Let PQR be a right angled isosceles triangle right angled at P (2, 1). If the equation of the line QR is 2x + y = 3, then the equation representing the pair of lines PQ and PR is
3x2 – 3y2 + 8xy – 20x – 10y + 25 = 0
Let slope of PQ and PR be m and –1/m respectively. Since PQR is an isosceles triangle
[âˆµ slope of QR = –2]
⇒ m + 2 = ± (1 – 2m) ⇒ m = 3 or –1/3
So the equations of PQ and PR are
Thus, joint equation representing PQ and PR is
[3(x – 2) – (y – 1)] [(x – 2) + 3(y – 1)] = 0
⇒ 3(x – 2)2 – 3(y – 1)2 + 8(x – 2) (y – 1) = 0
⇒ 3x2 – 3y2 + 8xy – 20x – 10y + 25 = 0.
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
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
The line x + y = 1 meets x-axis at A and y-axis at B, P is the mid-point ofAB (Fig). P1 is the foot of the perpendicular from P to OA; M1 is that from P1 to OP; P2 is that from M1 to OA; M2 is that from P2 to OP; P3is that from M2 to OA and so on. If Pn denotes the nth foot of the perpendicular on OA from Mn – 1, then OPn =
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 OAB, O being the origin, with right angle at N. M 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.
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
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
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
If pairs of lines x2 + 2xy + ay2 = 0 and ax2 + 2xy + y2 = 0 have exactly one line in common then the joint equation of the other two lines is given by
If the lines joining the origin to the intersection of the line y = mx + 2 and the curve x2 + y2 = 1 are at right angles, then
If θ is an angle between the lines given by the equation 6x2 + 5xy – 4y2 + 7x + 13y – 3 = 0, then equation of the line passing through the point of intersection of these lines and making an angle θ with the positive x-axis is