Show that the equations x^2-7x+6=0 and y^2-14y+40=0 form a rectangle. Also find the joint equations of diagonals.

1) The region between the four lines x = 6, x = 1, y = 4 and y = 10 describing both equations is a rectangle

2) The joint equations of diagonals are;

5·y = 56 - 6·x and 5·y = 6·x + 14.

Step-by-step explanation:

The equations are;

x² - 7·x + 6 = 0(1)

y² - 14·y + 40 = 0(2)

Factorizing equation (1) and equation (2) , we get

x² - 7·x + 6 = (x - 6)·(x - 1) = 0

Which are vertical lines at points x = 6 and x = 1

For equation (2) , we get

y² - 14·y + 40 = (y - 10)·(y - 4) = 0

Which are horizontal lines at point y = 4 and y = 10

The region between the four lines x = 6, x = 1, y = 4 and y = 10 describing both equations is a rectangle

2) The points of intersection of the equations are;

(1, 4), (1, 10), (6, 4), and (6, 10)

The end point of the diagonals are;

(1, 10), (6, 4) and  (1, 4), (6, 10)

The slope of the diagonals are;

(10 - 4)/(1 - 6) = -6/5 and (4 - 10)/(1 - 6) = 6/5

The equation of one of the diagonals are then, y - 10 = -6/5×(x - 1)

y = -6/5·x + 6/5 + 10 = -6/5·x + 56/5

5·y = 56 - 6·x

The other diagonal is therefore;

y - 4 = 6/5×(x - 1)

y = 6/5·x - 6/5 + 4 = 6/5·x + 14/5

5·y = 6·x + 14.

The joint equations of diagonals are therefore;

5·y = 56 - 6·x and 5·y = 6·x + 14.

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step-by-step explanation:

there is no question content here, but you for the free points! : d