Straight Lines

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

61. The given Eqⁿ of the line is $\frac{x}{4}+\frac{y}{6}$ = 1 ______ (1)

so, Slope of line = -$\frac{6}{4}=-\frac{3}{2}.$

The line ⊥ to line (1) say l₂ has

Slope of l₂ = $\frac{-1}{(-3/2)}=\frac{2}{3}.$

Let P (0, y) be the point of on y-axis where it is cut by the line (1)

Then,  $\frac{0}{4}+\frac{y}{6}=1$

y = 6

i.e, the point P has co-ordinate (0, 6)

Eqⁿ of line ⊥ to $\frac{x}{4}+\frac{y}{6}=1$ and cuts y-axis at P (0,6) is

y – 6 = $\frac{2}{3}$ (x – 0)

3y – 18 = 2x

2x – 3y + 18 = 0

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.

When a = 3, b = 1- a = 1 - 3 = - 2

When a = - 2, b = 1 - (-2) = 1 + 2 = 3

So, (a, b) = (3, -2) and (-2, 3)

Hence, eqⁿ (1) becomes,

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

2x – 3y = 6 and 2y - 3x = 6

Gives the read eqⁿ of lines

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 parallel to y-axis all y co-efficient = 0 then

- (4 -k)² = 0

=> – 4 + x² = 0

k² = 4

k = ± 2.

(iii) When the line pares through origin, (0, 0) need satisfy the given eqn then,

k² - 7k + 6 = 0

k² - k – 6k + 6 = 0

k (k- 1) - 6 (k - 1) = 0

(k = 1) (k - 6) = 0

k = 1 and k = 6

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 C(1, 2) is.

$y-(-1)=\frac{2-(-1)}{1-4}(x-4).$

$\Rightarrow y+1=\frac{2+1}{-3}(x-4)$

$\Rightarrow (y+1)=\frac{3}{-3}(x-4).$

$\Rightarrow y+1=-(x-4)$

$\Rightarrow xy+1=-x+4$

$\Rightarrow -x+y+1-4=0$

$\Rightarrow$ $x+y-3=0$

So, A=1, B=1 and C= $-$ 3.

Hence, length of AP=length of $\perp$ distance of A(2,3) from BC.