If d[(5,0), (x, 4)] = sqrt{17},then x = ldots | 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 $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.Given points are $(5, 0)$ and

Vedclass · Accepted Answer

$4, 6$

Vedclass · Answer

(C) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.Given points are $(5, 0)$ and $(x, 4)$ with distance $d = \sqrt{17}$.Substituting the values: $\sqrt{17} = \sqrt{(x - 5)^2 + (4 - 0)^2}$.Squaring both sides: $17 = (x - 5)^2 + 4^2$.$17 = (x - 5)^2 + 16$.$(x - 5)^2 = 17 - 16 = 1$.Taking the square root: $x - 5 = \pm 1$.Case $1$: $x - 5 = 1 \implies x = 6$.Case $2$: $x - 5 = -1 \implies x = 4$.Thus,$x = 4, 6$.

If $d[(5,0), (x, 4)] = \sqrt{17}$,then $x = \ldots \ldots \ldots$

Similar Questions

Find the distance between the following pairs of points: $(-5, 7)$ and $(-1, 3)$. (in $\sqrt{2}$)

$P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$ are the given points. If the line segment $\overline{PQ}$ is parallel to the $Y$-axis,then which of the following is true?

The distance of point $P(2, -3)$ from the $X$-axis is $\ldots \ldots \ldots \ldots .$.

The perpendicular distance of the $X-$ axis from the point $(5,3)$ is $\ldots \ldots \ldots \ldots .$

The triangle with the vertices $(0,0)$,$(4.2,0)$,and $(0,9.1)$ is a $\ldots \ldots \ldots$ triangle.

Vedclass Test Series

Exam Paper Generator

Online Exam Module