Question

$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \left(a\right)& \text{If}\frac{a_1}{a_2}=\frac{b_1}{b_2}\text{then}(a_{1}x+b_{1}y+c_1)(a_{2}x+b_{2}y+c_2)& \left(p\right)& \text{a parabola}\\ & +k=0,(k\ne 0)\text{represents}& & \\ \left(b\right)& \text{If}\frac{a_1}{a_2}\ne \frac{b_1}{b_2}\text{then}(a_{1}x+b_{1}y+c_1)(a_{2}x+b_{2}y+c_2)& \left(q\right)& \text{a pair of lines}\\ & +k=0,(k\ne 0)\text{represents}& & \\ \left(c\right)& \text{Locus of a point moving such that its distances from}& \left(r\right)& \text{a straight line}\\ & \text{the point (-13, 7) and the line 17x + 29y + 18 = 0}& & \\ & \text{are always equal}& & \\ \left(d\right)& \text{Locus of a point moving such that the ratio of its}& \left(s\right)& \text{a hyperbola}\\ & \text{distances from the point (3, 11) and the line}& & \\ & \text{14x - 5y + 13 = 0 is always 2}& & \end{array}$

• A. (A-q), (B-s), (C-r), (D-r)
• B. (A-r), (B-q), (C-p), (D-s)
• C. (A-r), (B-s), (C-p), (D-q)
• D. (A-q), (B-q), (C-p), (D-s)

Answer 1

The correct option is A

(A-q), (B-s), (C-r), (D-r)

(a) If $\frac{a_1}{a_2}=\frac{b_1}{b_2},(a_{1}x+b_{1}y+c_1)(a_{2}x+b_{2}y+c_2)=0$ represents a pair of parallel lines so $\Delta =0$ and $h^2=ab$. Now in $(a_{1}x+b_{1}y+c_1)(a_{2}x+b_{2}y+c_2)+k=0(k\ne 0),{\Delta }^{\prime }=0+abk-k{h}^2=k(ab-h^2)=0$, so it is again a pair of lines.

(b) If $\frac{a_1}{a_2}\ne \frac{b_1}{b_2},h^2>ab,{\Delta }^{\prime }\ne 0$, so it is a hyperbola

(c) The point is on the line. So the locus is a straight line perpendicular to the given line through the given point.

(d) The point is on the line. So the locus is a pair of lines making angle $si{n}^{-1}\left(\frac{1}{2}\right)$ i.e. ${30}^{\circ }$ with the given lines and passing through the given point.