AP EAPCET 18 May 2024 Shift 1 Solved Paper

We have the following equations representing the sides of a square:
‌3x+y−4=0
‌x−αy+10=0
‌βx+2y+4=0
‌3x+y+k=0
These lines form pairs of parallel lines: equations (i) with (iv) and (ii) with (iii).
Equations (i) and (ii) are perpendicular, so their slopes must multiply to -1 :
(m1×m2)=−1
From equation (i), the slope m1=−3. From equation (ii), the slope m2=‌

1

α

. Thus:
‌(−3)×‌

1

α

=−1
‌⇒α=3
Similarly, equations (iii) and (iv) are perpendicular:
‌(‌

−β

2

)×(−3)=−1
‌⇒β=−‌

2

3

Substituting α and β back into the equations, we have:
From equation (ii), x−3y+10=0.
From equation (iii), −‌

2

3

x+2y+4=0, simplifying gives:

x−3y−6=0
Since the distances between the parallel sides must be equal for the square, we equate:
‌

|k+4|

√32+12

=‌

|10+6|

√12+(−3)2

Which simplifies to:
|k+4|=16
Thus, the expression αβ(k+4)2 becomes:
3×(−‌

2

3

)(16)2=−512