The point of intersection of the diagonals of | vedclass

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

Question

(B) The sides of the rectangle are given by the lines $x=8, x=10, y=11$,and $y=12$. These lines form a rectangle with vertices at the intersection poi

Vedclass · Accepted Answer

$\left(9, \frac{23}{2}\right)$

Vedclass · Answer

(B) The sides of the rectangle are given by the lines $x=8, x=10, y=11$,and $y=12$. These lines form a rectangle with vertices at the intersection points: $(8, 11), (10, 11), (10, 12)$,and $(8, 12)$. The diagonals of a rectangle intersect at the midpoint of either diagonal. Taking the diagonal connecting $(8, 11)$ and $(10, 12)$,the midpoint is calculated as: $M = \left(\frac{8+10}{2}, \frac{11+12}{2}\right) = \left(\frac{18}{2}, \frac{23}{2}\right) = \left(9, \frac{23}{2}\right)$. Thus,the point of intersection is $\left(9, \frac{23}{2}\right)$.

The point of intersection of the diagonals of the rectangle whose sides are contained in the lines $x=8, x=10, y=11$ and $y=12$ is

Similar Questions

If the incentre and the circumcentre of the triangle formed by the lines $x=2$,$4x+3y+7=0$ and $y=3$ are $I$ and $S$ respectively,then $IS=$

The area of the parallelogram formed by the lines $a_1x + b_1y + c_1 = 0$,$a_1x + b_1y + d_1 = 0$,$a_2x + b_2y + c_2 = 0$,and $a_2x + b_2y + d_2 = 0$ is:

The area of the parallelogram formed by the lines $y = mx, y = mx + 1, y = nx, y = nx + 1$ is:

The points $(0, 8/3), (1, 3)$ and $(82, 30)$ are:

The in-centre of the triangle formed by the lines $x=0, y=0$ and $3x+4y=12$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module