Question ID: #1161
Consider the lines $x(3\lambda+1)+y(7\lambda+2)=17\lambda+5,$ $\lambda$ being a parameter, all passing through a point $P$. One of these lines (say $L$) is farthest from the origin. If the distance of $L$ from the point $(3, 6)$ is $d$, then the value of $d^{2}$ is
- (1) 20
- (2) 30
- (3) 10
- (4) 15
Solution:
$$L_1 + \lambda L_2 = 0$$
$$x(3\lambda+1)+y(7\lambda+2) – (17\lambda+5) = 0$$
$$(x+2y-5) + \lambda(3x+7y-17) = 0$$
$$x + 2y – 5 = 0$$
$$3x + 7y – 17 = 0$$
$$3(5 – 2y) + 7y – 17 = 0$$
$$15 – 6y + 7y – 17 = 0$$
$$y = 2$$
$$x + 2(2) – 5 = 0 \Rightarrow x = 1$$
$$P(1, 2)$$
The line $L$ farthest from the origin and passing through $P$ will be perpendicular to the line segment joining the origin $O(0,0)$ and $P(1,2)$.
$$m_{OP} = \frac{y_2 – y_1}{x_2 – x_1}$$
$$m_{OP} = \frac{2 – 0}{1 – 0} = 2$$
$$m_L = -\frac{1}{m_{OP}} = -\frac{1}{2}$$
$$y – y_1 = m_L(x – x_1)$$
$$y – 2 = -\frac{1}{2}(x – 1)$$
$$2y – 4 = -x + 1$$
$$x + 2y – 5 = 0$$
$$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$
$$d = \frac{|1(3) + 2(6) – 5|}{\sqrt{1^2 + 2^2}}$$
$$d = \frac{|3 + 12 – 5|}{\sqrt{5}}$$
$$d = \frac{10}{\sqrt{5}} = 2\sqrt{5}$$
$$d^2 = (2\sqrt{5})^2 = 20$$
Ans. (1)
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