AP EAMCET Engineering 20 Apr 2019 Shift 1 Paper

The point B(x1,y1) is the reflection of point A with respect to the line x−y+5=0 so,
‌

x1−1

1

=‌

y1+2

−1

‌‌=−2(‌

1+2+5

2

)
‌‌=−8
x1‌‌=−7
And,
y1=6
So, point is B(−7,6)
Similarly, for point C(x2,y2) is reflection of A w.r.t line x+2y=0 so,
‌

x2−1

1

‌‌=‌

y2+2

2

‌‌=−2(‌

1−4

5

)
‌‌=‌

6

5

x2‌‌=‌

11

5

And
y2=‌

2

5

So, point C is (‌

11

5

,‌

2

5

)
So equation of line BC is,
y−6=‌‌‌

6−‌ 2 5

−7−‌ 11 5

(x+7)
y−6=‌‌‌

28

−46

(x+7)
=‌‌‌

−14

23

(x+7)
Further simplify the above,
14x+23y‌‌=138−98
14x+23y−40‌‌=0