Class 11th

68. The given eqⁿ of line is

l1: x + y = 4

Let R divides the line joining two points P (?1,1) and Q (5,7) in ratio k:1. Then,

Co-ordinate of R = ($\left(\frac{5k-1}{k+1},\frac{7k+1}{k+1}\right)$)

As l₁ divides line joining PQ, then R lies on l₁

i e,  $\left(\frac{5k-1}{k+1},\frac{7k+1}{k+1}\right)$=4

5k ?1 + 7k + 1= 4 (k + 1)

12k = 4k + 4

8k = 4

k = $\frac{1}{2}$

The ratio in which x + y = 4 divides line joining (?1,1) ad (5,7) is $\frac{1}{2}$ :1 i.e., 1: 2.

30. Given, f (x)= $\left\{\left(x,\frac{x^2}{1+x^2}\right):x\in R\right\}$

We know that, for x $\in$ R.

So,  x²≥ 0 ⇒ $\frac{x^2}{x^2+1}\ge \frac{0}{x^2+1}\Rightarrow \frac{x^2}{x^2+1}\ge 0\Rightarrow f(x)\ge 0$

and x²+1>x²

⇒ $1>\frac{x^2}{x^2+1}$

⇒1 > f (x).

So, 0 ≤ f (x) < 1

∴ Range of f (x) = [0,1).

67. The given eqⁿ of line is.

l₁ : y = mx + c.

Slope of l₁ = m

Let m? be the slope of line passing through origin (0, 0) and making angle θ with l₁

Thus, (y 0) = m? (x 0)

y = m? x

m? =  $\mathrm{y}$
$\frac{\mathrm{}\mathrm{x}}{}$______ (1)

And tanθ = $\left|\frac{m^{\prime }-m}{1+m^{\prime }·m}\right|$ = $\frac{m^{\prime }-m}{1+m^{\prime }m}$

When, tanθ = $\frac{m^{\prime }-m}{1+m^{\prime }m}.$

tanθ + m? m tanθ = m' - m

m + tanθ = m? - m?m tanθ

m' = $\frac{m+tan\theta }{1-mtan\theta }.$

When tan θ = $-\left(\frac{m^{\prime }-m}{1+m^{\prime }m}\right)$

tan θ + m? m tanθ = -m? + m

m' = $\frac{m-tan\theta }{1+mtan\theta }.$

Hence combining the two we get,

$m^{\prime }=\frac{m\pm tan\theta }{1?mtan\theta }.$

$\Rightarrow \frac{y}{x}=\frac{m\pm tan\theta }{1?mtan\theta }.$ {-: eqⁿ (1) }

66. The given eqn of the line is.

4x + 7y – 3 = 0 _____ (1)

2x – 3y + 1 = 0 _______ (2)

Solving (1) and (2) using eqn (1) 2 x eqn (2) we get,

(4x + 7y – 3) 2 [ (2x – 3y + 1)] = 0

4x + 7y – 3 – 4x + 6y – 2 = 0

13y = 5

y = $\frac{5}{13}$

And 2x – 3 $\left(\frac{5}{13}\right)$ + 1 = 0

2x = $\frac{15}{13}$ – 1 = $\frac{2}{13}$

$\Rightarrow x=\frac{1}{13}$

Point of intersection of (1) and (2) is $\left(\frac{1}{13},\frac{5}{13}\right)$

Since, the line passing through $\left(\frac{1}{13},\frac{5}{13}\right)$ has equal intercept say c then it is of the form

$\frac{x}{c}+\frac{y}{c}=1.$

x + y = c

$\Rightarrow \frac{1}{13}+\frac{5}{13}=c$

c = $\frac{6}{13}$

the read eqn of line is x + y = $\frac{6}{13}$

13x + 13y – 6 = 0

65. x – 2y = 3

y = $\frac{\mathrm{x}}{2}$ - $\frac{3}{2}$______ (1)

Slope of line (1) is $\frac{1}{2}\cdot$

Let the line through P (3, 2) have slope m

Then, angle between the line = $\left|\frac{m-\frac{1}{2}}{1+m\frac{1}{2}}\right|$

$\Rightarrow tan45^{°}=\left|\frac{2m-1}{2+m}\right|$

$\Rightarrow 1=\left|\frac{2m-1}{2+m}\right|$

$\Rightarrow \frac{2m-1}{2+m}=\pm 1.$

When,  $\frac{2m-1}{2+m}=1$ =>2m – 1 = 2 + m=> m = 3.

The eqn of line through (3, 2) is

y – 2 = 3 (x – 3) 3x – y – 7 = 0.

When $\frac{2m-1}{2+m}$ = – 1=> 2m – 1 = – 2 – m =>3m = – 1 m = $-\frac{1}{3}$

The equation of line through (3,2) is,

y – 2 = $-\frac{1}{3}$ (x – 3) => 3y – 6 = – X + 3

x + 3y – 9 = 0

64. The given eqⁿ of the three lines are

y = m₁ x + c₁ ______ (1)

y = m₂ x + c₂ ______ (2)

y = m₃ x + c₃ ______ (3)

The point of intersection of (2) and (3) is given by.

y - y = (m₂x + c₂) - (m₃ x + c₃)

(m₂ - m₃) x = c₃ - c₂

$\Rightarrow x=\frac{c_3-c_2}{m_2-m_3}.$

Hence, y = $\frac{m_2\left(c_3-c_2\right)}{\left(m_2-m_3\right)}+c_2$

$=\frac{m_2\left(c_3-c_2\right)+c_2\left(m_2-m_3\right)}{m_2-m_3.}$

$=\frac{m_2{c}_3-m_3{c}_2}{m_2-m_3}.$

$ie,\left(\frac{c_3-c_2}{m_2-m_3},\frac{m_2{c}_3-m_3{c}_2}{m_2-m_3}\right)$

As the three lines are concurrent, the point of intersection of (2) and (3) lies on line (1) also

i e, $\frac{m_2{c}_3-m_3{c}_2}{m_2-m_3}=m_1\left(\frac{c_3-c_2}{m_2-m_3}\right)+c_1$

$\Rightarrow \frac{-m_1\left(c_2-c_3\right)+c_1\left(m_2-m_3\right)}{m_2-m_3}=\frac{-m_2{c}_3-m_3{c}_2}{m_2-m_3}.$

m₁ (c₂ - c₃) - c₁ (m₂ - m₃) + m₂ c₃ - m₃ c₂ = 0

m₁ (c₂ - c₃) - m₂ c₁ + m₃ c₁ + m₂ c₃ - m₃ c₂ = 0

m₁ (c₂ - c₃) + m₂ (c₃ - c₁) + m₃ (c₁ - c₂) = 0

63. The given eqⁿ of the lines are.

3x + y - 2 = 0 _____ (1)

Px + 2y - 3 = 0 ______ (2)

2x - y - 3 = 0 _____ (3)

Point of intersection of (1) and (3) is given by,

(3x + y - 2) + (2x - y - 3) = 0

=> 5x - 5 = 0

=> x = $\frac{5}{5}$

=> x = 1

So, y = 2 - 3x = 2 -3 (1) = 2 - 3 = 1.

i e, (x, y) = (1, -1).

As the three lines interests at a single point, (1, -1) should line on line (2)

i e, P * 1 + 2 * (-1)- 3 = 0

P - 2 - 3 = 0

P = 5

29. Given, f(x)=|x – 1|.

The given function is defined for all real number x.

Hence, domain of f(x)=R.

As f(x)=|x – 1|, x $\in$ R is a non-negative no.

Range of f(x)=[0, ?), if positive real numbers.

28. Given, f (x)=

The given fxⁿ is valid for all x such that x – 1 ≥ 0 ⇒x≥ 1

∴ Domain of f (x)= [1,∞)

As x ≥ 1

⇒ x – 1 ≥ 1 – 1

⇒ x – 1 ≥ 0

⇒ ≥ 0

⇒ f (x) ≥ 0

So, range of f (x)= [0,∞ )

62. 

The given eqⁿ of the lines are

 y - x = 0 _____ (1)

x + y = 0 ______ (2)

x - k = 0 ______ (3)

The point of intersection of (1) and (2) is given by

(y - x) - (x + y) = 0

⇒ y - x -x -y = 0

y = 0 and x = 0

ie, (0, 0)

The point of intersection of (2) and (3) is given by

(x + y) – (x – k) = 0

y + k = 0

y = –k and x = k

i.e, (k, –k)

The point of intersection of (3) and (1) is given by

x = k

and y = k

ie, (k, k).

Hence area of triangle whose vertex are (0, 0), (k, –k)

and (k, k) is