The distance between the points (6, 8) and (3 | vedclass

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

Question

(D) To find the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$,we use the distance formula:$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$Given

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) To find the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$,we use the distance formula:$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$Given points are $(6, 8)$ and $(3, 4)$.Here,$x_1 = 6, y_1 = 8$ and $x_2 = 3, y_2 = 4$.Substituting these values into the formula:$d = \sqrt{(3 - 6)^2 + (4 - 8)^2}$$d = \sqrt{(-3)^2 + (-4)^2}$$d = \sqrt{9 + 16}$$d = \sqrt{25}$$d = 5$Therefore,the distance between the points is $5$.

The distance between the points $(6, 8)$ and $(3, 4)$ is .........

Similar Questions

If the mid-point of the line segment joining the points $A (3, 4)$ and $B (k, 6)$ is $P (x, y)$ and $x+y-10=0,$ find the value of $k.$

Obtain the circumcentre of $\Delta ABC$ with vertices $A (-3, -1)$,$B (-1, 3)$ and $C (6, 2)$.

If $A(0, 0)$,$B(2, 0)$,$C(2, 2)$,and $D(0, 2)$,then $\square ABCD$ is a $\ldots \ldots \ldots$

State whether the following statement is true or false. Justify your answer.
Points $A(-6, 10)$,$B(-4, 6)$,and $C(3, -8)$ are collinear such that $AB = \frac{2}{9} AC$.

The point on the $X-$ axis which is equidistant from $A (-1, 2)$ and $B(5, 4)$ is $\ldots \ldots \ldots \ldots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module