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  •  Find the slopes of the sides of a square the co-ordinates of two opposites angular points of which are <img alt="(3, 4)" src="https://entrancecorner.oncodecogs.com/g

 Find the slopes of the sides of a square the co-ordinates of two opposites angular points of which are (3, 4) and (1, -1)

 

Option: 1

\mathrm{-\frac{3}{7}\ and \ -\frac{7}{3}}


Option: 2

\mathrm{-\frac{3}{7}\ and \ \frac{7}{3}}


Option: 3

\frac{3}{7}\ and\ -\frac{7}{3}


Option: 4

\frac{2}{7}\ and \ -\frac{7}{2}


Answers (1)

best_answer

Let angular point of the square be \mathrm{A, B, C, D} and Let the point \mathrm{\left ( 3,4 \right )} and \mathrm{\left ( 1,-1 \right )} be \mathrm{A} and \mathrm{C}. 
The equation of the line \mathrm{AC\ y-4=5/2\left ( x-3 \right )} i,e. \mathrm{5x - 2y -7=0}
\mathrm{AB} and \mathrm{DA} makes an angle \mathrm{45^{\circ}} with \mathrm{AC}
Any line through A is  \mathrm{ y - 4 = m(x - 3)}  and \mathrm{tan\ 45^{\circ}=\frac{m-\left ( 5/2 \right )}{1+\left ( 5m/2 \right )}}

\text { i.e. } 1=\frac{\mathrm{m}-(5 / 2)}{1+(5 \mathrm{~m} / 2)} \quad \text { i.e. } \mathrm{m}=-\frac{7}{3}
Equation to \mathrm{AB} \text { is } \mathrm{y}-4=-7 / 3(\mathrm{x}-3) \quad \therefore \mathrm{AD} \text { is } \mathrm{y}-4=3 / 7(\mathrm{x}-3)
\mathrm{ CB} and \mathrm{ CD} are lines through \mathrm{ \left ( 1,-1 \right )} parallel to \mathrm{ AD} and \mathrm{ AB} respectively 
\mathrm{\therefore } their slopes are \mathrm{ 3/7} and \mathrm{ -7/3} respectively 
\mathrm{\therefore \mathrm{y}+1=3 / 7(\mathrm{x}-1) \quad \text { and } \therefore \mathrm{y}+1=-7 / 3(\mathrm{x}-1) }
\mathrm{\therefore } The sides of the squares are \mathrm{ 7 x+3 y-33=0,\3 x-7 y+19=0,\3 x-7 x-10=0\ \&\ 7 x+3 y-4=0 }
 

 

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himanshu.meshram

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