# Question e2075

Oct 5, 2017

Point B has the coordinates $\left(- \left(\frac{7}{3}\right) , \left(- \frac{10}{3}\right)\right)$

#### Explanation:

Eqn of line AB is x—y=-1 & slope of AB$= 1$
Eqn of line AD is $7 x - y = 5$ & slope of AD $= 7$

Solving Eqn AB & AD, we get point A.
$6 x = 6$ or $x = 1$ & $y = 2$
Point A $\left(1 , 2\right)$

E is the midpoint of A & C
Point E $\left(- 1 , - 2\right)$

$\frac{x + 1}{2} = - 1$
$x = - 3$
$\frac{y + 2}{2} = - 2$
$y = - 6$
Point C$\left(- 3 , - 6\right)$

Slope of BC is same as slope of AD $= 7$
Eqn of line BC is $y + 6 = 7 \left(x + 3\right)$
$y - 7 x = 5$
Solving line Equations AB & BC, we get point B.
Line AB $x - y = - 1$
Line BC $- 7 x + 1 = 5$
$\therefore - 6 x = 14$ or $x = - \left(\frac{7}{3}\right)$ & $y = - \left(\frac{10}{3}\right)$
Point B (-(7/3),-(10/3)#

Slope of CD is same as slope of AB $= 1$
Eqn of line CD is $\left(y + 6\right) = \left(x + 3\right)$
$y - x = - 3$
Solving Eans AD & CD, we get point D
Line AD $7 x - y = 5$
Line CD $y - x = - 3$
$\therefore 6 x = 2$ or $x = \frac{1}{3}$ & $y = - \left(\frac{8}{3}\right)$
Point D$\left(\frac{1}{3} , - \left(\frac{8}{3}\right)\right)$