Question

The point $A(1,3)$ and $C(5,1)$ are the oppositive vertices of rectangle. The equation of line passing through other two vertices and of gradient $2$, is ?

• A. $2x+y-8=0$
• B. $2x-y-4=0$
• C. $2x-y+4=0$
• D. $2x+y=0$

Answer 1

The correct option is A $2x+y-8=0$

ABCD is a rectangular.

Let $A(1,3),B(x_1,y_1),C(5,1)andD(x_2,y_2)$ be the vertices of the rectangular.

We know that, diagonals of rectangular bisect each other.

Let O be the point of intersection of diagonal AC and BD.

∴ Mid point of AC = Mid point BD.

Now, O(3, 2) lies on $y=2x+c.$

∴ $2=2\times 3+c$

$c=2–6=–4$

So, the value of c is – 4.

$(x_1,y_1)$ lies on $y=2x–4.$

∴ $y_1=2x_1–4$

$(x_2,y_2)$ lies on $y=2x–4$

∴ $y_2=2x_2–4$

Coordinates of B $=(x_1,2x_1–4)$

Coordinates of D $=(x_2,2x_2–4)$

AD ⊥ AB,

∴ Slope of AD × Slope of AB = – 1.

When $x_1=4$ and $x_2=2$, we get

Coordinates of B $=(x_1,2x_1–4)=(4,2\times 4–4)=(4,4)$

Coordinates of D $=(x_2,2x_2–4)=(2,2\times 2–4)=(2,0)$

When x_1 = 2 and x_2 = 4, we get

Coordinates of B $=(x_1,2x_1–4)=(4,2\times 4–4)=(2,0)$

Coordinates of D $=(x_2,2x_2–4)=(4,2\times 4–4)=(4,4)$

Thus, the other two vertices of the rectangle are (2, 0) and (4, 4).

Therefore $2x+y-8=0$