If A(0, y_{1}) and B(0, y_{2}) are the points | vedclass

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

Question

(C) The distance formula between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is given by $d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$.G

Vedclass · Accepted Answer

$|y_{1} - y_{2}|$

Vedclass · Answer

(C) The distance formula between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is given by $d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$.Given the points $A(0, y_{1})$ and $B(0, y_{2})$,we substitute these coordinates into the formula:$AB = \sqrt{(0 - 0)^{2} + (y_{2} - y_{1})^{2}}$$AB = \sqrt{0^{2} + (y_{2} - y_{1})^{2}}$$AB = \sqrt{(y_{2} - y_{1})^{2}}$Since the distance must be non-negative,we take the absolute value:$AB = |y_{2} - y_{1}|$ or $|y_{1} - y_{2}|$.Therefore,the correct option is $C$.

If $A(0, y_{1})$ and $B(0, y_{2})$ are the points on the $Y$-axis,then the distance between them $AB = \ldots$

Similar Questions

Find the coordinates of the point $R$ on the line segment joining the points $P(-1, 3)$ and $Q(2, 5)$ such that $PR = \frac{3}{5} PQ$.

If the distance between the points $(2, -3)$ and $(5, b)$ is $5,$ then $b = \ldots \ldots \ldots .$

If the points are $A(0, 0)$, $B(0, 2)$, and $C(\sqrt{3}, 1)$, then $\Delta ABC$ is $\ldots \ldots \ldots \ldots$ triangle.

If the points $A(1, 2)$,$O(0, 0)$,and $C(a, b)$ are collinear,then:

If the points are $(12, 10)$ and $(0, 8)$,then the midpoint of the line segment joining these two points is $\ldots \ldots \ldots \ldots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module