Class 11th

38. (i) Given, 3x + 2y 12 = 0.

3x + 2y = 12

Dividing both sides by 12 we get,

$\Rightarrow \frac{3x}{12}+\frac{2y}{12}=\frac{12}{12}$

$\Rightarrow \frac{x}{4}+\frac{y}{6}=1.$

Comparing the above equation with $\frac{x}{a}+\frac{y}{b}=1$ = we get, x-intercept, a = 4 and y-intercept b = 6.

(ii) Given, 4x - 3y = 6

Dividing the both sides by 6.

$\frac{4x}{6}-\frac{3y}{6}=\frac{6}{6}$

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

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

Comparing above equation by $\frac{x}{a}+\frac{y}{b}=1$ we get, x-intercept a = $\frac{3}{2}$ and y-intercept, b = -2

(iii) Given, 3y + 2 = 0.

3y = -2

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

As the equation of line is of form y = constant, it is parallel to x-axis and has no x-intercept.

y-intercept = -$\frac{2}{3}$

13. Given, A= {1,2,3,5}

B= {4,6,9}

R= { (x, y) : the difference of x & y is odd; x $\in$ A, y $\in$ B}.

= { (x, y):|x – y| is odd and x $\in$ A, y $\in$ B}

= { (1,4), (1,6), (2,9), (3,4), (3,6), (5,4), (5,6)}.

12. Given, R = { (x, y): y = x + 5, x is a natural number less than 4; x, y $\in$ N}

= { (x, y): y = x + 5; x, y $\in$ N and x < 4}.

= { (1,1+5), (2,2+5), (3,3+5)}

= { (1,6), (2,7), (3,8)}

So, domain of R = {1,2,3}

range of R = {6,7,8}

11. Given, A = {1,2,3, …, 14}

R = { (x, y): 3x – y = 0; x, y $\in$ A}

= { (x, y): 3x = y; x, y $\in$ A}.

= { (1,3), (2,6), (3,9), (4,12)}

Domain of R is the set of all the first elements of the ordered pairs in R

So, domain of R= {1,2,3,4}

Codomain of R is the whole set A.

So, codomain of R= {1,2,3, …, 14}

Range of R is the set of all the second elements of the ordered pains in R.

So, range of R= {3,6,9,12}

Exercise 9.3

37.  (i) Given, x + 7y = 0.

7y = -x

y = $\frac{-1}{7}$ x + 0.

Comparing the above equation with y = mx + c we get, slope, m = -$\frac{1}{7}$ and c = 0, y-intercept

(ii) Given, 6x + 3y - 5 = 0

3y = -6x + 5

y = -$\frac{6}{3}$ x + $\frac{5}{3}$ = -2x + $\frac{5}{3}$

Comparing the above equation with y = mx + c we get, slope, m = -2 and $c=\frac{5}{3}$ , y-intercept

(iii) Given, y = 0

y = 0xx + 0

Comparing the above equation with y = mx + c we get, Slope, m = 0 and c = 0, y-intercept.

10. Given, n (A * A)=9

n (A) *n (A) = 9.

n (A)² = 3².

n (A) = 3 .

And (–1,0), (0,1) $\in$ A * A i.e., A * A = { (x, y), x $\in$ A, y $\in$ B}

? A= {–1,0,1}

And A * A= {–1,0,1} * {–1,0,1}

= { (–1, –1), ( –1,0), ( –1,1), (0, –1), (0,0), (0,1), (1, –1), (1,0), (1,1)}

9. Given, n (A)=3

n (B)= 2

So, n (Ax B)=n (A).n (B)=3x 2=6

as (x, 1), (y, 2), (z, 1) ∈Ax B= { (x, y), x∈Aand y∈B}.

A= {x, y, z} and B= {1,2}

As n (A) = 3as n (B) = 2

36. 

Let the given points be A (3, 0), B (–2, –2) and C (8, 2). Then by two point form we can write equation of line passing point A (3, 0) and B (–2, –2) as

$y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$

$\Rightarrow y-0=\frac{-2-0}{-2-3}\left(x-3\right)$

$\Rightarrow y=\frac{-2}{-5}\left(x-3\right)$

$\Rightarrow 5y=2\left(x-3\right)$

$\Rightarrow 5y=2x-6$

$\Rightarrow 2x-5y-6=0\left(1\right)$

If the three points A, B and C are co-linear, C will also lieonm the line formed by AB or satisfies equation (1).

Hence, putting x = 8 and y = 2 we have

L.H.S. = 2 * 8 – 5 * 2 – 6

= 16 – 10 – 6

= 0 = R.H.S.

The given three points are collinear.

8. A Given, A= {1,2}

B= {3,4}

So, A* B= { (1,3), (1,4), (2,3), (2, 4)}

i.e., n (A *B)=4

A *B will have subset =2⁴=16.They are,

Φ, { (1,3)}, { (1,4)}, { (2, 3)}, { (2,4)}, { (1,3), (1,4)}, { (1,3), (2,3)},

{ (1,3), (2, 4)}, { (1,4), (2, 3)}, { (1,4), (2, 4)}, { (2,3), (2, 4)},

{ (1,3), (1,4), (2, 3)}, { (1,3), (1,4), (2,4)}, { (1,3), (2,3), (2, 4)}, { (1,4), (2,3), (2,4)},

and { (1,3), (1,4), (2,3), (2,4)}