Find the coordinates of points on the x-axis which are at a distance of 17 units from the point (11,-8)

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Solution

Let the required point be (x,0).
$\sqrt{(x - 11)^{2} + (8)^{2}} = 17$

$(x - 11)^{2} + 64 = 289$

$x^{2} + 121 - 22 x + 64 = 289$

$x^{2} - 22 x - 104 = 0$

$x^{2} - 26 x + 4 x - 104 = 0$

$x (x - 26) + 4 (x - 26) = 0$

$(x - 26) (x + 4) = 0$

$x = 26, - 4$

Hence, the required points are $(26, 0)$ and $(- 4, 0)$ .

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units from the point

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Find the coordinates of points on the x-axis which are at a distance of 17 units from the point (11,-8)

Let the required point be (x,0).√(x−11)2+(8)2=17(x−11)2+64=289x2+121−22x+64=289x2−22x−104=0x2−26x+4x−104=0x(x−26)+4(x−26)=0(x−26)(x+4)=0x=26,−4Hence, the required points are (26,0) and (−4,0).