Class 11: The Straight Line – Exercise 23.6 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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.

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

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

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

Answer:

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 $\displaystyle 2 : 3$ .

Answer:

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.

Answer:

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 $\displaystyle 7$ .

Answer:

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 $\displaystyle 5 : 3$ .