A line L passes through points (1, 1) and (2, 0) and another line L' passes through left( {frac{

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9.Straight Line

medium

A line $L$ passes through the points $(1, 1)$ and $(2, 0)$ and another line $L'$ passes through $\left( {\frac{1}{2},0} \right)$ and perpendicular to $L$. Then the area of the triangle formed by the lines $L,L'$ and $y$- axis, is

$15\over8$

$25\over4$

$25\over8$

$25\over16$

Solution

(d) Here $L \equiv x + y = 2$and $L' \equiv 2x – 2y = 1$.

Equation of $y$-axis is $x = 0$

Hence the vertices of the triangle are $A(0,\,2),B\left( {0, – \frac{1}{2}} \right)$ and $C\left( {\frac{5}{4},\frac{3}{4}} \right)$. Therefore, the area of the triangle is $\frac{1}{2}\left| {\,\begin{array}{*{20}{c}}0&2&1\\0&{ – \frac{1}{2}}&1\\{\frac{5}{4}}&{\frac{3}{4}}&1\end{array}\,} \right| = \frac{{25}}{{16}}$.

Standard 11

Mathematics

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hard

(JEE MAIN-2019)

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