AP EAMCET201921 Apr 2019Evening ShiftMathematicsStraight LinesActual
A straight line 4 x + y - 1 = 0 through the point A ( 2 , - 7 ) meets the line B C whose equation is 3 x - 4 y + 1 = 0 at the point B . Then the equation of the line A C such that A B = A C , is
Options
- A89 x - 52 y - 162 = 0
- B52 x + 89 y + 519 = 0
- C4 x - y - 15 = 0
- D4 x + 3 y + 13 = 0
Correct answer
B. 52 x + 89 y + 519 = 0
Step-by-step solution
The equation of line given as 4 x + y - 1 = 0 ,   3 x - 4 y + 1 = 0 . The slopes of line are m 1 = - 4 and m 2 = 3 4 . Triangle A B C is an isosceles; A B = A C . A B and A C both pass through points ( 2 , - 7 ) . The slope is, m 3 = - 4 - 3 4 1 + ( - 4 ) 3 4 = 19 8 The required equation of line A C is, y + 7 x - 2 = 3 4 - 19 8 1 + 3 4 × 19 8 After simplification, 89 y + 623 = - 52 x + 104 ⇒ 52 x + 89 y + 519 = 0