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 $\displaystyle 3 \text{ and } 2$ from the axes.

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

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.

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.

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 $\displaystyle a \text{ and } b$ 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.

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 $\displaystyle 25$.

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.

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

Given $\displaystyle (-4,3)$ divides the $\displaystyle A(a,0) \text{ and } B( 0, b)$ 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 $\displaystyle \frac{x}{2\alpha} + \frac{y}{2\beta} = 1$.

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

Given $\displaystyle (\alpha, \beta)$ divides the $\displaystyle A(a,0) \text{ and } B( 0, b)$ 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 $\displaystyle 2 : 3$.

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

Given $\displaystyle P(3,4)$ divides the $\displaystyle A(a,0) \text{ and } B( 0, b)$ 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.

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

Given $\displaystyle P(h, k)$ divides the $\displaystyle A(a,0) \text{ and } B( 0, b)$ 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 $\displaystyle 7$.

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

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 $\displaystyle 5 : 3$.

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

Given $\displaystyle (-4,3)$ divides the $\displaystyle A(a,0) \text{ and } B( 0, b)$ 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 $\displaystyle 5$.

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

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 $\displaystyle A \text{ and } B$ respectively so that $\displaystyle 4 AP - 3 BP =0$.

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

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

Given $\displaystyle P(1,-7)$ divides the $\displaystyle A(a,0) \text{ and } B( 0, b)$ 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 $\displaystyle 9$.

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

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 $\displaystyle A \text{ and } B$ respectively so that $\displaystyle \frac{AP}{BP} = \frac{2}{3}$ .

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

Given $\displaystyle AP : BP = 2:3$

Given $\displaystyle P(2, 6)$ divides the $\displaystyle A(a,0) \text{ and } B( 0, b)$ 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 $\displaystyle a \text{ and } b$ respectively on x and y-axes such that $\displaystyle a -b =2$.

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

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.

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.

Given $\displaystyle 3x -5y =15$

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

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

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

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.

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

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

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

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

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