Class 11th

18.

The slope of the line is,  m = tan 75° – tan (45° + 30°)

17.

Given slope = m and $\left(x_0,y_0\right)=\left(0,0\right)$

By point slope from, equation of line is

$m=\frac{y-y_0}{x-x_0}$

$\Rightarrow m=\frac{y-0}{x-0}$

$\Rightarrow m=\frac{y}{x}$

$\Rightarrow mx=y$

$\Rightarrow$ $y=mx$

2.

Given,

$m=\frac{1}{2},\left(x_0,y_0\right)=\left(-4,3\right)$

By point slope form equation of line is

$m=\frac{y-y_0}{x-x_0}$

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

$\Rightarrow x+4=2\left(y-3\right)$

$\Rightarrow x+4=2y-6$

$\Rightarrow x-2y+4+6=0$

$\Rightarrow x-2y+10=0$

Exercise 9.2

15. 

For any points on x-axis the y-co-ordinate or the ordinate is always 0. Hence, the x-axis have the equation y = 0.

Similarly for xy points on y-axis the x-co-ordinate or the abscissa is always 0. Hence, the y-axis have the equation x = 0.

14.

From figure,

Slope of AB = $\frac{97-92}{1995-1985}$

$=\frac{5}{10}$

$=\frac{1}{2}$

As, A, B and C lie on same line

Slope of AB = Slope of BC

$\Rightarrow \frac{1}{2}=\frac{a-97}{2010-1995}=\frac{a-97}{15}$

$\Rightarrow 15=2\left(a-97\right)$

$\Rightarrow 15=2a-194$

$\Rightarrow 194+15=2a$

$\Rightarrow 209=2a$

$\Rightarrow a=\frac{209}{2}=104.5$

Hence, population in 2010 will be 104.5 crores.

13.

Since the three points say P (h, o), Q (a, b) and R (o, k) lie on a line.

Slope of PQ = slope of QR

$\Rightarrow \frac{b-o}{a-h}=\frac{k-b}{o-a}$

$\Rightarrow \frac{b}{a-h}=\frac{k-b}{-a}$

$\Rightarrow -ab=\left(k-b\right)\left(a-h\right)$

$\Rightarrow -ab=ka-kh-ab+bh$

$\Rightarrow ka+bh=kh$

Dividing both sides by kh we get,

$\Rightarrow \frac{ka}{kh}+\frac{bh}{kh}=\frac{kh}{kh}$

$\Rightarrow \frac{a}{h}+\frac{b}{k}=1$

12.

Let the line l passes through points (x₁, y₁) and (h, x)

Then slope of l,

$m=\frac{k-y_1}{h-x_1}$

$\Rightarrow \left(h-x_1\right)m=k-y_1$

Hence, proved

11.

Let the two slopes be m and 2m. And if θ is the angle between the two lines.

$ta\mathrm{n\theta }=\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$\Rightarrow \frac{1}{3}=\left|\frac{2m-m}{1+2m..m}\right|=\left|\frac{m}{1+2m^2}\right|$

$\Rightarrow \frac{m}{1+2m^2}=\pm \frac{1}{3}($$\left|x\right|=\pm x)$

Case I. When $\frac{m}{1+2m}=\frac{1}{3}$

$\Rightarrow 3m=2m^2+1$

$\Rightarrow 2m^2-3m+1=0$

$\Rightarrow 2m^2-2m-m+1=0$

$\Rightarrow 2m\left(m-1\right)-\left(m-1\right)=0$

$\Rightarrow \left(m-1\right)\left(2m-1\right)-0$

$\Rightarrow m=1m=\frac{1}{2}$

Case II. When $\frac{m}{1+2m^2}=-\frac{1}{3}$

$\Rightarrow 3m=-\left(1+2m^2\right)$

$\Rightarrow 3m=-1-2m^2$

$\Rightarrow 2m^2+3m+1=0$

$\Rightarrow 2m^2+2m+m+1=0$

$\Rightarrow 2m\left(m+1\right)+\left(m+1\right)=0$

$\Rightarrow \left(m+1\right)\left(2m+1\right)=0$

$\Rightarrow m=-1m=-\frac{1}{2}$

Hence, $m=1,\frac{1}{2},-1m=-\frac{1}{2}$

$\therefore$ The possible slopes of the two lines (m, 2m) are (1, 2), $\left(\frac{1}{2},1\right),\left(-1,-2\right)$ and $\left(-\frac{1}{2},-1\right)$

10.

 Slope of line joining points A (3, –1) and B (4, –2) is

$m=\frac{-3-\left(-1\right)}{4-3}=\frac{-2+1}{1}=-1$

As slope of AB is the angle made by the line-segment AB w.r.t. x-axis, the angle between them is $$ and is given by $ta\mathrm{n\Theta }=m$

$\Rightarrow ta\mathrm{n\Theta }=-1\Rightarrow ta\mathrm{n\Theta }=-tan45°=tan\left(180°-45°\right)$

$\Rightarrow tan=tan135°$

$\Rightarrow =135°$

9.

Let A (–2, –1), B (4, 0), C (3, 3) and D (–3, 2) be the given points.

Slope of DC = $\frac{3-2}{3-\left(-3\right)}=\frac{1}{3+3}=\frac{1}{6}$

As slope of AB = slope of DC

We conclude that AB | | DC.

Similarly, slope of BC = $\frac{3-0}{3-4}=\frac{3}{-1}=-3$

Slope of AD = $\frac{2-\left(-1\right)}{-3-\left(-2\right)}=\frac{2+1}{-3+2}=\frac{3}{-1}=-3$

As slope of BC = slope of AD we conclude that BC | | AD.

Hence, as the pair of opposite sides of ABCD are parallel we can conclude that the given points are the vertices of a parallelogram.