The number of points on the line x+y=4 which | vedclass

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(A) The given lines are $L_1: x+y-4=0$ and $L_2: 2x+2y-5=0$, which can be rewritten as $x+y-2.5=0$. Since the slopes of both lines are equal $(-1)$, t

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$0$

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(A) The given lines are $L_1: x+y-4=0$ and $L_2: 2x+2y-5=0$, which can be rewritten as $x+y-2.5=0$. Since the slopes of both lines are equal $(-1)$, the lines are parallel. The perpendicular distance $d$ between two parallel lines $ax+by+c_1=0$ and $ax+by+c_2=0$ is given by $d = \frac{|c_1-c_2|}{\sqrt{a^2+b^2}}$. Here, $d = \frac{|-4 - (-2.5)|}{\sqrt{1^2+1^2}} = \frac{|-1.5|}{\sqrt{2}} = \frac{1.5}{\sqrt{2}} = \frac{3}{2\sqrt{2}} \approx 1.06$. Since the distance between the two lines is $\approx 1.06$, which is greater than $1$, there is no point on the line $x+y=4$ that is at a distance of $1$ unit from the line $2x+2y=5$. Thus, the number of such points is $0$.

The number of points on the line $x+y=4$ which are at a unit distance from the line $2x+2y=5$ is

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