Given the family of lines, a (3x +4y +6) + b (x +y +2) =0 . The line of the family…
Mathematics · JEE Advanced · NTA Exams — Co-ordinate Geometry
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
- none
Answer
(A) 4x + 3y + 8 = 0
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- Let OX and OY be two fixed lines inclined at a constant angle α. A variable line cuts OX at P and OY at Q…
- A tangent to the parabola x 2 + 4ay = 0 cuts the parabola x 2 = 4by at A and B the locus of the mid point of…
- If the circle x 2 + y 2 + 2x + 2ky + 6 = 0 and x 2 + y 2 + 2ky + k = 0 intersect orthogonally, then k is :
- The length of the side of an equilateral triangle inscribed in the parabola, y 2 = 4x so that one of its…
- A line meets the co-ordinate axes in A & B. A circle is circumscribed about the triangle OAB. If d 1 and d 2…
- If the circle C 1 : x 2 + y 2 = 16 intersects another circle C 2 of radius 5 in such a manner that the common…
- Let OX and OY be two fixed lines inclined at a constant angle α. A variable line cuts OX at P and OY at Q…
- If the chord through the points whose eccentric angles are θ & φ on the ellipse, x^ 2 a^ 2 + y^ 2 b^ 2 =1…
More Co-ordinate Geometry questions · Browse all practice questions