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 the three lines.

Answer 1

Equation of the line is $y=x-2$

Slope of the given line is $m_1=1$

$\Rightarrow$ Given line is inclined to the x-axis at an angle of $45degrees$

So the inclination of the other two lines with the x-axis will be 0 degrees( 45-45) and 90 degrees $(45+45)$

Now, since the lines pass through the point $(3,4)$

So, the equation of other two lines by one point formula is $x=3$ and $y=4$

Now,

Line $y=x-2$ and $x=3$ will intersect at $P_1(3,1)$
Line $y=x-2$ and $y=4$ will intersect at $P_2(6,4)$
Line $x=3$ and $y=4$ will intersect at $P_3(3,4)$
Hence,the coordinates of the three vertices of the triangle are $P_1,P_2,P_3$

So,area of the bounded triangle=$\left|\frac{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}{2}\right|$

$\Rightarrow \left|\frac{3(4-4)+6(4-1)+3(1-4)}{2}\right|$

$\Rightarrow \left|\frac{18-9}{2}\right|$

$\Rightarrow \frac{9}{2}$