If the midpoint of a line segment joining (3, | vedclass

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

Question

(D) The midpoint formula for a line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$.

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) The midpoint formula for a line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$.Given the points $(3, 4)$ and $(a, 8)$,the midpoint is $\left(\frac{3 + a}{2}, \frac{4 + 8}{2}\right)$.We are given that the midpoint is $(4, 6)$.Equating the coordinates,we get $\left(\frac{3 + a}{2}, \frac{12}{2}\right) = (4, 6)$.This simplifies to $\left(\frac{3 + a}{2}, 6\right) = (4, 6)$.Comparing the $x$-coordinates: $\frac{3 + a}{2} = 4$.Multiplying by $2$,we get $3 + a = 8$.Therefore,$a = 8 - 3 = 5$.

If the midpoint of a line segment joining $(3, 4)$ and $(a, 8)$ is $(4, 6)$,then $a = \ldots$

Similar Questions

Show that $(2,2), (5,2), (5,5)$ and $(2,5)$ are the vertices of a square.

$A(6,1), B(8,2)$ and $C(9,4)$ are three vertices of a parallelogram $ABCD$. If $E$ is the midpoint of $DC$,find the area of $\triangle ADE$ (in sq. units).

Find the length of the median through the vertex $(-1, 3)$ of the triangle having vertices $(-1, 3)$,$(1, -1)$,and $(5, 1)$.

The points $A(2, 9)$,$B(a, 5)$,and $C(5, 5)$ are the vertices of a triangle $ABC$ right-angled at $B$. Find the value of $a$ and the area of $\triangle ABC$.

$\overline{AB}$ is a diameter of $\odot(P, r)$. If $A(4, 0)$ and $B(-2, 2)$ are given,then the coordinates of $P$ are.............

Vedclass Test Series

Exam Paper Generator

Online Exam Module