TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

Given line
3x+4y=5 and 4x−3y=15
Slope of line 3x+4y=5 is
i.e., ‌‌m1=‌

−3

4

Slope of line 4x−3y=15
i.e., ‌‌m2=‌

4

3

Let the slope of line BC=m
Given, AB=AC
∵‌‌tan‌α=tan‌β
‌

m−‌ 4 3

1+‌ 4m 3

=‌

− 3 4 −m

1−‌ 3m 4

‌

3m−4

3+4m

=‌

−3−4m

4−3m

12m2−9m2−16+12m=−9−12m−12m−16m2
⇒‌‌7m2+48m−7=0
&(7m−1)(m+7)=0
&m=‌

1

7

,m=−7
Equation of line passing through (1,2) and slope ‌

1

7

is,
=y−2=‌

1

7

(x−1)=x−7y+13=0
Equation of line passing through (1,2) and slope =−7 is
y−2‌‌=−7(x−1)
7x+y‌‌=9