Question

YA 5 3 33. In the given figure, if OABC is a rectangle whose diagonals BO and CA intersect at M (2, 1), then the equations of the diagonals BO and CA respectively are
(a) x = 2y, x + 2y = 4
(b) x = y, x + y = 0
(C) 2x = y, 2x + y = 0
(d) x = 3y, x + 3y = 0 ​

Answer 1

Correct answer is

(a) x = 2y, x + 2y = 4

Answer 2

Step-by-step explanation:

Given: In the given figure, if OABC is a rectangle whose diagonals BO and CA intersect at M (2, 1).

To find: The equations of the diagonals BO and CA respectively are

(a) x = 2y, x + 2y = 4

(b) x = y, x + y = 0

(C) 2x = y, 2x + y = 0

(d) x = 3y, x + 3y = 0

Solution:

Tip: Equation of a line passing through $(x_1,y_1)$ and $(x_2,y_2)$ is given by

$\bf y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

Step 1: Write end points of diagonal OB.

We can easily see from given figure;

O(0,0) and B(4,2)

Put the values in the formula and find the equation of OB

$y-0=\frac{2-0}{4-0}(x-0)$

$y = \frac{1}{2} x$

$2y = x$

or

$\bf x = 2y$

Step 2: Write end points of diagonal AC.

We can easily see from given figure;

A(4,0) and C(0,2)

But, as both diagonals are intersected at M(2,1), we can take one end points and another point M

Put the values in the formula and find the equation of AC

A(4,0) and M(2,1)

$y-0=\frac{1-0}{2-4}(x-4)$

$y = \frac{1}{ - 2} (x - 4)$

or

$y = \frac{ - 1}{2} (x - 4)$

$2y = - x + 4$

$\bf x + 2y = 4$

Final answer:

Equations of diagonals OB and AC are $x = 2y$ and $x + 2y = 4$

respectively.

Option A is correct.

Hope it helps you.

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