Straight Lines

42. 

(i) Given, equation of lines are

15x + 8y- 34 = 0

15x + 8y + 31 = 0

So, c₁ = 34 and c₂ = 31, A = 15 and B = 8

Kindly go through the solution

40.

The given equation of the line is.

12 (x + 6) = 5 (y- 2)

⇒ 12x + 72 = 5y- 9

⇒ 12x- 5y + 72 + 9 = 0

⇒ 12x- 5y + 82 = 0

The perpendicular distance of point (-1, 1) from the line is given by

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

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.

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.

35. Equation of line with intercept form is

$\frac{x}{a}+\frac{y}{b}=1\left(1\right)$

As R (h, x) divides line segment joining point A (a, 0) and B (0, b) in the ratio 1 : 2 we can write,

$\left(h,k\right)=\left(\frac{1*0+2*a}{1+2},\frac{1*b+2*0}{1+2}\right)$

So,  $h=\frac{0+2a}{3}k=\frac{b+0}{3}$

$\Rightarrow \mathrm{3h}=2ab=3k$

$\Rightarrow a=\frac{3h}{2}b=3k$

Hence, putting value of a and b in equation (1) we get,

$\frac{x}{3h/2}+\frac{y}{\mathrm{3k}}=1$

$\Rightarrow \frac{x}{\mathrm{3h}}+\frac{y}{\mathrm{3k}}=1$

34.

Since P (a, b) is the mid-point of the line segment say AB with points A (0, y) and B (x, 0) we can write,

$\left(a,b\right)=\left(\frac{x+0}{2},\frac{y+0}{2}\right)$

$\Rightarrow a=\frac{x}{2}\text{?}b=\frac{y}{2}$

$\Rightarrow x=2a\text{?}y=2b$

So, the equation of line with x and y intercept 2a and 2b using intercept form is

$\frac{x}{\mathrm{2a}}+\frac{y}{\mathrm{2b}}=1$

$\Rightarrow \frac{x}{a}+\frac{y}{b}=2$

Hence, proved

33. Assuming the price per litre say P in x-axis and the corresponding demand say D in y-axis, we have two point (14, 980) and (16, 1220) in xy plane. Then the points (P, D) will satisfy the equation.

$\Rightarrow D-980=\frac{1220-980}{16-14}(P-14)$

$\Rightarrow D-980=\frac{240}{2}(P-14)$

$\Rightarrow D-120(P-14)+980$

$\Rightarrow D-12\mathrm{0P}-1680+980$

$\Rightarrow D=12\mathrm{0P}-700$

Which is the required relation

Where P = 17, we have

D = 120 * 17 – 700

D = 1340

Hence, the owner can sell 1340 litres of milk weekly at? 17/litre