Question

(1) Find points on the x-axis which are at a distance of 5 units from points
(5,-4).​

Answer 1

Answer

The required point is either (2,0) or (8,0)

Given

• The point (5,-4) are at distance 5 units from X-axis.

To Find

• The point the point on X - axis

Formula Used

• Distance between two points (x₁ , y₁) and (x₂ , y₂) is given by
• $\sqrt{( x_{2} - x_{1} ) ^{2} + (y _{2} - y_{1})^{2} }$

Solution

Let us consider the point on X-axis be A(a,0)

and let B(5, -4)

Now by using distance formula

$AB = \sqrt{(5 - x) {}^{2} + ( - 4 - 0) ^{2} } \\ \implies 5 = \sqrt{ 25 + {x}^{2} - 10x + 16} \\ \implies 5 = \sqrt{41 + {x}^{2} - 10x } \\ \implies25 = {x}^{2} - 10x + 41 \\ \implies {x}^{2} - 10x + 16 = 0\\ \implies {x}^{2} - 2x - 8x + 16 = 0 \\ \implies x(x - 2) - 8(x - 2) = 0 \\ \implies(x - 2)(x - 8) = 0$

Now

$x - 2 = 0 \\ \implies \bold{ x = 2}$

and

$x - 8 = 0 \\ \implies \bold{x = 8}$

Answer 2

Solutions:

Let the consider the point on x axis be A(a,0) and (5,-4)

Now,

AB = √(5-x)² + (-4 - 0)²

AB = √25 + x² - 10x + 16

5 = √41 + x² - 10x

25 = x² - 10x + 10x

x² - 10x + 16 = 0

x² - 2x - 8x + 16 = 0

x(x - 2) - 8(x - 2) = 0

(x - 8) (x - 2) = 0

x - 2 = 0. and x - 8 = 0

x = 2 and x = 8

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