Question

Area of the triangle formed by the lines $x=0,x+y=2$, $x-y=\frac{1}{2}$ is

• A. $\frac{25}{16}$
• B. $\frac{15}{8}$
• C. $\frac{3}{4}$
• D. $\frac{3}{8}$

Answer 1

The correct option is A $\frac{25}{16}$

Equation of sides of triangle :

$x=0$ ---(1)

$x+y=2$ ---(2)

$x-y=\frac{1}{2}$ ---(3)

From (2)& (3),

$x=\frac{5}{4},y=\frac{3}{4}$

From (1) & (2),

$x=0,y=2$

From (1) & (3),

$x=0,y=\frac{-1}{2}$

$\therefore$ three vertices of trainlge $(\frac{5}{4},\frac{3}{4}),(0,2),(0,\frac{-1}{2})$

Area of triangle $=\left|\frac{1}{2}\right[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\left]\right|$

$=\left|\frac{1}{2}\right[0+0+\frac{5}{4}(-\frac{1}{2}-2)\left]\right|$

$=|-\frac{5}{8}\times \frac{5}{2}|$

$=\left|\frac{-25}{16}\right|=\frac{25}{16}$