Ncert Solutions Maths class 11th

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.

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

20. (i) Domain of the given relation = {2,5,8,11,14,17}

Since every element of the domain has one and only one image, the given relation is a fxⁿ.

So, domain = {2,5,8,11,14,17}

range = {1}

(ii) Domain of the given relation = {2,4,6,8,10,12,14}

Since every element of the domain has one and only one image, the given relation is a fxⁿ.

So, domain = {2, 4, 6, 8, 10, 12, 14}

range = {1,2,3,4,5,6,7}

(iii) Domain of the given relation = {1,2}

As element 1 has more than one image i.e., 3 and 5, the given relation is not a fxⁿ.

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.