Straight Lines

60. The given eqn of lines are

x - 7y + 5 = 0 ______ (1) ⇒ x = 7y - 5

and 3x + y = 0 _________ (2)

Solution (1) and (2) we get,

3 [7y – 5] + y = 0 .

⇒ 21y - 15 + y = 0

⇒ 22y = 15

Kindly go through the solution

58. Let (0, y) be the point on y-axis which is at a distance 4 unit from the line $\frac{x}{3}+\frac{y}{4}=1$

Then, the line $\frac{x}{3}+\frac{y}{4}=1$

4x + 3y = 12.

4x + 3y - 12 = 0

57. Let a and b be the x & y intercept. Then,

$\frac{x}{a}+\frac{y}{b}=1.$ _____ (1)

Given, a + b = 1. ______ (2) b = 1 -a _____ (3)

and ab = -6 _____ (4)

Putting eqn (B) in (iii) we get a

a (1- a) = - 6

a - a² = - 6

a² - a - 6 = 0

a² + 2a - 3a - 6 = 0

a (a + 2) - 3 (a + 2) = 0

(a + 2) (a -3) = 0

(a + 2) (a -3) = 0.

a = 3 or a = -2.

W

56. The given equation of the line is 3x + y + 2 = 0

55. We have (k - 3) x - (4 - k²) y + k² - 7 y + 6 = 0.

(i) When the line is parall to x-axis, all x coefficient = 0. then,

(k - 3)x - (4 -k²)y + k² - 7y + 6 = 0 x.x - a x y       where a = constant

Equating the co-efficient,

K – 3 = 0

=> k = 3

(ii) When the line is paral

54. The equation of line whose intercept on axes are a and b is given by,

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

Multiplying both sides by ab we get,

$\frac{abx}{a}+\frac{aby}{b}=ab$

$\Rightarrow bx+ay-ab=0.$

53. Let P be the point on the BC dropped from vertex A.

Slope of BC$=\mathrm{2\; -}$
$\frac{\mathrm{(-1)}}{1\; -4}$

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

$=\frac{3}{-3}$

= $-$ 1.

As A P $\perp$ BC,

Slope of AP= $\frac{-1}{\mathrm{slope\; of\; B}C}=\frac{-1}{-1}=1.$

Using slope-point form the equation of AP is,

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

$\Rightarrow$ x $-$ 2 = y $-$ 3

$\Rightarrow$ x – y – 2 + 3 = 0 $\Rightarrow$ x – y + 1 = 0

The equation of line segment through B(4, -1) and

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