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 circumcentre and circumradius of a triangle with vertices $(3,0), (-1,-6),$ and $(4,-1)$.

Name the type of triangle formed by the points $A (-5, 6)$,$B (-4, -2)$,and $C (7, 5)$.

$A (1, 2)$ and $B (2, 3)$ are the given points. Find the point $C \in \overleftrightarrow{AB}$ such that $C - A - B$ and $2 AC = 3 AB$.

The perpendicular bisector of the line segment joining the points $A(1, 5)$ and $B(4, 6)$ cuts the $y$-axis at

$\overline{ AB }$ 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 $\ldots \ldots \ldots \ldots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module