Question

The distance between (-3, -1) and (3, y) is square root of 45. Find y.​

Answer 1

Answer:

Given,

Distance between the points is root 45---(1)

$distance \: between \: two \: points = \sqrt{(x_{2} - x_{1} )^{2} + ( y_{2} - y_{1}) ^{2} }$

$here \: x_{1} = ( - 3) \\ x_{2} = 3 \\ y_{1} = ( - 1) \\ y_{2} = y$

By substituting the values in the formula we get,

$\sqrt{(3 - ( - 3))^{2} + (y - ( - 1)) ^{2} } \\ \sqrt{ {(3 + 3)}^{2} + (y + 1) ^{2} } \\ \sqrt{ {6}^{2} + {y}^{2} + {1}^{2} + 2(y)(1) } \\ ( {a}^{2} + {b}^{2} + 2ab = ({a + b})^{2} ) \\ \sqrt{36 + {y}^{2} + 1 + 2y } \\ \sqrt{ {y}^{2} + 2y + 37 }$

By equating the above equation with (1) we get,

$\sqrt{ {y}^{2} + 2y + 37} = \sqrt{45}$

By squaring on both the sides root will be cancelled.

${( \sqrt{ {y}^{2} + 2y + 37} })^{2} = ( { \sqrt{45} })^{2} \\ {y}^{2} + 2y + 37 = 45 \\ {y}^{2} + 2y + 37 - 45 = 0 \\ {y}^{2} + 2y - 8 = 0$

By factorization,

${y}^{2} + 4y - 2y - 8 = 0 \\ y(y + 4) - 2(y + 4) = 0 \\ (y - 2)(y + 4) = 0$

Here,

=》(y-2)=0 (or) (y+4)=0

=》 y=2 (or) y=(-4)

The value of y is 2 or (-4).

I hope it is helpful.

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