Question

find equation of this question..

Answer 1

Solution:

Given:One diagonal of a square is along the line 8x-15y=0 and it's vertex is at (1,2).
here, the vertex is the place where the diagonal and sides intersect,so,it is the solution for sides equation and diagonal equation.

       A    l----------/l B
              l           / l
              l         /   l
              l       /     l
              l     /       l
              l   /         l
         C  l /           l
- - - - - -<>---------l D
      (1,2)  (x,y)  
                   ↑(Rough diagram)           
"/" represents diagonal of square "-" represents base side of square, "l" represents height side of square.
(x,y)=(1,2)                                                                                                     /

so,if we use a graph paper and plot points for other points for equation 8x-15y=0,
x=1, 
8(1)-15y=0
8-15y=0
-15y=-8
y=-8/-15
y=8/15....(i),

∴It's slope=8/15

we require to find equations on the sides AC & DC.
So,BC is along the line(equation) 8x-15y=0,

Then, we can say that,
     The angle opposite to diagonal is 90°
     The angle formed by the sides with diagonal are equal(LBCD=LCBD)
     The 2 sides and the diagonal forms a triangle (BCD)

    LBCD+LCBD+90°=180°
                       2LBCD =90°
                         LBCD=45°
                      ∴LCBD=45°

as equation for straight line y=mx+b,
  tanθ =$(m_1-m_2)/(1+m_1 m_2)$
      
                     (m-8/15)
    tan45   =--------------
                     1+8m/15
  
                     (m-8/15)
              1  =--------------
                     1+8m/15
1+8m/15=m-8/15
(8m+15)/15 =(15m-8)/15
         8m+15=15m-8
       15m-8m=15+8
                7m=23
                  m=23/7

As AD and DC are perpendicular
$m_1 m_2 =-1$
As we know, $m_1=23/7$
hence, $m_2=-7/23$
as, for straight lines
$y-y_1=mx-mx_1$
So,far AD
y-2=-7(x-1)/23
23y-46=-7x+7
7x+23y-46=7
=>7x+23y-53=0.... The required equation