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A line segment is of length 10 units and one of its end is (-2, 3). If the ordinate of the other end is 9, find the abscissa of the other end.

Coordinate Geometry

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Answer

Given,

Ordinate of the point on the other end = 9.

Let abscissa = x.

Given,

Distance between the two ends (-2, 3) and (x, 9) = 10 units.

By distance formula,

d=(x2−x1)2+(y2−y1)2∴[x−(−2)]2+(9−3)2=10⇒[x+2]2+62=10⇒x2+4+4x+36=10⇒x2+4x+40=100d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2} \\[1em] \therefore \sqrt{[x - (-2)]^2 + (9 - 3)^2} = 10 \\[1em] \Rightarrow \sqrt{[x + 2]^2 + 6^2} = 10 \\[1em] \Rightarrow \sqrt{x^2 + 4 + 4x + 36} = 10 \\[1em] \Rightarrow x^2 + 4x + 40 = 100

On squaring both sides,

⇒x2+4x+40−100=0⇒x2+4x−60=0⇒x2+10x−6x−60=0⇒x(x+10)−6(x+10)=0⇒(x−6)(x+10)=0⇒x−6=0 or x+10=0⇒x=6 or x=−10.\Rightarrow x^2 + 4x + 40 - 100 = 0 \\[1em] \Rightarrow x^2 + 4x - 60 = 0\\[1em] \Rightarrow x^2 + 10x - 6x - 60 = 0 \\[1em] \Rightarrow x(x + 10) - 6(x + 10) = 0 \\[1em] \Rightarrow (x - 6)(x + 10) = 0 \\[1em] \Rightarrow x - 6 = 0 \text{ or } x + 10 = 0 \\[1em] \Rightarrow x = 6 \text{ or } x = -10.

Hence, abscissa of other end = 6 or -10.

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