Question

Two vertices of a rectangle are (6, -1), (6, 9) and
remaining two vertices are on Y-axis, then find the
area of rectangle

Answer 1

Step-by-step explanation:

Given:-

Two vertices of a rectangle are (6, -1), (6, 9) and

remaining two vertices are on Y-axis.

To find:-

find the area of rectangle

Solution:-

See the above attachment for understanding the concept

Given points are (6,-1) and (6,9)

and other two points on the y-axis

we know that

The equation of y-axis is x=0

So the other two pints be like (0,y1) and (0,y2)

The given four points are the vertices of the rectangle .

So the other two points should be (0,-1) and (0,9)

So the four points are (6, -1), (6, 9),(0,-1) and (0,9)

Let A(0,9) ; B(0,-1) ;C(6,-1) ;D(6,9)

ABCD is a rectangle

(x,y1)=(0,9)

(x,y2)=(0,-1)

Distance between A and B = | y2-y1 | units

=>AB =| -1-9 |

=>AB= | -10 | units

AB = 10 units

CD = 10 units

Since ,The opposite sides are equal

and

We have

(x,y1) = (0,-1)

(x,y2)=((6,-1)

Distance between B and C = | x2-x1 | units

=>BC = | 6-0 |

=>BC = 6 units

BC = AD = 6 units

Opposite sides are equal.

Area of a rectangle = length × breadth

=>Area of a rectangle ABCD = AB × BC sq.units

=>ar(ABCD) = 10×6 sq.units

ar(ABCD)=60 sq.units

Answer:-

Area of the given rectangle is 60 sq.units

Used formulae:-

• The equation of y-axis is x=0
• The opposite sides are equal
• Area of a rectangle = length × breadth
• The distance from a point on x-axis is |x2-x1 | units
• The distance from a point on y-axis is | y2-y1| units