KCET 2015 Math Solved Paper

Let points, A(0, 14), B(5, 0) and C(19, 14) Now, equation of side AB is ‌

x

5

+‌

y

14

=1
⇒14x+5y=70→ (1)
Equation of side AC is
y=14→ (2)
and equation of side BC is
x−y=5 → (3)
Therefore, the shaded region shown in figure is given by the inequation
14x+5y>70y<14 and x−y<5