Given the family of lines, a (3 x +4 y +6)+ b ( x + y +2)=0. The line of the family situated at the greatest distance from the point P (2,3) has equation -

Question

Given the family of lines, a (3 x +4 y +6)+ b ( x + y +2)=0. The line of the family situated at the greatest distance from the point P (2,3) has equation - (A) 4 x+3 y+8=0 (B) 5 x+3 y+10=0 (C) 15 x+8 y+30=0 (D) none

Tardigrade Admin · Accepted Answer

L 1: 3 x +4 y +6=0 L 2: x + y +2=0 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/289d861fff74d45fa650704dc9dd90b5-.png" /> Line situated at greatest distance from P (2,3) is the line passing through point of intersection (Q) of the given lines and perpendicular to PQ.

Q. Given the family of lines, $a (3 x + 4 y + 6) + b (x + y + 2) = 0$ . The line of the family situated at the greatest distance from the point $P (2, 3)$ has equation -

$4 x + 3 y + 8 = 0$

$5 x + 3 y + 10 = 0$

$15 x + 8 y + 30 = 0$

none

$L_{1} : 3 x + 4 y + 6 = 0$
$L_{2} : x + y + 2 = 0$

Line situated at greatest distance from $P (2, 3)$ is the line passing through point of intersection (Q) of the given lines and perpendicular to $PQ$ .

Q. Given the family of lines, a(3x+4y+6)+b(x+y+2)=0. The line of the family situated at the greatest distance from the point P(2,3) has equation -

4x+3y+8=0

5x+3y+10=0

15x+8y+30=0