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Mathematics

Find the value of a, if the distance between the points A (-3, -14) and B (a, -5) is 9 units.

Coordinate Geometry

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Answer

By distance formula,

d = (x2−x1)2+(y2−y1)2\sqrt{(x2 - x1)^2 + (y2 - y1)^2}

⇒AB=[a−(−3)]2+[(−5)−(−14)]2⇒9=(a+3)2+(−5+14)2⇒9=a2+9+6a+(9)2⇒9=a2+6a+9+81\Rightarrow AB = \sqrt{[a - (-3)]^2 + [(-5) - (-14)]^2} \\[1em] \Rightarrow 9 = \sqrt{(a + 3)^2 + (-5 + 14)^2} \\[1em] \Rightarrow 9 = \sqrt{a^2 + 9 + 6a + (9)^2} \\[1em] \Rightarrow 9 = \sqrt{a^2 + 6a + 9 + 81}

On squaring both sides,

⇒a2+6a+90=(9)2⇒a2+6a+90=81⇒a2+6a+90−81=0⇒a2+6a+9=0⇒a2+3a+3a+9=0⇒a(a+3)+3(a+3)=0⇒(a+3)(a+3)=0⇒a+3=0⇒a=−3.\Rightarrow a^2 + 6a + 90 = (9)^2 \\[1em] \Rightarrow a^2 + 6a + 90 = 81 \\[1em] \Rightarrow a^2 + 6a + 90 - 81 = 0 \\[1em] \Rightarrow a^2 + 6a + 9 = 0 \\[1em] \Rightarrow a^2 + 3a + 3a + 9 = 0 \\[1em] \Rightarrow a(a + 3) + 3(a + 3) = 0 \\[1em] \Rightarrow (a + 3)(a + 3) = 0 \\[1em] \Rightarrow a + 3 = 0 \\[1em] \Rightarrow a = -3.

Hence, value of a = -3.

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