Question

30. If points A(0, 3), B(-2, a) and C(-1, 4) are the vertices of a right triangle right-angled at
then (i) find the value of 'a', (ii) find the length of the longest side, and (iii) find the area o
AABC .​

Answer 1

Given :

• Coordinate of A = (0 , 3)

• Coordinate of B = (-2 , a)

• Coordinate of C = (-1 , 4)

To find :

• The value of a .

• Length of the longest side

• Area of the ∆ ABC.

Solution :

First let us find the distance between the points !

Distance between A and B :

Using the Distance formula and substituting the values in it, we get :

$\underline{\bf{D = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}}}$

Here,

• $\bf{x_{1} = 0}$

• $\bf{x_{2} = (-2)}$

• $\bf{y_{1} = 3}$

• $\bf{y_{2} = a}$

$:\implies \bf{D = \sqrt{[(-2) - 0]^{2} + (a - 3)^{2}}}$

Using the identity ,

(a - b)² = a² - 2ab + b²

$:\implies \bf{D = \sqrt{(-2)^{2} + (a^{2} - 2 \times 3 \times a + 3^{2})}}$

$:\implies \bf{D = \sqrt{4 + (a^{2} - 6a + 9)}}$

$:\implies \bf{D = \sqrt{a^{2} - 6a + 13}}$

$\boxed{\therefore \bf{D = \sqrt{a^{2} - 6a + 13}}}$

Hence, the distance between A and B is $\bf{\sqrt{a^{2} - 6a + 13}}$

Distance between B and C :

Using the Distance formula and substituting the values in it, we get :

$\underline{\bf{D = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}}}$

Here,

• $\bf{x_{1} = (-2)}$

• $\bf{x_{2} = (-1)}$

• $\bf{y_{1} = a}$

• $\bf{y_{2} = 4}$

$:\implies \bf{D = \sqrt{[(-1) - (-2)]^{2} + (a - 4)^{2}}}$

Using the identity ,

(a - b)² = a² - 2ab + b²

$:\implies \bf{D = \sqrt{[(-2) + 1]^{2} + (a^{2} - 2 \times 4 \times a + 4^{2})}}$

$:\implies \bf{D = \sqrt{(-1)^{2} + (a^{2} - 8a + 16)}}$

$:\implies \bf{D = \sqrt{1 + a^{2} - 8a + 16}}$

$:\implies \bf{D = \sqrt{a^{2} - 8a + 17}}$

$\boxed{\therefore \bf{D = \sqrt{a^{2} - 8a + 17}}}$

Hence, the distance between A and B is $\bf{\sqrt{a^{2} - 8a + 17}}$

Distance between A and C :

Using the Distance formula and substituting the values in it, we get :

$\underline{\bf{D = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}}}$

Here,

• $\bf{x_{1} = 0}$

• $\bf{x_{2} = (-1)}$

• $\bf{y_{1} = 3}$

• $\bf{y_{2} = 4}$

$:\implies \bf{D = \sqrt{[(-1) - 0]^{2} + (4 - 3)^{2}}}$

$:\implies \bf{D = \sqrt{(-1)^{2} + 1^{2}}}$

$:\implies \bf{D = \sqrt{1 + 1}}$

$\boxed{\therefore \bf{D = \sqrt{2}}}$

Hence, the distance between A and B is $\bf{\sqrt{2}}$

____________________________________________________________________

Since, the Given Triangle is right-angled Triangle , we can use the Pythagoras theorem , to find out the value of a .

Here ,

• Hypotenuse = $\bf{\sqrt{a^{2} - 8a + 17}}$ cm

• Height = $\bf{\sqrt{a^{2} - 6a + 13}}$ cm

• Base = 2 cm

Using the Pythagoras theorem and substituting the values in it, we get :

$\boxed{\bf{H^{2} = P^{2} + B^{2}}}$

Where :-

• H = Hypotenuse
• B = Base
• P = Height

$:\implies \bf{\sqrt{(a^{2} - 8a + 17})^{2} = \sqrt{(a^{2} - 6a + 13})^{2} + (\sqrt{2})^{2}}$

$:\implies \bf{a^{2} - 8a + 17 = a^{2} - 6a + 13 + 2}$

$:\implies \bf{0 = a^{2} - 6a + 13 + 2 - (a^{2} - 8a + 17)}$

$:\implies \bf{0 = a^{2} - 6a + 15 - (a^{2} - 8a + 17)}$

$:\implies \bf{0 = a^{2} - 6a + 15 - a^{2} + 8a - 17}$

$:\implies \bf{0 = \not{a}^{2} - 6a + 15 - \not{a}^{2} + 8a - 17}$

$:\implies \bf{0 = - 6a + 15 + 8a - 17}$

$:\implies \bf{0 = 2a - 2}$

$:\implies \bf{2 = 2a}$

$\boxed{\therefore \bf{a = 1}}$

Hence, the value of a is 1.

Length of the longest side :

Since, we know that the hypotenuse is the longest side of an Triangle , we will out the value of a , in the hypotenuse.

Using the hypotenuse of the triangle and substituting the value of a in it , we get :

$:\implies \bf{H = \sqrt{a^{2} - 8a + 17}}$

$:\implies \bf{H = \sqrt{1^{2} - 8 \times 1 + 17}}$

$:\implies \bf{H = \sqrt{1 - 8 + 17}}$

$:\implies \bf{H = \sqrt{-7 + 17}}$

$:\implies \bf{H = \sqrt{10}}$

$\boxed{\therefore \bf{Hypotenuse\:(h) = \sqrt{10}}}$

Hence, the longest side of the Triangle is √10 cm.

Area of the triangle :

We know the formula for area of a Triangle i.e,

$\bf{A = \dfrac{1}{2} \times Base \times height}$

$:\implies \bf{A = \dfrac{1}{2} \times \sqrt{2} \times\sqrt{a^{2} - 6a + 13}}$

Putting the value of a in the height , we get :

$:\implies \bf{A = \dfrac{1}{2} \times \sqrt{2} \times \sqrt{1^{2} - 6 \times 1 + 13}}$

$:\implies \bf{A = \dfrac{1}{2} \times \sqrt{2} \times \sqrt{8}}$

$:\implies \bf{A = \dfrac{1}{2} \times \sqrt{2} \times 2\sqrt{2}}$

$:\implies \bf{A = \sqrt{2} \times \sqrt{2}}$

$\boxed{\therefore \bf{Area\:(A) = 2\:cm^{2}}}$

Hence, the area of the triangle is 2 cm².