The line $L$ given by $\frac{x}{5}+\frac{y}{b} \doteq 1$ passes through the point $(13,32)$. The line K is…

The line $L$ given by $\frac{x}{5}+\frac{y}{b} \doteq 1$ passes through the point $(13,32)$. The line K is parallel to line L and has the equation $\frac{x}{c}+\frac{y}{3}=1$. Then the distance between L and K is $\qquad$ units.
  1. $\frac{23}{15}$
  2. $\sqrt{17}$
  3. $\frac{17}{\sqrt{15}}$
  4. $\frac{23}{\sqrt{17}}$

Solution

Line L passes through ( 13,32 ). $\begin{array}{rr} \therefore \quad & \frac{13}{5}+\frac{32}{b}=1 \\ & \Rightarrow b=-20 \end{array}$
So, equation of $L$ is $\frac{x}{5}-\frac{y}{20}=1 \Rightarrow 4 x-y=20$ Slope of $L$ is $m_1=4$. Slope of $\frac{x}{\mathrm{c}}+\frac{y}{3}=1$ is $\mathrm{m}_2=\frac{3}{\mathrm{c}}$ $\begin{aligned} & \Rightarrow-\frac{3}{c}=4 \\ & \Rightarrow c=\frac{3}{4} \end{aligned}$
Equation of line K is $-\frac{4 x}{3}+\frac{y}{3}=1$ $\Rightarrow 4 x-y=-3$ Distance between $L$ and $K$ is $\left|\frac{20+3}{\sqrt{16+1}}\right|=\frac{23}{\sqrt{17}}$

Asked in: MHT CET 2024 (04 May Shift 1)

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