The acute angle between the lines x-3=0 and x | vedclass

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(D) The given lines are $x-3=0$ $(i)$ and $x+y=19$ (ii).For line $(i)$,$x=3$,which is a vertical line parallel to the $y$-axis. The angle it makes wit

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(D) The given lines are $x-3=0$ $(i)$ and $x+y=19$ (ii).For line $(i)$,$x=3$,which is a vertical line parallel to the $y$-axis. The angle it makes with the positive $x$-axis is $	heta_1 = 90^{\circ}$.For line (ii),$x+y=19$,we can rewrite it as $y = -x + 19$. Comparing this with the slope-intercept form $y = mx + c$,the slope $m_2 = -1$.Since $m_2 = 	an 	heta_2 = -1$,the angle $	heta_2 = 135^{\circ}$.The angle between the two lines is $|	heta_2 - 	heta_1| = |135^{\circ} - 90^{\circ}| = 45^{\circ}$.Since $45^{\circ}$ is an acute angle,the required angle is $45^{\circ}$.

The acute angle between the lines $x-3=0$ and $x+y=19$ is: (in $^{\circ}$)

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