The line $\frac{x}{a}+\frac{y}{b}=1$ moves in such a way that $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$, where $c$ is a constant. The locus of the foot of the perpendicular from the origin on the given line is $x^2+y^2=c^2$.
The lines $a x+2 y+1=0, \quad b x+3 y+1=0$ and $c x+4 y+1=0$ are concurrent, if $a, b$ and $c$ are in GP.
Line joining the points $(3,-4)$ and $(-2,6)$ is perpendicular to the line joining the points $(-3,6)$ and $(9,-18)$.
Match the following.
| Column I | Column II | ||
|---|---|---|---|
| (i) | The coordinates of the points $P$ and $Q$ on the line $x+5 y=13$ which are at a distance of 2 units from the line $12 x-5 y+26=0$ are | (a) | $(3,1),(-7,11)$ |
| (ii) | The coordinates of the point on the line $x+y=4$, which are at a unit distance from the line $4 x+3 y-10=0$ are | (b) | $\left(-\frac{1}{3}, \frac{11}{3}\right),\left(\frac{4}{3}, \frac{7}{3}\right)$ |
| (iii) | The coordinates of the point on the line joining $A(-2,5)$ and $B(3,1)$ such that $A P=P Q=Q B$ are | (c) | $\left(1, \frac{12}{5}\right),\left(-3, \frac{16}{5}\right)$ |
(i) Let the coordinate of point $P\left(x_1, y_1\right)$ on the line $x+5 y=13 i$.e.,
$$P\left(13-5 y_1, y_1\right)$$
$\therefore$ Distance of $P$ from the line $12 x-5 y+26=0$,
$$\begin{array}{l} & 2 =\left|\frac{12\left(13-5 y_1\right)-5 y_1+26}{\sqrt{144+25}}\right| \\ \Rightarrow \quad & 2 = \pm \frac{156-60 y_1-5 y_1+26}{13} \\ \Rightarrow \quad & -65 y_1 =-156 \quad \text{[taking positive sign]}\\ \Rightarrow \quad& y_1 =\frac{156}{65}=\frac{12}{5} \\ \because \quad & x_1=13-5 y_1 \\ & =13-12=1 \end{array}$$
So, the coordinate of is $P\left(1, \frac{12}{5}\right)$.
Similarly, the coordinates of $Q$ are $\left(-3, \frac{16}{5}\right)\quad \text{[taking negative sign]}$.
(ii) Let coordinates of the point on the line $x+y=4$ be $\left(4-y_1, y_1\right)$.
Distance from the line $4 x+3 y-10=0$.
$$\begin{aligned} & 1=\left|\frac{4\left(4-y_1\right)+3 y_1-10}{\sqrt{16+9}}\right| \\ \Rightarrow & 1= \pm \frac{16-4 y_1+3 y_1-10}{5} \quad\text{[taking negative sign]}\\ \Rightarrow & 5=6-y_1 \\ \Rightarrow & y_1=1 \end{aligned}$$
If $y_1=1$, then $x_1=3$
So, the point is $(3,1)$.
Similarly, taking negative sign the point is $(-7,11)$.
(iii) Given point $A(-2,5)$ and $B(3,1)$.
Now, the point $P$ divides line joining the point $A$ and $B$ in $1: 2$.
$$\begin{array}{ll} \because & x_1=\frac{1 \cdot 3+2(-2)}{1+2}=\frac{3-4}{3}=\frac{-1}{3} \\ \text { and } & y_1=\frac{1 \cdot 1+2 \cdot 5}{1+2}=\frac{11}{3} \end{array}$$
So, the coordinates of $P$ are $\left(\frac{-1}{3}, \frac{11}{3}\right)$.
Thus, the point $Q$ divided the line joining $A$ to $B$ in $2: 1$.
$$\begin{array}{ll} \because & x_2=\frac{2 \cdot 3+1(-2)}{2+1}=\frac{4}{3} \\ \text { and } & y_2=\frac{2 \cdot 1+1 \cdot 5}{2+1}=\frac{7}{3} \end{array}$$
Hence, the coordinates of $Q$ are $\left(\frac{4}{3}, \frac{7}{3}\right)$.
Hence, the correct matches are (i) $\rightarrow$ (c), (ii) $\rightarrow$ (a), (iii) $\rightarrow$ (b).
The value of the $\lambda$, if the lines $(2+3 y+4)+\lambda(6 x-y+12)=0$ are
| Column I | Column II | ||
|---|---|---|---|
| (i) | parallel to Y-axis is | (a) | $\lambda=-\frac{3}{4}$ |
| (ii) | perpendicular to $7x+y-4=0$ is | (b) | $\lambda=-\frac{1}{3}$ |
| (iii) | passes through $(1,2)$ is | (c)` | $\lambda=-\frac{17}{41}$ |
| (iv) | parallel to $x$-axis is | (d) | $\lambda=3$ |
(i) Given equation of the line is
$$(2 x+3 y+4)+\lambda(6 x-y+12)=0\quad \text{.... (i)}$$
If line is parallel to $Y$-axis i.e., it is perpendicular to $X$-axis
$$\therefore \quad \text { Slope }=m=\tan 90 \Upsilon=\infty$$
From line (i), $x(2+6 \lambda)+y(3-\lambda)+4+12 \lambda=0$
and slope $$=\frac{-(2+6 \lambda)}{3-\lambda}$$
$$\begin{array}{ll} \Rightarrow & \frac{-2-6 \lambda}{3-\lambda}=\infty \\ \Rightarrow & \frac{-2-6 \lambda}{3-\lambda}=\frac{1}{0} \Rightarrow \lambda=3 \end{array}$$
(ii) If the line (i) is perpendicular to the line $7 x+y-4=0$ or $y=-7 x+4$
$$\begin{aligned} \because \quad & \frac{-(2+6 \lambda)}{(3-\lambda)}(-7) =-1 \\ \Rightarrow \quad & 14+42 \lambda =-3+\lambda \\ \Rightarrow \quad & 41 \lambda =-17 \\ \Rightarrow \quad & \lambda =-\frac{17}{41} \end{aligned}$$
(iii) If the line (i) passes through the point $(1,2)$.
$$\begin{aligned} \text{Then, }\quad (2+6+4)+\lambda(6-2+12) & =0 \\ 12+16 \lambda=0 \Rightarrow \lambda & =-\frac{3}{4} \end{aligned}$$
(iv) If the line is parallel to $X$-axis the slope $=0$.
Then, $$\frac{-(2+6 \lambda)}{3-\lambda}=0$$
$$\Rightarrow \quad-(2+6 \lambda)=0 \Rightarrow \lambda=-\frac{1}{3}$$
So, the correct matches are (i) $\rightarrow$ (d), (ii) $\rightarrow$ (c), (iii) $\rightarrow$ (a), (iv) $\rightarrow$ (b).