Two lines are drawn through (3, 4),each of wh | vedclass

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(B) The slope of the line $x - y = 2$ is $m_1 = 1$. Let the slope of the required lines be $m$. The angle between the lines is $45^\circ$,so $	an 45^

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$9/2$

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(B) The slope of the line $x - y = 2$ is $m_1 = 1$. Let the slope of the required lines be $m$. The angle between the lines is $45^\circ$,so $	an 45^\circ = |\frac{m - 1}{1 + m(1)}|$.$1 = |\frac{m - 1}{m + 1}|$.This gives $m - 1 = m + 1$ (which is impossible) or $m - 1 = -(m + 1)$,which implies $2m = 0$,so $m = 0$. The other line must be perpendicular to $y = 4$ (since the angle is $45^\circ$),so $x = 3$.The lines are $y = 4$,$x = 3$,and $x - y = 2$.The intersection points are:$1$) $y = 4$ and $x = 3 \implies (3, 4)$.$2$) $y = 4$ and $x - y = 2 \implies x = 6 \implies (6, 4)$.$3$) $x = 3$ and $x - y = 2 \implies y = 1 \implies (3, 1)$.The area of the triangle with vertices $(3, 4), (6, 4), (3, 1)$ is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes |6 - 3| 	imes |4 - 1| = \frac{1}{2} 	imes 3 	imes 3 = \frac{9}{2}$.

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

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