Question

A rectangle on a coordinate plane has vertices L(0, 6), M(8, 6), N(8, 0), and O(0, 0). What are the dimensions of the rectangle?

Answer 1

Based on the given, there are two ways to know the dimensions of the rectangle. First, by simply drawing the coordinates given in a cartesian coordinate plane, P1(0,6),P2(8,6) by looking at the ordinate the length is 6 units, thru P3 (0,0) and P4 (8,0) and looking at the abscissa, the width is 8 units. The second way is through the distance formula, d=sqrt (X2-X1)2+(Y2-Y1)2. The dimensions are L=8 and W=6

Answer 2

The rectangle on a coordinate plane has vertices L(0, 6), M(8, 6), N(8, 0), and O(0, 0) has the dimensions as 8 units and 6 units.

How to measure distance between two points?

To measure the distance between two points of a coordinate system, the formula can be given as,

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The rectangle given in the problem is LMNO and the coordinate of vertices of rectangle are L(0, 6), M(8, 6), N(8, 0), and O(0, 0).

To find the distance between the rectangle, we have to find the distance of the LM and MN.

As the LM is the length of the rectangle and the coordinates of the L and M are (0, 6), and (8, 6). Thus by the distance formulas,

$LM=\sqrt{(8-0)^2+(6-6)^2}\\LM=\sqrt{64+0}\\LM=8$

Thus the length of the triangle is 8 units long.

As the MN is the width of the rectangle and the coordinates of the M and N are (8, 6) and (8, 0) .Thus by the distance formula,

$MN=\sqrt{(8-8)^2+(6-0)^2}\\MN=\sqrt{0+36}\\MN=6$

Thus the width of the triangle is 6 units long.

Hence, the rectangle on a coordinate plane has vertices L(0, 6), M(8, 6), N(8, 0), and O(0, 0) has the dimensions as 8 units and 6 units.

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