Two lines are drawn through (3, 4) , each of which makes angle of 45° with line x

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9.Straight Line

medium

Two lines are drawn through $(3, 4)$, each of which makes angle of $45^\circ$ with the line $x - y = 2$, then area of the triangle formed by these lines is

$9$

$9\over2$

$2$

$2\over9$

Solution

(b) The equation of lines are $y – {y_1} = \frac{{m \pm \tan \alpha }}{{1 \mp m\tan \alpha }}(x – {x_1})$

==> $y – 4 = \frac{{1 \pm \tan 45^\circ }}{{1 \mp \tan 45^\circ }}(x – {x_1})$

==> $y – 4 = \frac{{1 \pm 1}}{{1 \mp 1}}(x – 3)$ ==> $y = 4$ or $x = 3$

Hence, the lines which make the triangle are $x – y = 2,$ $x = 3$ and $y = 4$. The intersection points of these lines are $(6,\,4),\,(3,\,1)$ and $(3,\,4)$

$\Delta = \frac{1}{2}[6( – 3) + 3(0) + 3(3)]$ $ = \frac{9}{2}$.

Standard 11

Mathematics

In a $\triangle A B C$, points $X$ and $Y$ are on $A B$ and $A C$, respectively, such that $X Y$ is parallel to $B C$. Which of the two following equalities always hold? (Here $[P Q R]$ denotes the area of $\triangle P Q R)$.

$I$. $[B C X]=[B C Y]$

$II$. $[A C X] \cdot[A B Y]=[A X Y] \cdot[A B C]$

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Two lines are drawn through $(3, 4)$, each of which makes angle of $45^\circ$ with the line $x - y = 2$, then area of the triangle formed by these lines is

Solution

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