AP EAPCET 24 May 2025 Shift 1 Paper

Given, L1=ax−3y+5=0,
L2=4x−6y+8=0
To find x-intercept of line L1, set y=0, then
‌ax−3(0)+5‌=0
⇒‌x‌=−‌

5

a

So, x-intercept (p)=−‌

5

a

To find y-intercept (q), set x=0, then
‌a(0)−3y+5=0
⇒−3y=−5
So, q=y=‌

5

3

Now, to find y-intercept (n), set x=0
So, 4(0)−6y+8=0
⇒‌‌y=‌

−8

−6

So, n=y=‌

4

3

to find x-intercept (m), set y=0 in equation L2
‌4(x)−6(0)+8‌=0
⇒‌4x‌=−8
∴‌m‌=x=−2
Also, L1:ax−3y+5=0
⇒‌‌y=‌

a

3

x+‌

5

3

∴ Slope of L1=‌

a

3

And L2:4x−6y+8=0
⇒‌‌y=‌

4

6

x+‌

8

6

=‌

2

3

x+‌

4

3

∴ Slope of L2=‌

2

3

Since, L1 and L2 are parallel, so their slopes are equal.
∴‌‌

a

3

=‌

2

3

⇒a=2
Now, equation of the line passing through (p,q)=(‌

−5

2

,‌

5

3

)
And (m,n)=(−2,‌

4

3

) is
‌y−‌

4

3

=‌

‌ 4 3 − 5 3

−2−(‌ −5 2 )

(x−(−2))
⇒y−‌

4

3

=‌

−1∕3

1∕2

(x+2)
⇒‌

3y−4

3

=‌

−2

3

(x+2)
⇒3y−4=−2x−4
⇒3y+2x=0