The length of the intercept of the line x+1=0 | vedclass

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(B) The line $x+1=0$ is equivalent to $x=-1$.To find the intersection point with $3x+2y=5$,substitute $x=-1$:$3(-1)+2y=5 \implies -3+2y=5 \implies 2y=

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$1$

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(B) The line $x+1=0$ is equivalent to $x=-1$.To find the intersection point with $3x+2y=5$,substitute $x=-1$:$3(-1)+2y=5 \implies -3+2y=5 \implies 2y=8 \implies y=4$.So,the first point is $(-1, 4)$.To find the intersection point with $3x+2y=3$,substitute $x=-1$:$3(-1)+2y=3 \implies -3+2y=3 \implies 2y=6 \implies y=3$.So,the second point is $(-1, 3)$.The length of the intercept is the distance between $(-1, 4)$ and $(-1, 3)$:Distance $= \sqrt{(-1 - (-1))^2 + (4 - 3)^2} = \sqrt{0^2 + 1^2} = \sqrt{1} = 1$.

The length of the intercept of the line $x+1=0$ between the lines $3x+2y=5$ and $3x+2y=3$ is

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