If the vertices of Delta ABC are A(9,5), B(6, | vedclass

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

Question

(D) Since $\overline{BE}$ is the median of $\Delta ABC$ to the side $\overline{AC}$,$E$ is the midpoint of $\overline{AC}$.The coordinates of $A$ are

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Since $\overline{BE}$ is the median of $\Delta ABC$ to the side $\overline{AC}$,$E$ is the midpoint of $\overline{AC}$.The coordinates of $A$ are $(9,5)$ and $C$ are $(3,3)$.Using the midpoint formula,the coordinates of $E = \left(\frac{9+3}{2}, \frac{5+3}{2}\right) = \left(\frac{12}{2}, \frac{8}{2}\right) = (6,4)$.The length of median $\overline{BE}$ is the distance between $B(6,7)$ and $E(6,4)$.Using the distance formula,$BE = \sqrt{(6-6)^2 + (7-4)^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.

If the vertices of $\Delta ABC$ are $A(9,5), B(6,7)$ and $C(3,3)$,then the length of median $\overline{BE}$ is.......

Similar Questions

If the distance between $A(-4, -3)$ and $B(6, a)$ is $10$,then $a = \ldots \ldots \ldots . . .$

Find the point on the $X$-axis which is equidistant from $(5, 4)$ and $(-2, 3)$.

The points $A(9,0), B(9,6), C(-9,6)$ and $D(-9,0)$ are the vertices of a

$P(3, 2)$ and $Q(7, k)$ are the given points. If $d(P, Q) = 5$,then $k = \dots \dots \dots$

If $Q(0, 1)$ is equidistant from $P(5, -3)$ and $R(x, 6)$,find the values of $x$. Also,find the distances $QR$ and $PR$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module