Question

The distance between two points (X,7)and (1,15) is 10 , find the value of x​

Answer 1

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

• Given that •

• In coordinate two points have been given ( x,7) and (1,15).

• The distance between these two points is 10 unit .

• Let points are A(x,7) and B(1,15)

• Formula to find the distance between any two points (x1,y1) and (x2,y2)

$\sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} }$

• Now solution •

$\sqrt{ {(1 - x)}^{2} + {(15 - 7)}^{2} } = 10 \\ = > {(1 - x)}^{2} + {(8)}^{2} = {(10)}^{2} \\ = > 1 + {(x)}^{2} - 2x + 64 - 100 = 0 \\ = > {x}^{2} - 2x - 35 = 0$

• Here we got quadratic equations which can't be solved by normally

• now to compare this equation to standred Equation we get

a = 1 , b = -2 ,c = -35

Formula of shree dharacharya

$x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac } }{2a}$

• Solution •

$x = \frac{ - ( - 2) + - \sqrt{ {( - 2)}^{2} - 4 \times 1 \times ( - 35) } }{2 \times 1} \\ \\ x = \frac{4 + - \sqrt{4 + 140} }{2} \\ x = \frac{4 + - \sqrt{144} }{4}$

Hence we got the value of x

•••I hope it may help •••