A Line Cuts The x-axis At A (7, 0) And The y-axis At B(0, –5). A Variable LinePQ is Draw Perpendicular To AB cutting The x-axis At P and The y-axis At In θ. If AQ and BP intersect At R, The Locus Of R is

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Question

A line cuts the x-axis at (7, 0) and the y-axis at B(0, –5). A variable linePQ is draw perpendicular to AB cutting the x-axis at P and the y-axis at in θ. If AQ and BP intersect at R, the locus of R is

Solution

Correct option is

x2 + y2 – 7x + 5y = 0

Let (a, 0) and (0, b).  

   

                               

   

  

  

    

⇒ x (7 – x) = y (5 + y)   

⇒ x2 + y2 – 7x + 5y = 0  

Which is the locus of R (xy)

SIMILAR QUESTIONS

Q1

The area enclosed by 2|x| + 3|y≤ 6 is

Q2

Let O be the origin, A (1, 0) and B (0, 1) and P (xy) are points such thatxy > 0 and x + y < 1, then

Q3

If a line joining two points A (2, 0) and B (3, 1) is rotated about A in anticlockwise direction through and angle 15o, then equation of the line is the new position is

Q4

An equation of a line through the point (1, 2) whose distance from the point (3, 1) has the greatest value is

Q5

Let 0 < α < π/2 be a fixed angle. If P = (cos θ, sin θ) and Q = (cos (α – θ), sin (α – θ) then Q is obtained from P by

Q6

On the portion of the straight line x + y = 2 which is intercepted between the axes, a square is constructed, away from the origin, with this portion as one of its side. If p denotes the perpendicular distance of a side of this square from the origin, then the maximum value of p is 

Q7

The line x + y = 1 meets x-axis at A and y-axis at BP is the mid-point ofAB (Fig). P is the foot of the perpendicular from P to OAM1 is that from P1 to OPP2 is that from M1 to OAM2 is that from P2 to OPP3is that from M2 to OA and so on. If Pn denotes the nth foot of the perpendicular on OA from Mn – 1, then OPn =  

Q8

The line x + y = a, meets the axis of x and y at A and B respectively. A triangle AMN is inscribed in the triangle OABO being the origin, with right angle at NM and N lie respectively on OB and AB. If the area of the triangle AMN is 3/8 of the area of the triangle OAB, then AN/BN is equal to.

Q9

The point (4, 1) undergoes the following transformation successively.

(i) Reflection about the line y = x

(ii) Translation through a distance 2 units along the positive direction of x-axis.

(iii) Rotation through an anlge π/4 about the origin in the anticlockwise direction  

(iv) Reflection about x = 0

The final position of the given point is

 

Q10

Equation of a line which is parallel to the line common to the pair of lines given by 6x2 – xy – 12y2 = 0 and 15x2 + 14xy – 8y2 = 0 and the sum of whose intercepts on the axes is 7, is