Co ordinate

1. (a) Find the equation of a straight line passing through the point $(-2,2)$$(-2,2)$(-2,2)(-2,2)$(-2,2)$ and slope -3 .

$\Rightarrow$$\Rightarrow$=>quad\Rightarrow \quad$\Rightarrow$ Here, passing point $({\mathrm{x}}_1,{\mathrm{y}}_1)=(-2,2)$$\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=(-2,2)$(x_(1),y_(1))=(-2,2)\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(-2,2)$({\mathrm{x}}_1,{\mathrm{y}}_1)=(-2,2)$

Slope $(m)=-3$$(m)=-3$(m)=-3(m)=-3$(m)=-3$

We know that;

Equation of line in point slope form is;

$$\begin{array}{ll}& y-y_1=m(x-x_1)\\ \text{or,}& y-2=-3(x+2)\\ \text{or,}& y-2=-3x-6\\ \therefore & 3x+y+4=0\text{is the required equation.}\end{array}$$$$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}y-y_1=m\left(x-x_1\right)\\ \text{ or, }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}y-2=-3(x+2)\\ \text{ or, }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}y-2=-3x-6\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3x+y+4=0\text{ is the required equation. }\end{array}$${:[,y-y_(1)=m(x-x_(1))],[" or, ",y-2=-3(x+2)],[" or, ",y-2=-3x-6],[:.quad,3x+y+4=0" is the required equation. "]:}\begin{array}{ll} & y-y_{1}=m\left(x-x_{1}\right) \\ \text { or, } & y-2=-3(x+2) \\ \text { or, } & y-2=-3 x-6 \\ \therefore \quad & 3 x+y+4=0 \text { is the required equation. } \end{array}$$\begin{array}{ll}& y-y_1=m(x-x_1)\\ \text{ or, }& y-2=-3(x+2)\\ \text{ or, }& y-2=-3x-6\\ \therefore & 3x+y+4=0\text{ is the required equation. }\end{array}$$

(b) Find the equation of a straight line passing through the point $(4,5)$$(4,5)$(4,5)(4,5)$(4,5)$ and slope $=7$$=7$=7=7$=7$.

$\Rightarrow$$\Rightarrow$=>quad\Rightarrow \quad$\Rightarrow$ Here, passing point $(x_1,y_1)=(4,5)$$\left(x_1,y_1\right)=(4,5)$(x_(1),y_(1))=(4,5)\left(x_{1}, y_{1}\right)=(4,5)$(x_1,y_1)=(4,5)$

Slope $(m)=7$$(m)=7$(m)=7(m)=7$(m)=7$

We know that;

Equation of line in point slope form is;

$$\begin{array}{ll}& y-y_1=m(x-x_1)\\ \text{or,}& y-5=7(x-4)\\ \text{or,}& y-5=7x-28\\ \therefore & 7x-y-23=0\text{is the required equation.}\end{array}$$$$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}y-y_1=m\left(x-x_1\right)\\ \text{ or, }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}y-5=7(x-4)\\ \text{ or, }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}y-5=7x-28\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}7x-y-23=0\text{ is the required equation. }\end{array}$${:[,y-y_(1)=m(x-x_(1))],[" or, ",y-5=7(x-4)],[" or, "quad,y-5=7x-28],[:.quad,7x-y-23=0" is the required equation. "]:}\begin{array}{ll} & y-y_{1}=m\left(x-x_{1}\right) \\ \text { or, } & y-5=7(x-4) \\ \text { or, } \quad & y-5=7 x-28 \\ \therefore \quad & 7 x-y-23=0 \text { is the required equation. } \end{array}$$\begin{array}{ll}& y-y_1=m(x-x_1)\\ \text{ or, }& y-5=7(x-4)\\ \text{ or, }& y-5=7x-28\\ \therefore & 7x-y-23=0\text{ is the required equation. }\end{array}$$

2. (a) Find the equation of line passing through $(2,-3)$$(2,-3)$(2,-3)(2,-3)$(2,-3)$ and inclined at an angle of ${135}^{\circ }$${135}^{\circ }$135^(@)135^{\circ}${135}^{\circ }$ with the positive direction of $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$-axis.

$\Rightarrow$$\Rightarrow$=>quad\Rightarrow \quad$\Rightarrow$ Here, passing point $({\mathrm{x}}_1,{\mathrm{y}}_1)=(2,-3)$$\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=(2,-3)$(x_(1),y_(1))=(2,-3)\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(2,-3)$({\mathrm{x}}_1,{\mathrm{y}}_1)=(2,-3)$

Inclination with $+\mathrm{v}\mathrm{e}x$$+\mathrm{v}\mathrm{e}x$+vex+\mathrm{ve} x$+\mathrm{v}\mathrm{e}x$-axis $(\theta )={135}^{\circ }$$(\theta )={135}^{\circ }$(theta)=135^(@)(\theta)=135^{\circ}$(\theta )={135}^{\circ }$

$\therefore$$\therefore$:.quad\therefore \quad$\therefore$ slope $(\mathrm{m})=\tan \theta =\tan {135}^{\circ }=-1$$(\mathrm{m})=\tan \theta =\tan {135}^{\circ }=-1$(m)=tan theta=tan 135^(@)=-1(\mathrm{m})=\tan \theta=\tan 135^{\circ}=-1$(\mathrm{m})=\tan \theta =\tan {135}^{\circ }=-1$

We know that,

Equation of straight line passing through $({\mathrm{x}}_1,{\mathrm{y}}_1)$$\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$(x_(1),y_(1))\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$({\mathrm{x}}_1,{\mathrm{y}}_1)$ and having slope $\mathrm{m}$$\mathrm{m}$m\mathrm{m}$\mathrm{m}$ is;

$$\mathrm{y}-{\mathrm{y}}_1=\mathrm{m}(\mathrm{x}-{\mathrm{x}}_1)$$$$\mathrm{y}-{\mathrm{y}}_1=\mathrm{m}\left(\mathrm{x}-{\mathrm{x}}_1\right)$$y-y_(1)=m(x-x_(1))\mathrm{y}-\mathrm{y}_{1}=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_{1}\right)$$\mathrm{y}-{\mathrm{y}}_1=\mathrm{m}(\mathrm{x}-{\mathrm{x}}_1)$$

or, $y-(-3)=-1(x-2)$$y-(-3)=-1(x-2)$quad y-(-3)=-1(x-2)\quad y-(-3)=-1(x-2)$y-(-3)=-1(x-2)$

or, $y+3=-x+2$$y+3=-x+2$quad y+3=-x+2\quad y+3=-x+2$y+3=-x+2$

$\therefore \mathrm{x}+\mathrm{y}+1=0$$\therefore \mathrm{x}+\mathrm{y}+1=0$:.quadx+y+1=0\therefore \quad \mathrm{x}+\mathrm{y}+1=0$\therefore \mathrm{x}+\mathrm{y}+1=0$ is the required equation. (b)

Find the equation of line passing through $(5,7)$$(5,7)$(5,7)(5,7)$(5,7)$ and inclined at an angle of ${45}^{\circ }$${45}^{\circ }$45^(@)45^{\circ}${45}^{\circ }$ with the positive direction of $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$-axis.

$\Rightarrow$$\Rightarrow$=>quad\Rightarrow \quad$\Rightarrow$ Here, passing point $({\mathrm{x}}_1,{\mathrm{y}}_1)=(5,7)$$\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=(5,7)$(x_(1),y_(1))=(5,7)\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(5,7)$({\mathrm{x}}_1,{\mathrm{y}}_1)=(5,7)$

Inclination with +ve $\mathrm{x}$$\mathrm{x}$x\mathrm{x}$\mathrm{x}$-axis $(\theta )={45}^{\circ }$$(\theta )={45}^{\circ }$(theta)=45^(@)(\theta)=45^{\circ}$(\theta )={45}^{\circ }$

$\therefore$$\therefore$:.quad\therefore \quad$\therefore$ slope $(\mathrm{m})=\tan \theta =\tan {45}^{\circ }=1$$(\mathrm{m})=\tan \theta =\tan {45}^{\circ }=1$(m)=tan theta=tan 45^(@)=1(\mathrm{m})=\tan \theta=\tan 45^{\circ}=1$(\mathrm{m})=\tan \theta =\tan {45}^{\circ }=1$

We know that,

Equation of straight line passing through $({\mathrm{x}}_1,{\mathrm{y}}_1)$$\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$(x_(1),y_(1))\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$({\mathrm{x}}_1,{\mathrm{y}}_1)$ and having slope $\mathrm{m}$$\mathrm{m}$m\mathrm{m}$\mathrm{m}$ is;

$$\begin{array}{ll}& y-y_1=m(x-x_1)\\ \text{or,}& y-7=1(x-5)\\ \text{or,}& y-7=x-5\\ \therefore & x-y+2=0\text{is the required equation.}\end{array}$$$$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}y-y_1=m\left(x-x_1\right)\\ \text{ or, }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}y-7=1(x-5)\\ \text{ or, }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}y-7=x-5\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x-y+2=0\text{ is the required equation. }\end{array}$${:[,y-y_(1)=m(x-x_(1))],[" or, "quad,y-7=1(x-5)],[" or, "quad,y-7=x-5],[:.quad,x-y+2=0" is the required equation. "]:}\begin{array}{ll} & y-y_{1}=m\left(x-x_{1}\right) \\ \text { or, } \quad & y-7=1(x-5) \\ \text { or, } \quad & y-7=x-5 \\ \therefore \quad & x-y+2=0 \text { is the required equation. } \end{array}$$\begin{array}{ll}& y-y_1=m(x-x_1)\\ \text{ or, }& y-7=1(x-5)\\ \text{ or, }& y-7=x-5\\ \therefore & x-y+2=0\text{ is the required equation. }\end{array}$$

(c) Find the equation of line passing through $(2,2\sqrt{3})$$(2,2\sqrt{3})$(2,2sqrt3)(2,2 \sqrt{3})$(2,2\sqrt{3})$ and inclined at angle of ${30}^{\circ }$${30}^{\circ }$30^(@)30^{\circ}${30}^{\circ }$ with the positive direction of $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$-axis.

$\Rightarrow$$\Rightarrow$=>quad\Rightarrow \quad$\Rightarrow$ Here, passing point $({\mathrm{x}}_1,{\mathrm{y}}_1)=(2,2\sqrt{3})$$\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=(2,2\sqrt{3})$(x_(1),y_(1))=(2,2sqrt3)\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(2,2 \sqrt{3})$({\mathrm{x}}_1,{\mathrm{y}}_1)=(2,2\sqrt{3})$

Inclination with +ve $\mathrm{x}$$\mathrm{x}$x\mathrm{x}$\mathrm{x}$-axis $(\theta )={30}^{\circ }$$(\theta )={30}^{\circ }$(theta)=30^(@)(\theta)=30^{\circ}$(\theta )={30}^{\circ }$

$\therefore$$\therefore$:.quad\therefore \quad$\therefore$ slope $(\mathrm{m})=\tan \theta =\tan {30}^{\circ }=\frac{1}{\sqrt{3}}$$(\mathrm{m})=\tan \theta =\tan {30}^{\circ }=\frac{1}{\sqrt{3}}$(m)=tan theta=tan 30^(@)=(1)/(sqrt3)(\mathrm{m})=\tan \theta=\tan 30^{\circ}=\frac{1}{\sqrt{3}}$(\mathrm{m})=\tan \theta =\tan {30}^{\circ }=\frac{1}{\sqrt{3}}$

We know that,

Equation of straight line passing through $({\mathrm{x}}_1,{\mathrm{y}}_1)$$\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$(x_(1),y_(1))\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$({\mathrm{x}}_1,{\mathrm{y}}_1)$ and having slope $\mathrm{m}$$\mathrm{m}$m\mathrm{m}$\mathrm{m}$ is;

$$\mathrm{y}-{\mathrm{y}}_1=\mathrm{m}(\mathrm{x}-{\mathrm{x}}_1)$$$$\mathrm{y}-{\mathrm{y}}_1=\mathrm{m}\left(\mathrm{x}-{\mathrm{x}}_1\right)$$y-y_(1)=m(x-x_(1))\mathrm{y}-\mathrm{y}_{1}=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_{1}\right)$$\mathrm{y}-{\mathrm{y}}_1=\mathrm{m}(\mathrm{x}-{\mathrm{x}}_1)$$

or, $y-2\sqrt{3}=\frac{1}{\sqrt{3}}(x-2)$$y-2\sqrt{3}=\frac{1}{\sqrt{3}}(x-2)$quad y-2sqrt3=(1)/(sqrt3)(x-2)\quad y-2 \sqrt{3}=\frac{1}{\sqrt{3}}(x-2)$y-2\sqrt{3}=\frac{1}{\sqrt{3}}(x-2)$

or, $\sqrt{3}\mathrm{y}-2\times 3=\mathrm{x}-2$$\sqrt{3}\mathrm{y}-2\times 3=\mathrm{x}-2$quadsqrt3y-2xx3=x-2\quad \sqrt{3} \mathrm{y}-2 \times 3=\mathrm{x}-2$\sqrt{3}\mathrm{y}-2\times 3=\mathrm{x}-2$

$\therefore x-\sqrt{3}y+4=0$$\therefore x-\sqrt{3}y+4=0$:.quad x-sqrt3y+4=0\therefore \quad x-\sqrt{3} y+4=0$\therefore x-\sqrt{3}y+4=0$ is the required equation.

(d) Find the equation of the line passing through the point $(1,2)$$(1,2)$(1,2)(1,2)$(1,2)$ and making an angle of ${\tan }^{-1}\left(\frac{1}{3}\right)$${\tan }^{-1}\left(\frac{1}{3}\right)$tan^(-1)((1)/(3))\tan ^{-1}\left(\frac{1}{3}\right)${\tan }^{-1}\left(\frac{1}{3}\right)$ with the $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$-axis.

$\Rightarrow$$\Rightarrow$=>quad\Rightarrow \quad$\Rightarrow$ Here, passing point $({\mathrm{x}}_1,{\mathrm{y}}_1)=(1,2)$$\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=(1,2)$(x_(1),y_(1))=(1,2)\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(1,2)$({\mathrm{x}}_1,{\mathrm{y}}_1)=(1,2)$

Inclination with $+\mathrm{v}\mathrm{e}x$$+\mathrm{v}\mathrm{e}x$+vex+\mathrm{ve} x$+\mathrm{v}\mathrm{e}x$-axis $(\theta )={\tan }^{-1}\left(\frac{1}{3}\right)$$(\theta )={\tan }^{-1}\left(\frac{1}{3}\right)$(theta)=tan^(-1)((1)/(3))(\theta)=\tan ^{-1}\left(\frac{1}{3}\right)$(\theta )={\tan }^{-1}\left(\frac{1}{3}\right)$

or, $\tan \theta =\frac{1}{3}$$\tan \theta =\frac{1}{3}$quad tan theta=(1)/(3)\quad \tan \theta=\frac{1}{3}$\tan \theta =\frac{1}{3}$

$\therefore$$\therefore$:.quad\therefore \quad$\therefore$ slope $(\mathrm{m})=\frac{1}{3}$$(\mathrm{m})=\frac{1}{3}$(m)=(1)/(3)(\mathrm{m})=\frac{1}{3}$(\mathrm{m})=\frac{1}{3}$

We know that,

Equation of straight line passing through $({\mathrm{x}}_1,{\mathrm{y}}_1)$$\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$(x_(1),y_(1))\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$({\mathrm{x}}_1,{\mathrm{y}}_1)$ and having slope $\mathrm{m}$$\mathrm{m}$m\mathrm{m}$\mathrm{m}$ is;

$$y-y_1=m(x-x_1)$$$$y-y_1=m\left(x-x_1\right)$$y-y_(1)=m(x-x_(1))y-y_{1}=m\left(x-x_{1}\right)$$y-y_1=m(x-x_1)$$

or, $y-2=\frac{1}{3}(x-1)$$y-2=\frac{1}{3}(x-1)$quad y-2=(1)/(3)(x-1)\quad y-2=\frac{1}{3}(x-1)$y-2=\frac{1}{3}(x-1)$

or, $3y-6=x-1$$3y-6=x-1$quad3y-6=x-1\quad 3 y-6=x-1$3y-6=x-1$

$\therefore \mathrm{x}-3\mathrm{y}+5=0$$\therefore \mathrm{x}-3\mathrm{y}+5=0$:.quadx-3y+5=0\therefore \quad \mathrm{x}-3 \mathrm{y}+5=0$\therefore \mathrm{x}-3\mathrm{y}+5=0$ is the required equation.