AP EAPCET 24-Aug-2021 Shift 1 Paper

In Δ‌ABC
Equation of AB, BC and CA are 2x+y=0,
x+Py=q and x−y=3 respectively.
and P(2,3) is orthocentre. Solving 2x+y=0 and x−y=3
y=−2x and x+2x=3
⇒3x=3
⇒x=1,y=−2
∴ A(1,−2)
Slope of AP=

3−(−2)

2−1

=5
Slope of BC=−

1

p

AP⊥BC
⇒5×(−

1

p

)=−1
p = 5
Now for vertex B
2x+y=0,y=−2x
x+5y=9
x−10x=q
⇒x=−

9

9

,y=

21

9

B is (−

q

9

,

2q

9

)
Slope of BP=

2q 9 −3

− q 9 −2

=

2q−27

−q−18

Slope of AC=1
BP⊥AC
⇒(

q−27

−q−18

)(l)=−1
⇒2q−27=q+18
2q−q=18+27
q=45
p+q=5+45=50