If P(4, -7) and Q(-1, 5),then PQ = dots | vedclass

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

Question

(A) To find the distance between two points $P(x_1, y_1)$ and $Q(x_2, y_2)$,we use the distance formula:$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$Giv

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(A) To find the distance between two points $P(x_1, y_1)$ and $Q(x_2, y_2)$,we use the distance formula:$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$Given $P(4, -7)$ and $Q(-1, 5)$,we have $x_1 = 4, y_1 = -7, x_2 = -1, y_2 = 5$.Substituting these values into the formula:$PQ = \sqrt{(-1 - 4)^2 + (5 - (-7))^2}$$PQ = \sqrt{(-5)^2 + (5 + 7)^2}$$PQ = \sqrt{(-5)^2 + (12)^2}$$PQ = \sqrt{25 + 144}$$PQ = \sqrt{169}$$PQ = 13$Therefore,the distance $PQ$ is $13$.

If $P(4, -7)$ and $Q(-1, 5)$,then $PQ = \dots$

Similar Questions

If the centroid of the triangle with vertices $(-5, 3), (a, -1), (6, b)$ is $(1, -1)$,then find the values of $a$ and $b$.

Prove that $A (1, 7)$,$B (2, 4)$,and $C (5, 5)$ are the vertices of an isosceles right triangle.

Find the points on the $X$-axis at a distance of $13$ from the point $(11, 12)$.

$\overline{ AB }$ is given. If $P$ is a point on $\overline{ AB }$ such that $A - P - B$,then $P$ divides $\overline{ AB }$ from $B$ in the division ratio $\ldots \ldots \ldots \ldots$

The distance between $(7, 5)$ and $(2, 5)$ is........

Vedclass Test Series

Exam Paper Generator

Online Exam Module