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

Question 1: Find the equation to the straight line:

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

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

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

Therefore the equation  of the line:

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

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

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

Therefore the equation  of the line:

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

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

$\\$

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

Here $a = b$

Therefore the equation  of the line:

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

$\Rightarrow x+y = a$

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

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

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

$\\$

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

i)       Here $a = b$

Therefore the equation  of the line:

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

$\Rightarrow x+y = a$

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

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

Hence the equation of the line is $x+y = 11$

ii)      Here $b=-a$

Therefore the equation  of the line:

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

$\Rightarrow x-y = a$

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

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

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

$\\$

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

Given $2x - 3y + 6 = 0$

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

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

Therefore the x-intercept $= a = -3$ and y-intercept $= b = 2$

We also have $ax+by+8=0$

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

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

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

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

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

$\\$

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

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

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

Since the intercepts are positive, we get $a = 5$

Therefore the equation  of the line:

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

$\Rightarrow x+y = 5$

$\\$

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

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

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

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

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

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

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

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

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

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

$\\$

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

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

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

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

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

Hence the equation of line is:

$\frac{x}{2\alpha}$ $+$ $\frac{y}{2\beta}$ $= 1$. Hence proved.

$\\$

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

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

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

i.e. $AP:BP = 2:3$

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

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

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

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

$\Rightarrow 2x+y = 10$

$\\$

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

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

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

i.e. $AP:BP = 1:2$

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

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

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

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

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

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

$\\$

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

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

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

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

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

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

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

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

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

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

Since the intercepts are positive we get $a = 3$

$\therefore b = 7 - 3 = 4$

Hence the equation of the line is:

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

$\Rightarrow 4x + 3y = 12$

$\\$

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

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

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

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

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

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

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

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

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

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

$\\$

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

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

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

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

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

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

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

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

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

$\Rightarrow a = 11 \ or \ \ \ \ a = 10$

$\Rightarrow b = 11-5 = 6 \ \ \ or \ \ \ \ b = 10-5 = 5$

Hence the equation of the lines are:

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

$\Rightarrow 6x+11y = 66$    or    $\Rightarrow x+2y = 10$

$\\$

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

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

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

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

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

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

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

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

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

$\Rightarrow 28x-3y=49$

$\\$

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

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

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

The line also passes through $(2,2 )$. Therefore

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

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

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

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

$\Rightarrow a = 3 \ or \ \ \ \ a = 6$

$\Rightarrow b = 9-3 = 6 \ \ \ or \ \ \ \ b = 9-6 = 3$

Hence the equation of the lines are:

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

$\Rightarrow 2x+y = 6$    or    $\Rightarrow x+2y = 6$

$\\$

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

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

Given $AP : BP = 2:3$

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

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

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

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

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

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

$\Rightarrow 9x+2y=30$

$\\$

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

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

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

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

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

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

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

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

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

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

Hence the equation of the lines are:

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

$\Rightarrow x-y = 1$    or    $\Rightarrow 2x+3y=12$

$\\$

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

Given line $2x + 3y = 6$

x-intercept $= A ( 3,0)$

y-intercept $= B ( 0,2)$

$P$ divides $AB$ in the ratio of $2:1$

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

$Q$ divides $AB$ in the ratio of $1:2$

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

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

Therefore equation of $OP$:

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

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

Similarly, equation of $OQ$:

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

$\\$

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

Given $3x -5y =15$

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

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

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 $OP$:

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

$\\$

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

Given $ax +by + c = 0$

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

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

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

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

Similarly, equation of $OQ$:

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