the diagonal of a square lies along the line 8x-15y=0 one vertex is (1,2) find equations to the side - Maths - Straight Lines

Question

the diagonal of a square lies along the line 8x-15y=0 one vertex
is (1,2) find equations to the side of square passing through
it?

Pronay · Accepted Answer

Let ABCD be the given square and the coordinates of the vertex D be (1, 2).We are required to find the equations of its sides DC and AD.Given that BD is along the line 8x – 15y = 0?y= $\frac{8}{15} x + 0$ , so its slope is $\frac{8}{15}$ [equation of straight line,y=mx+c,where m is slope]

The anglesmade by BD with sides AD and DC is 45° . Let the slope of DC be m. Then $\tan θ = \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}$ $\tan 45 = \frac{m - \frac{8}{15}}{1 + \frac{8 m}{15}} 15 + 8 m = 15 m - 8 7 m = 23 \Rightarrow m = \frac{23}{7}$

Since AD and DC are perpendicular,therefore $m_{1} {\times m}_{2} = - 1$

$m_{1} = \frac{23}{7}, H e n c e, m_{2} = \frac{- 7}{23}$ Therefore, the equation of the side DC is given by

$y - 2 = \frac{23}{7} (x - 1) \Rightarrow 23 x - 7 y - 9 = 0$ {equation of straight line, $y - y_{1} = m (x - x_{1})$ }

Similarly, the equation of another side AD is given by

$y - 2 = \frac{- 7}{23} (x - 1) \Rightarrow 7 x + 23 y - 53 = 0$

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the diagonal of a square lies along the line 8x-15y=0 one vertex is (1,2). find equations to the side of square passing through it?