A line is of length 10 units and one of its ends is (-2,3). If the ordinate of the order end is 9, prove that the abscissa of the other end is 6 or -10.

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Solution

The length of line is $10 c m$
let the abscissa be $x$
So the points are $(- 2, 3); (x, 9)$
Distance formula is $\sqrt{(x + 2)^{2} + (9 - 3)^{2}} = 10 x^{2} + 4 x + 4 + 36 = 100 x^{2} + 4 x - 60 = 0 x^{2} + 10 x - 6 x - 60 = 0 x (x + 10) - 6 (x + 10) (x + 10) (x - 6) x = 6, - 10$

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The length of a line segment is of 10 units and the coordinates of one end-point are (2, -3). If the abscissa of the other end is 10, find the ordinate of the other end.

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A line is of length 10 units and one of its ends is (-2,3). If the ordinate of the order end is 9, prove that the abscissa of the other end is 6 or -10.

The length of line is 10cmlet the abscissa be xSo the points are (−2,3);(x,9)Distance formula is √(x+2)2+(9−3)2=10x2+4x+4+36=100x2+4x−60=0x2+10x−6x−60=0x(x+10)−6(x+10)(x+10)(x−6)x=6,−10

The length of line is $10 c m$
let the abscissa be $x$
So the points are $(- 2, 3); (x, 9)$
Distance formula is $\sqrt{(x + 2)^{2} + (9 - 3)^{2}} = 10 x^{2} + 4 x + 4 + 36 = 100 x^{2} + 4 x - 60 = 0 x^{2} + 10 x - 6 x - 60 = 0 x (x + 10) - 6 (x + 10) (x + 10) (x - 6) x = 6, - 10$