Straight Lines

52. The given equation lines are.

line 1: xcosθ-y sin θcos 2θ

⇒ xcosθ-y sin θ - kcos 2θ = 0

The perpendicular distance from origin (0,0) to line 1 is

51. 

Let 0 (o, o) be the origin and P (-1, 2) be the given point on the line y = mx + c.

Then, slope of OP, = $=\frac{2-0}{-1-0}$

Slope of OP = -2

As the line y = mx + c is ⊥ to OP we can write

49.

Let P(3, 9) and Q (-1, 2) be the point. Let M bisects PQ at M so,

Co-ordinate of M = $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

$=\left(\frac{3+(-1)}{2},\frac{4+2}{2}\right)$

$=\left(\frac{3-1}{2},\frac{6}{2}\right)=\left(\frac{2}{2},\frac{6}{2}\right)=(1,3)$

Now, slope of AB, m = $\frac{y_2-y_1}{x_2-x_1}=\frac{2-4}{-1-3}=\frac{-2}{-4}=\frac{1}{2}.$

As the bisects AB perpendicular it has slope $-\frac{1}{m}$ and it passes through M(1, 3) it has the equation of the form,

$\frac{-1}{m}=\frac{y-y{}_{\mathrm{\sigma }}}{x-x{}_0}$

$\frac{-1}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{y-3}{x-1}$

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

$\Rightarrow -2(x-1)=y-3$

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

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

2x + y - 5 = 0

48.

Given, slope of line 1, m₁ = 2.

Let m₂ be the slope of line 2.

If θ is the angle between the two lines then we can write,

47. 

The slope of line 4x + by + c = 0 is

                                    m = -A/B

As the required line is parallel to the line Ax + by + c = 0

They have the same slope ie, m = -A/B

So, equation of line with slope m and passing through (x₁, y₁)

is given by point-slope from as,

⇒ - A (x-x₁) B (y -y₁)

⇒ A (x-x₁) + B (y-y₁)= 0.

Hence proved.

46. Slope of line 1 (passing) through (h. 3) and (4, 1) is

Kindly go through the solution

44. 

The slope of line x- 7y + 5 = 0 is

So, equation of line with slope m₂ and having x-intercept 3 is

y = m (x-d), d = x- intercept

⇒y = - 7 (x- 3)

⇒y = - 7x + 21

⇒ 7x + y- 21 = 0.

43. 

The slope of the line 3x - 4y + 2 = 0 is,

$m=-\frac{}{}$ \frac{3}{4}$$
$=\; -\frac{3}{-4}=\frac{3}{4}$

So, slope of line parallel to 3x - 4y + 2 = 0 is also m= $\frac{3}{4}$ .

Hence, this parallel line passes through ( -2, 3) we can write,

$m=\frac{y-y_0}{x-x_0}$

$\frac{3}{4}=\frac{y-3}{x-(-2)}=\frac{y-3}{x+2}$

3 (x + 2) = 4 (y - 3)

3x + 6 = 4y - 12

3x - 4y + 6 + 12 = 0.

3x - 4y + 18 = 0. Which is the required equation of line.