Question

Hello everyone!
Question is in the attachment!
Concept : Straight lines​

Answer 1

Explanation:

hope it helps you

thank you

Answer 2

Concept :

- To solve this question we will assume the vertices to be P(3,4) and Q(-2,3) and R(x,y)

- All the sides in an equilateral triangle are equal. Therefore PQ = PR = QR, we will find distance between them by using the distance formula

- Then we will equate PR = QR to form an equation in 2 variables [1] after this we will equate PR = PQ and will substitute the value of y from the above equation [1] to get a quadratic equation in one variable, and will solve it! :D

Distance formula =
`\sf \sqrt{ {( x_2 - x_1)}^(2) + {(y_2 - y_1)}^(2)}`

Solution :

Let, the coordinates of vertices of the equilateral triangle be P(3,4) and Q(-2,3) and R(x,y)

Also, PQ = PR = QR

Finding PQ using distance formula,

•
`x_(2)` = -2

•
`x_(1)` = 3

•
`y_(2)` = 3

•
`y_(1)` = 4

Putting the values,

`\begin{gathered} \sf PQ = \sqrt{ {( - 2 - 3)}^(2) + {(3- 4)}^(2)} \end{gathered}`

`\sf PQ = \sqrt{ {( - 5)}^(2) + {( - 1)}^(2)}`

`\sf PQ = √( 25 + 1)`

`\sf PQ = √( 26)`

Finding PR using distance formula,

•
`x_(2)` = x

•
`x_(1)` = 3

•
`y_(2)` = y

•
`y_(1)` = 4

Putting the values,

`\begin{gathered} \sf PR = \sqrt{ {( x - 3)}^(2) + {( y - 4)}^(2)} \end{gathered}`

Using (a - b)² = a² + b² - 2ab

`\begin{gathered} \sf PR = \sqrt{ {x}^(2) + { (3)}^(2) - 2(x)( 3)+ {y}^(2) + { (4)}^(2) - 2(y)( 4)} \end{gathered}`

`\begin{gathered} \sf PR = \sqrt{ {x}^(2) + 9 - 6x+ {y}^(2) + 16 - 8y} \end{gathered}`

`\begin{gathered} \sf PR = \sqrt{ {x}^(2) + {y}^(2) - 6x - 8y + 25} \end{gathered}`

Finding QR using distance formula,

•
`x_(2)` = x

•
`x_(1)` = -2

•
`y_(2)` = y

•
`y_(1)` = 3

`\begin{gathered} \sf QR = \sqrt{ {( x - (- 2))}^(2) + {( y - 3)}^(2)} \end{gathered}`

`\begin{gathered} \sf QR = \sqrt{ {( x + 2)}^(2) + {( y - 3)}^(2)} \end{gathered}`

Using (a - b)² = a² + b² - 2ab and (a + b)² = a² + b² + 2ab

`\begin{gathered} \sf QR = \sqrt{ {x}^(2) + {(2)}^(2) + 2(x)(2) + {y}^(2) + {(3)}^(2) - 2(y)(3)} \end{gathered}`

`\begin{gathered} \sf QR = \sqrt{ {x}^(2) + 4 + 4x + {y}^(2) + 9 - 6y} \end{gathered}`

`\begin{gathered} \sf QR = \sqrt{ {x}^(2) + {y}^(2) + 4x- 6y + 13} \end{gathered}`

Now, PR = QR

`\begin{gathered} \sf \sqrt{ {x}^(2) + {y}^(2) - 6x - 8y + 25} = \sqrt{ {x}^(2) + {y}^(2) + 4x- 6y + 13} \end{gathered}`

By squaring both sides, the root will get cancel

`\begin{gathered} \sf \cancel{ {x}^(2)} + \cancel{{y}^(2) } - 6x - 8y + 25 = \cancel{ {x}^(2)} + \cancel{{y}^(2) } + 4x- 6y + 13 \end{gathered}`

`\begin{gathered} \sf - 10x - 2y + 12 = 0 \end{gathered}`

`\begin{gathered} \sf - 5x - y + 6 = 0 \end{gathered}`

`\begin{gathered} \sf 6 - 5x = y \: \: (1)\end{gathered}`

Also PR = PQ

`\begin{gathered} \sf \sqrt{ {x}^(2) + {y}^(2) - 6x - 8y + 25} \sf = √( 26)\end{gathered}`

By squaring both sides, the root will get cancel

`\begin{gathered} \sf {x}^(2) + {y}^(2) - 6x - 8y \sf = 26 - 25\end{gathered}`

`\begin{gathered} \sf {x}^(2) + {y}^(2) - 6x - 8y \sf = 1\end{gathered}`

Put value of y from equation (1)

`\begin{gathered} \sf {x}^(2) + { (6 - 5x)}^(2) - 6x - 8( 6 - 5x) \sf = 1\end{gathered}`

Using (a - b)² = a² + b² - 2ab

`\begin{gathered} \sf {x}^(2) +25 {x}^(2) + 36 - 2(6)(5x) - 6x - 48+ 40x \sf = 1\end{gathered}`

`\begin{gathered} \sf {x}^(2) +25 {x}^(2) + 36 - 60x - 6x - 48+ 40x \sf = 1\end{gathered}`

`\begin{gathered} \sf 26 {x}^(2) - 26x - 12 \sf = 1\end{gathered}`

`\begin{gathered} \sf 26 {x}^(2) - 26x - 13 \sf =0 \end{gathered}`

`\begin{gathered} \sf 2 {x}^(2) - 2x - 1 \sf =0 \end{gathered}`

Using quadratic formula to solve the quartic equation,

`\begin{gathered} \sf x = \frac{ - b \pm \sqrt{ {b}^(2) - 4(a)(c)} }{2a} \end{gathered}`

`\begin{gathered} \sf x = \frac{ - ( - 2) \pm \sqrt{ {( - 2)}^(2) - 4(2)( - 1)} }{4} \end{gathered}`

`\begin{gathered} \sf x = (2 \pm √( 4 + 8) )/(4) \end{gathered}`

`\begin{gathered} \sf x = (2 \pm √( 12) )/(4) \end{gathered}`

`\begin{gathered} \sf x = (2 \pm √( 2 * 2 * 3) )/(4) \end{gathered}`

`\begin{gathered} \sf x = (2 \pm 2√( 3) )/(4) \end{gathered}`

`\begin{gathered} \sf x = (2(1 \pm √( 3) ))/(4) \end{gathered}`

`\begin{gathered} \sf x = (1 \pm √( 3) )/(2) \: \: (2)\end{gathered}`

Put value of x in equation (1)

`\begin{gathered} \sf 6 - 5((1 \pm √( 3) )/(2)) = y \end{gathered}`

`\begin{gathered} \sf (12 - 5 ( 1 \pm √( 3)) )/(2) = y \end{gathered}`

`\sf (7 \mp 5 √( 3) )/(2) = y`

The coordinates of third vertex is
`\sf ( (1 + √(3) )/(2) ) , ( (7 - 5 √(3) )/(2) )` or
`\sf ( (1 - √(3) )/(2) ) , ( (7 + 5 √(3) )/(2) )`

Correct Answer is - Option (b)! :)