqrt{a^2+b^2}} \Bigg| = \Bigg|  \frac{3 \times 4 - 5 \times 5 + 7}{\sqrt{(3)^2+(-5)^2}} \Bigg| = \frac{6}{\sqrt{34}}

Thus the required distance is \frac{6}{\sqrt{34}}

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Question 2: Find the perpendicular distance of the line joining the points ( \cos \theta , \sin \theta ) and ( \cos \theta , \sin \theta ) from the origin.

Answer:

The equation of the line joining ( \cos \theta, \sin \theta) and ( \cos \phi, \sin \phi) :

y - \sin \theta = \frac{\sin \phi - \sin \theta}{\cos \phi - \cos \theta} ( x - \cos \theta)

\Rightarrow (\cos \phi - \cos \theta ) y  - \sin \theta (\cos \phi - \cos \theta )  = (\sin \phi - \sin \theta) x - \cos \theta (\sin \phi - \sin \theta ) 

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

Comparing with ax + by + c = 0   we get a = (\sin \phi - \sin \theta), \ \ \ \ b = - (\cos \phi - \cos \theta ) , \ \ \ \ c = (\sin \theta \cos \phi - \cos \theta \sin \phi)

Therefore,

d = \Bigg|  \frac{(\sin \phi - \sin \theta)(0)- (\cos \phi - \cos \theta )(0) + (\sin \theta \cos \phi - \cos \theta \sin \phi)}{\sqrt{(\sin \phi - \sin \theta)^2+(- (\cos \phi - \cos \theta ))^2}}   \Bigg|

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

= \Bigg|  \frac{\sin ( \theta - \phi)}{ \sqrt{2 - 2 \cos ( \theta - \phi)} }   \Bigg|

= \frac{1}{\sqrt{2}} \Bigg|  \frac{\sin ( \theta - \phi)}{ \sqrt{1 -  \cos ( \theta - \phi)} }   \Bigg|

= \frac{1}{\sqrt{2}} \Bigg|  \frac{\sin ( \theta - \phi)}{ \sqrt{2 \sin^2 ( \frac{\theta - \phi}{2} )} }   \Bigg|

= \frac{1}{\sqrt{2}} \Bigg|  \frac{2 \sin ( \frac{\theta - \phi}{2} ) \cos ( \frac{\theta - \phi}{2} )}{ \sqrt{2} \sin ( \frac{\theta - \phi}{2} ) }   \Bigg|

=  \Big|  \cos \Big( \frac{\theta - \phi}{2} \Big) \Big|

Thus the required distance is \cos \Big( \frac{\theta - \phi}{2} \Big)

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Question 3: Find the length of the perpendicular from the origin to the straight line joining the two points whose coordinates are (a \cos \alpha, a \sin \alpha) and (a \cos \beta, a \sin \beta) .

Answer:

The equation of the line joining ( a \cos \alpha, a \sin \alpha) and ( a \cos \beta, a \sin \beta ) :

y - a \sin \alpha = \frac{ a \sin \beta - a \sin \alpha}{a \cos \beta - a \cos \alpha} ( x - a \cos \alpha)

y - a \sin \alpha = \frac{ 2 \cos ( \frac{\beta + \alpha }{2} ) \sin (\frac{\beta - \alpha }{2}) }{ 2 \sin (\frac{\beta + \alpha }{2}) \sin (\frac{\alpha - \beta }{2}) } ( x - a \cos \alpha)

\Rightarrow y - a \sin \alpha = - \cos \Big( \frac{\beta + \alpha }{2} \Big ) ( x - a \cos \alpha)

\Rightarrow x \cot \Big( \frac{ \alpha + \beta }{2} \Big) + y - a \sin \alpha - a \cos \alpha \cot \Big( \frac{ \alpha + \beta }{2} \Big) = 0

Comparing with ax + by + c = 0   we get

a = \cot \Big( \frac{ \alpha + \beta }{2} \Big) , \ \ \ \ b = - 1 , \ \ \ \ c = - a \sin \alpha - a \cos \alpha \cot \Big( \frac{ \alpha + \beta }{2} \Big)

Therefore the perpendicular distance from (0,0) is:

d = \Bigg| \frac{ \cot (\frac{ \alpha + \beta }{2}) ( 0) -1 (0 )  - a \sin \alpha - a \cos \alpha \cot ( \frac{ \alpha + \beta }{2} ) }{ \sqrt{ \cot^2 ( \frac{\alpha + \beta}{2} )  + 1 }  } \Bigg|

= \Bigg| \frac{  a \sin \alpha + a \cos \alpha \cot ( \frac{ \alpha + \beta }{2} ) }{ \sqrt{ \mathrm{cosec}^2 ( \frac{\alpha + \beta}{2} )   }  } \Bigg|

= \Big|   \sin (\frac{ \alpha + \beta }{2}) \sin \alpha + \cos \alpha \cos (\frac{ \alpha + \beta }{2})    \Big|

= a \cos \Big( \frac{ \alpha + \beta }{2} - \alpha \Big)

= a \cos \Big( \frac{ \beta - \alpha   }{2} \Big)

= a \cos \Big( \frac{\alpha - \beta   }{2} \Big)

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Question 4: Show that the perpendiculars let fall from any point on the straight line 2 x + 11 y -5 = 0 upon the two straight lines 24 x + 7 y = 20 and 4 x - 3 y - 2 = 0 are equal to each other.

Answer:

Given equations:

24 x + 7 y = 20      … … … … … i)

4 x - 3 y - 2 = 0      … … … … … ii)

Let P(a, b) be the point on 2 x + 11 y -5 = 0

Therefore, the distance d_1 of P( a, b) from line i)

d_1 = \Bigg|  \frac{24a+7b-20}{\sqrt{24^2 + 7^2}} \Bigg|= \Bigg|  \frac{24a+7b-20}{25} \Bigg|      … … … … … iii)

Similarly, the distance d_2 of P( a, b) from line ii)

d_2 = \Bigg|  \frac{4a-3b-2}{\sqrt{3^2 + (-4)^2}} \Bigg|= \Bigg|  \frac{4a-3b-2}{5} \Bigg|      … … … … … iv)

Since P(a, b) is on 2x + 11y - 5 = 0

\Rightarrow 2a + 11b - 5 = 0 \Rightarrow b = \frac{5-2a}{11}

Substituting the value of b in iii) and iv) we get

d_1 = \Big|  \frac{24a+7 \times (\frac{5-2a}{11})-20}{25} \Big| = \Big| \frac{50a-37}{55} \Big|

d_2 = \Big|  \frac{4a-3 \times (\frac{5-2a}{11}) - 2}{5} \Big| = \Big| \frac{50a-37}{55} \Big|

\therefore d_1 = d_2

Hence the perpendicular drawn from any point on the straight line 2 x + 11 y -5 = 0 upon the two straight lines 24 x + 7 y = 20 and 4 x - 3 y - 2 = 0 are equal to each other.

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Question 5: Find the distance of the point of intersection of the lines 2 x + 3 y =21 and 3 x - 4y +11 =0 from the line 8 x + 6 y + 5 = 0 .

Answer:

Given lines:

2 x + 3 y =21      … … … … … i)

3 x - 4y +11 =0      … … … … … ii)

Solving i) and ii) we get the point of intersection as ( 3, 5)

Comparing 8 x + 6 y + 5 = 0 with ax + by + c = 0   we get a = 8, \ \ \ \ b = 6 , \ \ \ \ c = 5

Therefore perpendicular distance from point ( 3, 5)   from 8 x + 6 y + 5 = 0

d = \Bigg|  \frac{ax_1+ by_1 + c}{\sqrt{a^2+b^2}} \Bigg| = \Bigg|  \frac{8 \times 3 + 6 \times 4 + 5}{\sqrt{(8)^2+(6)^2}} \Bigg| = \frac{59}{\sqrt{10}}

Thus the required distance is \frac{59}{\sqrt{10}}

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Question 6: Find the length of the perpendicular from the point (4, -7) to the line joining the origin and the point of intersection of the line  2x -3y + 14=0 and 5 x + 4y -7 =0 .

Answer:

Given lines:

2x -3y + 14=0      … … … … … i)

5 x + 4y -7 =0      … … … … … ii)

Solving i) and ii) we get the point of intersection as \Big( \frac{-35}{23} , \frac{84}{23} \Big)

Therefore the equation of line passing between (0,0) and \Big( \frac{-35}{23} , \frac{84}{23} \Big) :

y - 0 = \Big( \frac{\frac{84}{23} - 0}{\frac{-35}{23}-0} \Big) ( x - 0)

\Rightarrow y = \frac{-12}{5} x

Question 7: What are the points on x-axis whose perpendicular distance from the straight line \frac{x}{a} + \frac{y}{b} =1 is a ?

Answer:

Let the point P(t, 0) be on the x-axis.

Given line: \frac{x}{a} + \frac{y}{b} =1

Comparing with ax + by + c = 0   we get a = \frac{1}{a} , \ \ \ \ b = \frac{1}{b} , \ \ \ \ c = -1

Therefore perpendicular distance from point ( t,0)

d = \Bigg|  \frac{ax_1+ by_1 + c}{\sqrt{a^2+b^2}} \Bigg| = \Bigg|  \frac{\frac{1}{a} \times t + \frac{1}{b} \times (0) -1}{\sqrt{(\frac{1}{a})^2+(\frac{1}{b})^2}}   \Bigg| = a 

Squaring both sides we get

a^2 ( \frac{1}{a^2} + \frac{1}{b^2} ) = \frac{t^2}{a^2} +1 - \frac{2t}{a}

1 + \frac{a^2}{b^2} = \frac{t^2}{a^2} +1 - \frac{2t}{a}

a^2b^2 + a^2 = b^2t^2 + a^2b^2 - 2ab^2 t

b^2 t^2 - 2ab^2 t - a^2 = 0

t = \frac{2ab^2 \pm \sqrt{a^2b^4 + b^2 a^4}}{2b^2}

= \frac{a}{b} ( b \pm \sqrt{a^2 +b^2})

Hence the required points on x-axis are

\Big( \frac{a}{b} ( b + \sqrt{a^2 +b^2}), 0 \Big ) and \Big( \frac{a}{b} ( b - \sqrt{a^2 +b^2}), 0 \Big )

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Question 8: Show that the product of perpendiculars on the line                            \frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta =1 from the points (\pm \sqrt{a^2 - b^2}, 0) is b^2 .

Answer:

Let d_1 be the perpendicular distance from (\sqrt{a^2 - b^2}, 0) on \frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta =1

d_1 = \Bigg|  \frac{ \frac{\cos \theta}{a} (\sqrt{a^2 - b^2}) + \frac{\sin \theta}{b} (0) - 1 }{  \sqrt{ \frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta }{b^2} }  } \Bigg|   = b  \Bigg| \frac{(\sqrt{a^2 - b^2}) ( \cos \theta) - a}{ \sqrt{b^2\cos^2 \theta + a^2 \sin^2 \theta}  } \Bigg|

Let d_2 be the perpendicular distance from ( - \sqrt{a^2 - b^2}, 0) on \frac{x}{a} \cos \theta +\frac{y}{b} \sin \theta =1

d_1 = \Bigg|  \frac{ \frac{\cos \theta}{a} ( - \sqrt{a^2 - b^2}) + \frac{\sin \theta}{b} (0) - 1 }{  \sqrt{ \frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta }{b^2} }  } \Bigg|   = b  \Bigg| \frac{(\sqrt{a^2 - b^2}) ( \cos \theta) + a}{ \sqrt{b^2\cos^2 \theta + a^2 \sin^2 \theta}  } \Bigg|

\therefore d_1 \times d_2 = \Bigg|  \frac{(\sqrt{a^2 - b^2}) ( \cos \theta) - a}{ \sqrt{b^2\cos^2 \theta + a^2 \sin^2 \theta}  }   \Bigg| \times \Bigg|  \frac{(\sqrt{a^2 - b^2}) ( \cos \theta) + a}{ \sqrt{b^2\cos^2 \theta + a^2 \sin^2 \theta}  }   \Bigg|

= b^2  \Bigg|  \frac{(a^2 - b^2) \cos^2 \theta - a^2}{b^2 \cos^2 \theta + a^2 \sin^2 \theta} \Bigg|

= b^2  \Bigg|  \frac{a^2 ( \cos^2 \theta - 1) - b^2 \cos^2 \theta}{b^2 \cos^2 \theta + a^2 \sin^2 \theta}    \Bigg|

= b^2  \Bigg|  \frac{- a^2 \sin^2 \theta - b^2 \cos^2 \theta}{b^2 \cos^2 \theta + a^2 \sin^2 \theta}  \Bigg|

= b^2

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Question 9: Find the perpendicular distance from the origin of the perpendicular from the point (1, 2) upon the straight line x-\sqrt{3}y + 4 = 0 .

Answer:

Given line: x-\sqrt{3}y + 4 = 0      … … … … … i)

Therefore line perpendicular to line i) is \sqrt{3} x + y + \lambda = 0

The perpendicular passes through ( 1,2 ) . Therefore \lambda = -\sqrt{3}-2

Therefore the equation of the perpendicular line is

\sqrt{3} x + y -\sqrt{3}-2 = 0      … … … … … ii)

Now the perpendicular distance from (0,0) to line ii)

d = \Big| \frac{\sqrt{3}(0) + 1 (0) - ( \sqrt{3}+2)}{\sqrt{3+1}}  \Big| = \Big| \frac{\sqrt{3}+2}{2} \Big|

Hence the required distance is \frac{\sqrt{3}+2}{2} .

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Question 10: Find the distance of the point (1, 2) from the straight line with slope 5 and passing through the point of intersection of x + 2y =5 and x - 3 y =7 .

Answer:

Given lines:

x + 2y =5      … … … … … i)

x - 3 y =7      … … … … … ii)

Solving i) and ii) we get the point of intersection as \Big( \frac{29}{5} , \frac{-2}{5} \Big )

The equation of line passing through \Big( \frac{29}{5} , \frac{-2}{5} \Big ) and slope of 5 :

y - \Big( \frac{-2}{5} \Big) = 5 \Big( x - \frac{29}{5} \Big)

\Rightarrow 5y + 2 = 25 x - 145

\Rightarrow 25x - 5y - 147 = 0

Therefore perpendicular distance from ( 1,2) on 25x - 5y - 147 = 0

\Rightarrow d = \Big| \frac{25 \times ( 1) - 5 ( 2) - 147}{ \sqrt{25^2 + 5^2}} \Big| = \frac{132}{5\sqrt{26}}

Hence the required distance is \frac{132}{5\sqrt{26}}

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Question 11: What are the points on y-axis whose distance from the line       \frac{x}{3} + \frac{y}{4} = 1 in 4 units?

Answer:

Let P(0, t) be the point on y-axis.

Given line: \frac{x}{3} + \frac{y}{4} = 1

Comparing with ax + by + c = 0   we get a = \frac{1}{3} , \ \ \ \ b = \frac{1}{4} , \ \ \ \ c = -1

\therefore \Bigg| \frac{\frac{1}{3}( 0) + \frac{1}{4}(t) - 1}{\sqrt{\frac{1}{3^2}+\frac{1}{4^2}}} \Bigg| = 4

\Rightarrow \Big| \frac{t-4}{5} \Big| = \frac{4}{3}

\Rightarrow t-4 = \pm \frac{20}{3}

\Rightarrow t = 4 \pm \frac{20}{3}

Hence the required points on y-axis are \Big( 0, 4  + \frac{20}{3} \Big) and \Big( 0, 4 - \frac{20}{3} \Big)

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Question 12: In the \triangle ABC with vertices A (2, 3), B (4, -1) and C (1, 2) find the equation and the length of the altitude from vertex A .

Answer:

Given vertices: A (2, 3), B (4, -1) and C (1, 2)

Equation of BC:

y - ( -1) = \frac{2-(-1)}{1 - 4} (x-4)

\Rightarrow x + y - 3  = 0      … … … … … i)

Therefore equation perpendicular to i) is

x - y + \lambda = 0      … … … … … ii)

Since this line passes thought A ( 2, 3), \lambda = 1

Therefore the equation of the perpendicular is x - y + 1 = 0

Perpendicular distance of A(2,3) from BC :

d = \Big| \frac{1(2) + 1( 3) - 3}{\sqrt{1^2+1^2}} \Big| = \sqrt{2}

Hence the required distance is \sqrt{2}

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Question 13: Show that the path of a moving point such that its distances from two lines 3x - 2y = 5 and 3x+2y +7 = 0 are equal is a straight line.

Answer:

Let P( h, k) be the point such that its distances from two lines 3x - 2y = 5 and 3x+2y +7 = 0 are equal. Therefore

\Big| \frac{3h-2k-5}{\sqrt{3^2+(-2)^2}} \Big| = \Big| \frac{3h+2k-5}{\sqrt{3^2+2^2}} \Big|

\Rightarrow | 3h - 2k - 5 | = | 3h+ 2k - 5|

\Rightarrow ( 3h - 2k - 5 ) = \pm (  3h+ 2k - 5)

Consider + ve sign

3h - 2k - 5 = 3h+ 2k - 5 \hspace{0.5cm} \Rightarrow 4k = 0 \hspace{0.5cm} \Rightarrow k = 0

Consider - ve sign

3h - 2k - 5 = -3h- 2k + 5 \hspace{0.5cm} \Rightarrow 6h=10 \hspace{0.5cm} \Rightarrow 3h = 5

Therefore the equations are 3x=5 and y = 0

These are also straight lines.

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Question 14: If sum of perpendicular distances of a variable point P (x, y) from the lines x+y - 5 = 0 and 3x-2y+7=0 is always 10 . Show that P must move on a line.

Answer:

Let P(h, k)   be the point such that sum of the perpendicular distance from P to the given lines is 10 . Therefore

\Big| \frac{h+k-5}{\sqrt{1^2+1^2}} \Big| + \Big| \frac{3h-2k+7}{\sqrt{3^2+2^2}} \Big| = 10

\Big| \frac{h+k-5}{\sqrt{2}} \Big| + \Big| \frac{3h-2k+7}{\sqrt{13}} \Big| = 10

When both are + ve

\frac{h+k-5}{\sqrt{2}} + \frac{3h-2k+7}{\sqrt{13}} = 10

\Rightarrow ( \sqrt{13} + 3\sqrt{2}) h + ( \sqrt{13}- 2\sqrt{2})k + 7\sqrt{2}- 5\sqrt{13} - 10\sqrt{26} = 0

This is a straight line.

Similarly, when both  are - ve

- \frac{h+k-5}{\sqrt{2}} - \frac{3h-2k+7}{\sqrt{13}} = 10

\Rightarrow ( \sqrt{13} + 3\sqrt{2}) h - ( \sqrt{13}+ 2\sqrt{2})k + 7\sqrt{2}- 5\sqrt{13} + 10\sqrt{26} = 0

This is a straight line as well.

Similarly, the other two combinations are also straight lines.

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Question 15: If the length of the perpendicular from the point (1, 1) to the line ax -by + c = 0 be unity, show that \frac{1}{c} + \frac{1}{a} - \frac{1}{b} = \frac{c}{2ab}

Answer:

Given the perpendicular distance from point ( 1, 1) to the straight line ax -by + c = 0 is 1 . Therefore,

\Big| \frac{a-b+c}{\sqrt{a^2+b^2}} \Big| = 1

Squaring both sides

a^2 + b^2 + c^2 - 2ab - 2bc + 2ac = a^2 + b^2

\Rightarrow ab + bc - ac = \frac{c^2}{2}

\Rightarrow \frac{1}{c} + \frac{1}{a} - \frac{1}{b} = \frac{c}{2ab}