The equation of the diagonal of a square is 3x-4y+5=0 and one of its vertices is (1,2). Find the equation of the side of the square that meet at this point.
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Solution:
Given,
Let ABCD be a square with a diagonal AC and vertices A,B,C and D.

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Square ABCD with Diagonal AC



We know,

In a square, the diagonal bisects each angle. 
So, $\measuredangle$DCB = $\measuredangle$DCA + $\measuredangle$ACB
or, 90° = $\measuredangle$DCA + $\measuredangle$DCA [$\because$ angle is bisected]
or, 90° = 2$\measuredangle$DCA
So, $\measuredangle$DCA = 45° = $\measuredangle$ACB

According to the question,

Equation of the diagonal AC is 
3x-4y+5 = 0---(i)

Slope of the diagonal AC ($m_1$) = $- \frac{coefficient\;of \;x\; 'a'}{coefficient \;of\; y \;'b'}$
= $- \frac{a}{b}$
= $- \frac{3}{-4}$
$\therefore m_1 \;=\; \frac{3}{4}$

One of the vertices of the side of the square is (1,2). 

Testing whether the vertex lies on the diagonal or not!

Let coordinate (1,2) = (x,y)
Put value of (x,y) in equation 1. 
3x-4y+5 = 0
or, 3.1 -4.2 +5 = 0
or, 3-8+5 = 0
or, 0=0 which is true. 

So, the vertex is either the coordinate of A or C.

Let the vertex (1,2) be the coordinate of C. Now, we need to find the equation of BC.

For side BC,

coordinate of C = (1,2)
slope ($m_2$) = ?
y-intercept (c) = ?

We know, $\measuredangle$ABC = 45°

And,
Angle between two lines (AC and BC) is:

$ \theta = tan^{-1} \left \{ +_- \dfrac{m_1 -m_2}{1+m_1 \cdot m_2} \right \}$
or, $tan \theta =  \left \{ +_- \dfrac{m_1 -m_2}{1+m_1 \cdot m_2} \right \}$
or, $tan 45° =  \left \{ +_- \dfrac{\dfrac{3}{4} -m_2}{1+\dfrac{3}{4} \cdot m_2} \right \}$
or, $1 = \dfrac{ +_- \left ( \dfrac {3-4m_2}{4} \right ) }{\dfrac{4+3m_2}{4}}  $
or $4+3m_2 = +_-(3-4m_2)$

Taking positive sign;
$4+3m_2 = 3-4m_2$
or, $4m_2+3m_2 = 3-4$
or, $7m_2 = -1$
or, $m_2 = \frac{-1}{7}$

Taking negative sign;
$4+3m_2 = -(3-4m_2)$
or, $4+3m_2 = 4m_2-3$
or, $4+3 = 4m_2 -3m_2$
$ \therefore m_2 = 7$

Either $m_2 = - \frac {1}{7} \;or \;7$

When (1,2) = (x,y) and $m_2 = - \frac {1}{7}$

Using the formula,
$y = m_2 x+c$
or, $2 = - \frac {1}{7} \cdot 1 +c$
or, $2 = - \frac {1}{7} +c$
or, $c = 2+ \frac{1}{7}$
$\therefore c = \frac{15}{7}$

When (1,2) = (x,y) and $m_2 = 7$

Using the formula,
$y = m_2x+c$
or, $2 = 7 \cdot 1 +c$
or, $c = 2-7$
$\therefore c = -5$

Now, 
equation of line BC when $m_2 = - \frac{1}{7}\; and \;c \;= \frac{15}{7}$

Using the formula,
$y = m_2 x+c$
or, $y = - \frac{1}{7} x + \frac{15}{7}$
or, $y =  \frac{-x+15}{7}$
or, $7y = -x +15$
or, $x+7y-15 = 0$

Also,

equation of the line BC when $m_2 = 7 \;and \;c\;= -5$

Using the formula,
$y = m_2x +c$
or, $y = 7x -5$
or, $7x-y-5 = 0$

Hence, the equation of the side of having the vertex (1,2) is either $x+7y-15 = 0$ or $7x-y-5 = 0$.

This is a class 10 question from 'Angle between two lines' chapter of Unit Coordinate Geometry. 

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Related Notes and Solutions:

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

Link: Class 10 Notes
Link: Class 9 Notes


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