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The points A(3, 0), B(a, -2) and C(4, -1) are the vertices of triangle ABC right-angled at vertex A. Find the value of a.

Distance Formula

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Answer

Distance between the given points = (x2−x1)2+(y2−y1)2\sqrt{(x2 - x1)^2 + (y2 - y1)^2}

The length of AC:

=(4−3)2+((−1)−0)2=12+(−1)2=1+1=2= \sqrt{(4 - 3)^2 + ((-1) - 0)^2}\\[1em] = \sqrt{1^2 + (-1)^2}\\[1em] = \sqrt{1 + 1}\\[1em] = \sqrt{2}

The length of BC:

=(a−4)2+((−2)−(−1))2=(a−4)2+(−1)2=a2+16−8a+1=a2−8a+17= \sqrt{(a - 4)^2 + ((-2) - (-1))^2}\\[1em] = \sqrt{(a - 4)^2 + (-1)^2}\\[1em] = \sqrt{a^2 + 16 - 8a + 1}\\[1em] = \sqrt{a^2 - 8a + 17}

The length of AB:

=(a−3)2+((−2)−0)2=(a−3)2+(−2)2=a2+9−6a+4=a2−6a+13= \sqrt{(a - 3)^2 + ((-2) - 0)^2}\\[1em] = \sqrt{(a - 3)^2 + (-2)^2}\\[1em] = \sqrt{a^2 + 9 - 6a + 4}\\[1em] = \sqrt{a^2 - 6a + 13}

Using Pythagoras theorem in triangle ABC,

BC2 = AB2 + AC2

⇒(2)2+(a2−6a+13)2=(a2−8a+17)2⇒2+a2−6a+13=a2−8a+17⇒a2−6a+15=a2−8a+17⇒−6a+15+8a−17=0⇒2a−2=0⇒2a=2⇒a=22⇒a=1⇒ (\sqrt2)^2 + (\sqrt{a^2 - 6a + 13})^2 = (\sqrt{a^2 - 8a + 17})^2\\[1em] ⇒ 2 + a^2 - 6a + 13 = a^2 - 8a + 17\\[1em] ⇒ a^2 - 6a + 15 = a^2 - 8a + 17 \\[1em] ⇒ - 6a + 15 + 8a - 17 = 0\\[1em] ⇒ 2a - 2 = 0\\[1em] ⇒ 2a = 2 \\[1em] ⇒ a = \dfrac{2}{2} \\[1em] ⇒ a = 1

Hence, the value of a = 1.

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