In the adjoining figure, ABCD is a square. If the equation of its diagonal is x-3y=2, find the equations of AB and BC. | Coordinate Geometry | SciPiPupil

Question: In the adjoining figure, ABCD is a square. If the equation of its diagonal is x-3y=2, find the equations of AB and BC.

Solution:

Given: ABCD is a square. BD is a diagonal.

We know, each angle in a square is 90° and diagonal bisects the right angle in a square. So, <ABD = <CBD = 45°

To find: equations of line AB and BC.

We have,

Equation of diagonal BD is x -3y = 2

slope of the line (m) = $- \dfrac{coefficient of x}{coefficient of y}$

= $- \dfrac{1}{-3}$

= $\dfrac{1}{3}$

So, slope of BD (m) = 1/3

Using the formula of angle between two lines;

$tan \theta = +_- \left ( \dfrac{m_1 -m_2}{1 +m_1m_2} \right)$

Taking $m_1 = m = \dfrac{1}{3} \; and \;\theta = 45°$

$tan 45° = +_- \left ( \dfrac{\frac{1}{3} -m_2}{1 +\frac{1}{3}m_2} \right)$

Taking positive sign;

$tan 45° = \dfrac{\frac{1}{3} -m_2}{1 +\frac{1}{3}m_2}$

$1 = \dfrac{\frac{1 -3m_2}{3}}{\frac{3 +m_2}{2}}$

$1 = \dfrac{1 - 3 m_2}{3 + m_2}$

$3+m_2 = 1 -3m_2$

$or, 4m_2 = -2$

$\therefore, m_2 = -\frac{1}{2}$

So, slope of line AB = - (1/2)

Taking negative sign;

$or, tan 45° =-\left( \dfrac{\frac{1}{3} -m_2}{1 +\frac{1}{3}m_2} \right)$

$or, - 1 = \dfrac{\frac{1 -3m_2}{3}}{\frac{3 +m_2}{2}}$

$or, -1 (3+ m_2) = 1 - 3 m_2$

$-3-m_2 = 1- 3m_2$

$2m_2 = 4$

$\therefore, m_2 = 2$

So, slope of line BC = 2

Solving the equation of diagonal BD: x-3y=2

Taking testing points (5,1) = (x,y)

or, 5 - 3(1) = 2

or, 5 - 2 = 2

So, 2 = 2 which is true.

It means that point (5,1) lies on the diagonal and is the coordinate of one of the vertex.

Now,

Equation of line AB when $(x_1, y_1) = (5,1) \; and \; m = - \frac{1}{2}$

or, $(y - y_1) = m(x - x_1)$

or, $(y - 1) = - \frac{1}{2} (x -5)$

or, $2(y -1) = -x +5$

or, $2y -2 = -x +5$

or, $x -2y = -5-2$

$\therefore, x -2y = -7$ is the required equation of line AB.

Equation of line BC when $(x_1, y_1) = (5,1) \; and \; m = 2$

or, $(y -y_1) = m(x -x_1)$

or, $(y -1) = 2(x -5)$

or, $(y -1) = 2x -10$

or, $2x -y = 10 -1$

$\therefore, 2x -y = 9$ is the required equation of line BC.

Hence,

Required equations are (x -2y = 7) and (2x -y = 9).

Related Notes and Solutions:

Here is the website link to all the important formulae of Coordinate Geometry of Class 10.

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