AP EAMCET Engineering 22 Sep 2020 Shift 1 Paper

Equation of given lines
The point of intersection of line is same as the point of intersection of lines ‌

∂f

∂x

=0 and ‌

∂f

∂y

=0
∴‌‌‌

∂f

∂x

=6x−11y−7=0
and ‌‌‌

∂f

∂y

=−11x+20v+13=0
by cross multiplication method, we have
‌

x

−143+140

=‌

−y

78−77

=‌

1

120−121

⇒‌‌‌

x

3

=‌

y

1

=‌

1

1

∴ Point of intersection is (3,1).