Note: We know that the equation of the line is \displaystyle \frac{x}{a} + \frac{y}{b} = 1 where \displaystyle a is the x-intercept and \displaystyle b is the y-intercept.

Question 1: Find the equation to the straight line:

(i) cutting off intercepts \displaystyle 3 and \displaystyle 2 from the axes.

(ii) cutting off intercepts \displaystyle - 5 and \displaystyle 6 from the axes.

Answer:

i) Here \displaystyle a = 3 \hspace{0.3cm} b = 2

Therefore the equation of the line:

\displaystyle \frac{x}{3} + \frac{y}{2} = 1

\displaystyle \Rightarrow 2x + 3y - 6 = 0

ii) Here \displaystyle a = -5 \hspace{0.3cm} b = 6

Therefore the equation of the line:

\displaystyle \frac{x}{-5} + \frac{y}{6} = 1

\displaystyle \Rightarrow 6x - 5y + 30 = 0

\displaystyle \\

Question 2: Find the equation of the straight line which passes through \displaystyle (1,-2) and cuts off equal intercepts on the axes.

Answer:

Here \displaystyle a = b

Therefore the equation of the line:

\displaystyle \frac{x}{a} + \frac{y}{a} = 1

\displaystyle \Rightarrow x+y = a

Since the line passes through \displaystyle ( 1, -2) we get

\displaystyle 1+(-2) = a \hspace{0.5cm} \Rightarrow a = -1

Hence the equation of the line is \displaystyle x+y +1 = 0

\displaystyle \\

Question 3: Find the equation to the straight line which, passes through the point \displaystyle (5,6) and has intercepts on the axes (i) equal in magnitude and both positive. (ii) equal in magnitude but opposite in sign.

Answer:

i)       Here \displaystyle a = b

Therefore the equation of the line:

\displaystyle \frac{x}{a} + \frac{y}{a} = 1

\displaystyle \Rightarrow x+y = a

Since the line passes through \displaystyle ( 5,6) we get

\displaystyle 5+6 = a \hspace{0.5cm} \Rightarrow a = -11

Hence the equation of the line is \displaystyle x+y = 11

ii)      Here \displaystyle b=-a

Therefore the equation of the line:

\displaystyle \frac{x}{a} + \frac{y}{-a} = 1

\displaystyle \Rightarrow x-y = a

Since the line passes through \displaystyle ( 5,6) we get

\displaystyle 5-6 = a \hspace{0.5cm} \Rightarrow a = -1

Hence the equation of the line is \displaystyle x+y = -1

\displaystyle \\

Question 4: For what values of \displaystyle a and \displaystyle b the -intercepts cut off on the coordinate axes by the line \displaystyle ax+by+8=0 are equal in length but opposite in signs to those cut off by the line \displaystyle 2x - 3y + 6 = 0 on the axes.

Answer:

Given \displaystyle 2x - 3y + 6 = 0

\displaystyle \Rightarrow \frac{2x}{-6} - \frac{3y}{-6} + \frac{6}{-6} = 0

\displaystyle \Rightarrow \frac{x}{-3} + \frac{y}{2} = 1

\displaystyle \text{ Therefore the x-intercept } = a = -3 \displaystyle \text{ and y-intercept } = b = 2

We also have \displaystyle ax+by+8=0

\displaystyle \Rightarrow \frac{ax}{-8} + \frac{by}{-8} + \frac{8}{-8} = 0

\displaystyle \Rightarrow \frac{x}{(-8/a)} + \frac{y}{(-8/b)} = 1

\displaystyle \text{ Therefore the x-intercept } = \frac{-8}{a} \displaystyle \text{ and y-intercept } = \frac{-8}{b}

\displaystyle \therefore \frac{-8}{a} = -(-3) \Rightarrow a = \frac{-8}{3}

\displaystyle \text{ and } \frac{-8}{b} = -(2) \Rightarrow b = 4

\displaystyle \\

Question 5: Find the equation to the straight line which cuts off equal positive intercepts on the axes and their product is \displaystyle 25 .

Answer:

Here \displaystyle a = b \hspace{0.5cm} ab = 25

Solving: \displaystyle ab = 25 \hspace{0.5cm} \Rightarrow a^2 = 25 \hspace{0.5cm} \Rightarrow a = \pm 5

Since the intercepts are positive, we get \displaystyle a = 5

Therefore the equation of the line:

\displaystyle \frac{x}{5} + \frac{y}{5} = 1

\displaystyle \Rightarrow x+y = 5

\displaystyle \\

Question 6: Find the equation of the line which passes through the point \displaystyle (- 4, 3) and the portion of the line intercepted between the axes is divided internally in the ratio \displaystyle 5 : 3 by this point.

Answer:

Let the intercepts be \displaystyle A(a,0) and \displaystyle B(0,b) .

Given \displaystyle (-4,3) divides the \displaystyle A(a,0) and \displaystyle B( 0, b) in the ratio of \displaystyle 5:3

\displaystyle \therefore -4 = \frac{3 \times a + 5 \times 0}{5+3} \displaystyle \Rightarrow a = \frac{-32}{3}

\displaystyle \text{ Similarly } 3 = \frac{3 \times 0 + 5 \times b}{5+3} \displaystyle \Rightarrow b = \frac{24}{5}

Since the equation of line passing through \displaystyle ( -4, 3) , therefore

\displaystyle \frac{x}{\frac{-32}{3}} + \frac{y}{\frac{24}{5}} = 1

\displaystyle \Rightarrow \frac{-3x}{32} + \frac{5y}{24} = 1

\displaystyle \Rightarrow \frac{-3x}{4} + \frac{5y}{3} = 8

\displaystyle \Rightarrow 9x-20y+96=0

\displaystyle \\

Question 7: A straight line passes through the point \displaystyle (\alpha , \beta ) and this point bisects the portion of the line intercepted between the axes. Show that the equation of the straight ling is \displaystyle \frac{x}{2\alpha} + \frac{y}{2\beta} = 1 .

Answer:

Let the intercepts be \displaystyle A(a,0) and \displaystyle B(0,b) .

Given \displaystyle (\alpha, \beta) divides the \displaystyle A(a,0) and \displaystyle B( 0, b) in the ratio of \displaystyle 1:1

\displaystyle \therefore \alpha = \frac{a+0}{2} \Rightarrow a = 2 \alpha

\displaystyle \text{ Similarly } \beta = \frac{0+b}{2} \Rightarrow b = 2 \beta

Hence the equation of line is:

\displaystyle \frac{x}{2\alpha} + \frac{y}{2\beta} = 1

Hence proved.

\displaystyle \\

Question 8: Find the equation of the line which passes through the point \displaystyle (3,4) and is such that the portion of it intercepted between the axes is divided by the point in the ratio \displaystyle 2 : 3 .

Answer:

Let the intercepts be \displaystyle A(a,0) and \displaystyle B(0,b) .

Given \displaystyle P(3,4) divides the \displaystyle A(a,0) and \displaystyle B( 0, b) in the ratio of \displaystyle 2:3

i.e. \displaystyle AP:BP = 2:3

\displaystyle \therefore 3 = \frac{2 \times 0 + 3 \times a}{2+3} \Rightarrow a = 5

\displaystyle \text{ Similarly } 4 = \frac{2 \times b + 3 \times 0}{2+3} \Rightarrow b = 10

Since the equation of line passing through \displaystyle P( 3,4) , therefore

\displaystyle \frac{x}{5} + \frac{y}{10} = 1

\displaystyle \Rightarrow 2x+y = 10

\displaystyle \\

Question 9: Point \displaystyle R (h, k) divides a line segment between the axes in the ratio \displaystyle 1 : 2 . Find the equation of the line.

Answer:

Let the intercepts be \displaystyle A(a,0) and \displaystyle B(0,b) .

Given \displaystyle P(h, k) divides the \displaystyle A(a,0) and \displaystyle B( 0, b) in the ratio of \displaystyle 1:2

i.e. \displaystyle AP:BP = 1:2

\displaystyle \therefore h = \frac{1 \times 0 + 2 \times a}{1+2} \displaystyle \Rightarrow a = \frac{3h}{k}

\displaystyle \text{ Similarly } k = \frac{1 \times b + 2 \times 0}{1+2} \displaystyle \Rightarrow b = 3k

Since the equation of line passing through \displaystyle P( h,k) , therefore

\displaystyle \frac{x}{(3h/2)} + \frac{y}{3k} = 1

\displaystyle \frac{2x}{3h} + \frac{y}{3k} = 1

\displaystyle \Rightarrow 2kx+hy - 3hk=0

\displaystyle \\

Question 10: Find the equation of the straight line which passes through the point \displaystyle (- 3, 8) and cuts off positive intercepts on the coordinate axes whose sum is \displaystyle 7 .

Answer:

Let the intercepts be \displaystyle A(a,0) and \displaystyle B(0,b) .

Given: \displaystyle a+b = 7 \hspace{0.5cm} b = 7 - a .

The line also passes through \displaystyle ( -3, 8 ) . Therefore

\displaystyle \frac{-3}{a} + \frac{8}{7-a} = 1

\displaystyle \Rightarrow -3( 7-a) + 8a =a ( 7-a)

\displaystyle \Rightarrow -21 + 3a + 8a = 7a - a^2

\displaystyle \Rightarrow a^2 + 4a - 21 = 0

\displaystyle \Rightarrow ( a-3)(a+7) = 0

\displaystyle \Rightarrow a = 3 \ or \ a = -7

Since the intercepts are positive we get \displaystyle a = 3

\displaystyle \therefore b = 7 - 3 = 4

Hence the equation of the line is:

\displaystyle \frac{x}{3} + \frac{y}{4} = 1

\displaystyle \Rightarrow 4x + 3y = 12

\displaystyle \\

Question 11: Find the equation to the straight line which passes through the point \displaystyle (- 4, 3) and is such that the portion of it between the axes is divided by the point in the ratio \displaystyle 5 : 3 .

Answer:

Let the intercepts be \displaystyle A(a,0) and \displaystyle B(0,b) .

Given \displaystyle (-4,3) divides the \displaystyle A(a,0) and \displaystyle B( 0, b) in the ratio of \displaystyle 5:3

\displaystyle \therefore -4 = \frac{3 \times a + 5 \times 0}{5+3} \displaystyle \Rightarrow a = \frac{-32}{3}

\displaystyle \text{ Similarly } 3 = \frac{3 \times 0 + 5 \times b}{5+3} \displaystyle \Rightarrow b = \frac{24}{5}

Since the equation of line passing through \displaystyle ( -4, 3) , therefore

\displaystyle \frac{x}{\frac{-32}{3}} + \frac{y}{\frac{24}{5}} = 1

\displaystyle \Rightarrow \frac{-3x}{32} + \frac{5y}{24} = 1

\displaystyle \Rightarrow \frac{-3x}{4} + \frac{5y}{3} = 8

\displaystyle \Rightarrow 9x-20y+96=0

\displaystyle \\

Question 12: Find the equation of a line which passes through the point \displaystyle (22, - 6) and is such that the intercept on x-axis exceeds the intercept on y-axis by \displaystyle 5 .

Answer:

Let the intercepts be \displaystyle A(a,0) and \displaystyle B(0,b) .

Given: \displaystyle a=b+5 \hspace{0.5cm} b = a - 5

The line also passes through \displaystyle (22, -6 ) . Therefore

\displaystyle \frac{22}{a} + \frac{-6}{a-5} = 1

\displaystyle \Rightarrow 22(a-5)-6a=a(a-5)

\displaystyle \Rightarrow 22a - 110 - 6a = a^2 - 5a

\displaystyle \Rightarrow a^2 - 21 a + 110=0

\displaystyle \Rightarrow (a-11)(a-10) = 0

\displaystyle \Rightarrow a = 11 \ \text{ or } \ \ \ \ a = 10

\displaystyle \Rightarrow b = 11-5 = 6 \ \ \ \text{ or } \ \ \ \ b = 10-5 = 5

Hence the equation of the lines are:

\displaystyle \frac{x}{11} + \frac{y}{6} = 1 \text{ or } \frac{x}{10} + \frac{y}{5} = 1

\displaystyle \Rightarrow 6x+11y = 66 \text{ or } \Rightarrow x+2y = 10

\displaystyle \\

Question 13: Find the equation of the line, which passes through \displaystyle P (1, - 7) and meets the axes at \displaystyle A and \displaystyle B respectively so that \displaystyle 4 AP - 3 BP =0 .

Answer:

Let the intercepts be \displaystyle A(a,0) and \displaystyle B(0,b) .

Given \displaystyle 4 AP - 3 BP =0 \hspace{0.5cm} \Rightarrow AP : BP = 3: 4

Given \displaystyle P(1,-7) divides the \displaystyle A(a,0) and \displaystyle B( 0, b) in the ratio of \displaystyle 3:4

\displaystyle \therefore 1 = \frac{3 \times 0 + 4 \times a}{3+4} \displaystyle \Rightarrow a = \frac{7}{4}

\displaystyle \text{ Similarly } -7 = \frac{3 \times b + 4 \times 0}{3+4} \displaystyle \Rightarrow b = \frac{-49}{3}

Since the equation of line passing through \displaystyle P( 3,4) , therefore

\displaystyle \frac{x}{(7/4)} + \frac{y}{(-49/3)} = 1

\displaystyle \Rightarrow \frac{4x}{7} - \frac{3y}{49} = 1

\displaystyle \Rightarrow 28x-3y=49

\displaystyle \\

Question 14: Find the equation of the line passing through the point \displaystyle (2,2) and cutting off intercepts on the axes whose sum is \displaystyle 9 .

Answer:

Let the intercepts be \displaystyle A(a,0) and \displaystyle B(0,b) .

Given: \displaystyle a+b=9 \hspace{0.5cm} b = 9-a

The line also passes through \displaystyle (2,2 ) . Therefore

\displaystyle \frac{2}{a} + \frac{2}{9-a} = 1

\displaystyle \Rightarrow 18-2a+2a=9a-a^2

\displaystyle \Rightarrow a^2-9a+18=0

\displaystyle \Rightarrow (a-3)(a-6)=0

\displaystyle \Rightarrow a = 3 \ \text{ or } \ \ \ \ a = 6

\displaystyle \Rightarrow b = 9-3 = 6 \ \ \ \text{ or } \ \ \ \ b = 9-6 = 3

Hence the equation of the lines are:

\displaystyle \frac{x}{3} + \frac{y}{6} = 1 \text{   or   } \frac{x}{6} + \frac{y}{3} = 1

\displaystyle \Rightarrow 2x+y = 6 \text{ or } \Rightarrow x+2y = 6

\displaystyle \\

Question 15: Find the equation of the straight line which passes through the point \displaystyle P(2, 6) and cuts the coordinate axes at the point \displaystyle A and \displaystyle B respectively so that \displaystyle \frac{AP}{BP} = \frac{2}{3} .

Answer:

Let the intercepts be \displaystyle A(a,0) and \displaystyle B(0,b) .

Given \displaystyle AP : BP = 2:3

Given \displaystyle P(2, 6) divides the \displaystyle A(a,0) and \displaystyle B( 0, b) in the ratio of \displaystyle 2:3

\displaystyle \therefore 2 = \frac{2 \times 0 + 3 \times a}{2+3} \displaystyle \Rightarrow a = \frac{10}{3}

\displaystyle \text{ Similarly } 6 = \frac{2 \times b + 3 \times 0}{2+3} \displaystyle \Rightarrow b = 15

Since the equation of line passing through \displaystyle P( 2,6) , therefore

\displaystyle \frac{x}{(10/3)} + \frac{y}{15} = 1

\displaystyle \Rightarrow \frac{3x}{10} + \frac{y}{15} = 1

\displaystyle \Rightarrow 9x+2y=30

\displaystyle \\

Question 16: Find the equations of the straight lines each of which passes through the point \displaystyle (3,2) and cuts off intercepts \displaystyle a and \displaystyle b respectively on x and y-axes such that \displaystyle a -b =2 .

Answer:

Let the intercepts be \displaystyle A(a,0) and \displaystyle B(0,b) .

Given: \displaystyle a-b=2 \hspace{0.5cm} a= b+2

The line also passes through \displaystyle (3,2 ) . Therefore

\displaystyle \frac{3}{b+2} + \frac{2}{b} = 1

\displaystyle \Rightarrow 3b+ 2b + 4 = b^2 + 2b

\displaystyle \Rightarrow b^2 - 3b - 4 = 0

\displaystyle \Rightarrow (b+1)(b-4) =0

\displaystyle \Rightarrow b = -1 \ or \ \ \ \ b = 4

\displaystyle \Rightarrow a = -1+2=1 \ \ \ or \ \ \ \ a = 4+2 = 6

Hence the equation of the lines are:

\displaystyle \frac{x}{1} + \frac{y}{-1} = 1 \text{ or } \frac{x}{6} + \frac{y}{4} = 1

\displaystyle \Rightarrow x-y = 1 \text{ or } \Rightarrow 2x+3y=12

\displaystyle \\

Question 17: Find the equations of the straight lines which pass through the origin and trisect the portion of the straight line \displaystyle 2x + 3y = 6 which is intercepted between the axes.

2021-01-11_10-45-56Answer:

Given line \displaystyle 2x + 3y = 6 

\displaystyle \text{ x-intercept } = A ( 3,0)

\displaystyle \text{ y-intercept } = B ( 0,2)

\displaystyle P divides \displaystyle AB in the ratio of \displaystyle 2:1

\displaystyle \text{ Coordinates of  } P = \Big( \frac{2 \times 3 + 1 \times 0}{2+1} , \frac{2 \times 0 + 1 \times 2}{2+1} \Big) = \Big( 2 , \frac{2}{3} \Big)

\displaystyle Q divides \displaystyle AB in the ratio of \displaystyle 1:2

\displaystyle \text{ Coordinates of  } Q = \Big( \frac{1 \times 3 + 2 \times 0}{1+2} , \frac{1 \times 0 + 1 \times 2}{1+2} \Big) = \Big( 1 , \frac{4}{3} \Big)

\displaystyle \text{ Slope of } OP = \frac{\frac{2}{3}-0}{2-0} = \frac{1}{3}

Therefore equation of \displaystyle OP :

\displaystyle y - 0 = \frac{1}{3} ( x - 0) \hspace{0.5cm} \Rightarrow x-3y=0

\displaystyle \text{ Slope of } OQ = \frac{\frac{4}{3}-0}{1-0} = \frac{1}{3}

\displaystyle \text{ Similarly, equation of } OQ :

\displaystyle y - 0 = \frac{4}{3} ( x - 0) \hspace{0.5cm} \Rightarrow 4x-3y=0

\displaystyle \\

Question 18: Find the equation of the straight line passing through the point \displaystyle (2, 1) and bisecting the portion of the straight line \displaystyle 3x -5y =15 lying between the axes.

Answer:

Given \displaystyle 3x -5y =15

Therefore intercepts of \displaystyle x and \displaystyle y axis are \displaystyle A ( 5,0) and \displaystyle B ( 0, -3)

\displaystyle \text{ Mid point M of  } AB = \Big( \frac{5+0}{2} , \frac{0-3}{2} \Big) = \Big( \frac{5}{2} , \frac{-3}{2} \Big)

\displaystyle \text{ Slope of line passing through M and ( 2, 1)} = \frac{1 - ( \frac{-3}{2}) }{2-(\frac{5}{2})} = \frac{\frac{5}{2}}{\frac{-1}{2}} = -5

Therefore equation of \displaystyle OP :

\displaystyle y - 0 = -5 ( x - 2) \hspace{0.5cm} \Rightarrow y-1=-5x+10 \hspace{0.5cm} \Rightarrow 5x+y = 11

\displaystyle \\

Question 19: Find the equation of the straight tine passing through the origin and bisecting the portion of the line \displaystyle ax +by + c = 0 intercepted between the coordinate axes.

Answer:

Given \displaystyle ax +by + c = 0

When \displaystyle x = 0, y = \frac{-c}{b} \displaystyle \Rightarrow \displaystyle \text{ y-intercept } B( 0, \frac{-c}{b} )

When \displaystyle y = 0 , x = \frac{-c}{a} \displaystyle \Rightarrow \displaystyle \text{ y-intercept } A( \frac{-c}{a} , 0)

\displaystyle \text{ Therefore midpoint of  } AB = ( \frac{ \frac{-c}{a}+0 }{2} , \frac{ 0+ \frac{-c}{b} }{ 2 } ) = ( \frac{-c}{2a} , \frac{-c}{2b} )

\displaystyle \text{ Slope of } AB = \frac{ \frac{-c}{2b}-0 }{ \frac{-c}{2a}-0 } = \frac{a}{b}

\displaystyle \text{ Similarly, equation of } OQ :

\displaystyle y - 0 = \frac{a}{b} ( x - 0) \hspace{0.5cm} \Rightarrow ax-by = 0=0