Question

Through the point $(3,4)$ are drawn two straight lines each inclined at ${45}^{\circ }$ to the straight line $x-y=2$. Find their equations and also find the area of triangle bounded by three lines.

Answer 1

Given point: $(3,4)$

Given line: $x-y=2$

i.e., $y=x-2$ .........$\left(A1\right)$

Solution:

Plotting the given point & line

we have to find the equations of line A & B

The angle line A makes with x axis $=45+a^o$

$\therefore \theta a=45+a^o$

but $a={45}^o$- angle of inclination given in quetsion

$\therefore \theta a=45+45={90}^o$

The angle line B makes with x axis$={45}^o-b^o$

$\therefore \theta b=45-45$

$\therefore \theta b=0$

$\therefore$ Slope of line $A\left(m_A\right)={\tan }^{-1}\left(\theta a\right)$

$={\tan }^{-1}\left(90\right)=\infty$ ........(i)

Slope of line $B\left(m_b\right)={\tan }^{-1}\left(\theta b\right)$

$={\tan }^{-1}\left(0\right)$ ..........(ii)

$=0$

Equation of a line in slope point form is given as

$(y-y_1)=m(x-x_1)$

$\therefore$ equation of line A is

$(y-4)=m_A(x-3)$

$\therefore (x-3)=\frac{y-4}{m_A}$

$\therefore x-3=\frac{y-4}{\infty }$ by (i)

$\therefore x-3=0$

$\therefore x=3$

$\therefore$ equation of line A: $x=3$

i.e., $x-3=0$ ........(ii)

Equation of line B is

$(y-4)=m_B(x-3)$

$\therefore (y-4)=0(x-3)$ by (ii)

$\therefore$ equation of line B: $y-4=0$

or $y=4$ ......(iv)

Equation of line A: $x-3=0$

Equation of line B: $y-4=0$

replotting the graph

Solving equation (iv) and $A1$

we get point $p=6,4$

$\therefore$ Area of the triangle $=\frac{1}{2}\times$ base$\times$ height

$=\frac{1}{2}\times (4-1)\times (6-3)$

$=\frac{1}{2}\times 3\times 3=\frac{9}{2}$

$\therefore$ The area of the triangle$=4.5$ sq. units.