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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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=

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(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

Similar Questions

The positions of the points $(3, 4)$ and $(2, -6)$ with respect to the line $3x - 4y = 8$ are:

Find the distance between the lines $3x + 2y + 7 = 0$ and $6x + 4y + 3 = 0$.

Let $R$ be the point $(3,7)$ and let $P$ and $Q$ be two points on the line $x+y=5$ such that $\triangle PQR$ is an equilateral triangle. Then the area of $\triangle PQR$ is

Which pair of points lie on the same side of the line $3x - 8y - 7 = 0$?

The system of equations $x+2y=3$ and $3x+6y=a-2$ has no solution if:

Vedclass Test Series

Exam Paper Generator

Online Exam Module