Question

Question :-
☆ If one diagonal of a square is along the line $\bf{8x - 15y = 0}$ and one of its vertex is at (1, 2), then find the equation of sides of the square passing through the vertex.

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Answer 1

$\cal \huge \underline{\underline{\color{blue}ANSWER}}$

Equation of diagonal is:

8x-15y=0

So first we have to find value of y

8x-15y=0

8x=15y

8x/15=y

It means that slope of CA is 8/15

Now let the slope of AB be k.

Also we know that diagonals of square make 45° Angle with side.

Then

$\sf \large Tan \: 45°= \dfrac{k - \dfrac{8}{5} }{1 + \dfrac{8k}{15} }$

$\sf \large \: 1= \dfrac{k- \dfrac{8}{5} }{1 + \dfrac{8k}{15} }$

Now cross multiply

$\sf1 + \dfrac{8k}{15} = k - \dfrac{8}{15}$

$\sf \dfrac{8k}{15} - k = - \dfrac{8}{15} - 1$

$\sf \dfrac{8k - 15k}{15} = \dfrac{ - 8 - 15}{15}$

8k-15k=-8-15

-7k=-23

k=23/7

So the equation for side AB is:

$\sf y-2= \dfrac{23}{7} (x-1)$

$\sf\to \: 23x - 7y - 9=0$

Similarly the equation for side AD is:

$\sf y-2= \frac{ - 7}{23} (x-1)$

$\sf \to 7x+23y-53=0$

So the equation of sides are:-

• 23x-7y-9=0
• 7x+23y-53=0

Answer 2

Given:

One diagonal of a square is along the line 8x - 15y = 0.

One of its vertex is at (1,2).

To find:

The equation of sides of the square passing through the vertex.

Solution:

Let ABCd is the required square.

The coordinates of the vertex of square be (1,2).

We need to find the equation of side C, D and AD.

Given that : BD is alone line 8x - 15y = 0.

⟹ $\sf 8x - 15y = 0$

⟹ $\sf \dfrac {8}{15} x - 15y = 0$

∴ It's slope is $\sf \dfrac {8}{15}$

Now,

The angle made by BD with sides DC and AD is 45°.

Let the slope of side DC be m.

Therefore,

⟹ $\sf tan \theta = \dfrac {m_1 - m_2}{1 + m_1 m_2}$

⟹ $\sf tan 45^{\circ} = \dfrac {m - \dfrac {8}{15}}{1 + \dfrac {8m}{15}}$

⟹ $\sf 15 + 8m = 15m - 8$

⟹ $\sf m = \dfrac {23}{7}$

Therefore,

AD and DC are perpendicular.

⟹ $\sf m_1 \times m_2 = -1$

⟹ $\sf m_1 = \dfrac {23}{7} , m_2 = - \dfrac {23}{7}$

The equation of the side DC,

⟹ $\sf y - 2 = - \dfrac {7}{23} (x-1)$

⟹ $\sf 23x - 7y - 9 = 0$

The equation of the side AD,

⟹ $\sf y - 2 = - \dfrac {7}{23} (x-1)$

⟹ $\sf 7x + 23y - 53 = 0$

Hence, The required equations are :

• $\bf 23x - 7y - 9 = 0$
• $\bf 7x + 23y - 53 = 0$