AP EAPCET 20-Aug-2021 Shift 2 Paper

Let ABCD be square, then AC be diagonal Equation of AC be

y−3

5−3

=

x−1

7−1

⇒y−3=

2(x−1)

6

⇒y−3=

x−1

3

⇒3y−9=x−
⇒13y=x+8 ...(i)
Slope of line AC=

1

3

Let the equation of side passes through A be y=mx+c ...(ii)
Since, angle between Eqs. (i) and (ii) is 45°
tan‌45°=|

1 3 −m

1+ m 3

|
⇒1=|

1−3m

3+m

|
i.e.

1−3m

3+m

=1 and

3m−1

3+m

=1
⇒1−3m=3+m
⇒m=

−1

2

⇒3m−1=3+m
⇒m=2
Equation of line become,
y=

−1

2

x+C and y=2x+C
Since, it passes through A(1,3) , we obtain
3=

−1

2

+C⇒C=

−7

2

3=2+C⇒C=1
∴ Equation of side become
y=

−1

2

x+

7

2

⇒2y+x−7=0
y=2x+1⇒2x−y+1=0