Question

The points

A (2, 9), B (a, 5)

and

C (5, 5)

are the vertices of a triangle $A B C$ right angled at $B$ . Find the values of $a$ and hence the area of

△ A B C

.

Accepted Answer

The distance $d$ between two points $(x_{1}, y_{1}) and (x_{2}, y_{2})$ is given by the formula

$d = \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}}$

Consider the above figure,

Using distance formula,

$AC = \sqrt{(2 - 5)^{2} + (9 - 5)^{2}}$

$AC = \sqrt{(- 3)^{2} + (4)^{2}}$

$AC = \sqrt{25}$

${AC}^{2} = 25 units$ ...(i)

$AB = \sqrt{(2 - a)^{2} + (9 - 5)^{2}}$

$AB = \sqrt{(4 + a^{2} - 4 a) + (16)}$

$AB = \sqrt{a^{2} - 4 a + 20}$

${AB}^{2} = (a^{2} - 4 a + 20) units$ ...(ii)

$BC = \sqrt{(5 - a)^{2} + (5 - 5)^{2}}$

$BC = \sqrt{(25 + a^{2} - 10 a) + 0}$

$BC = \sqrt{a^{2} - 10 a + 25}$

${BC}^{2} = (a^{2} - 10 a + 25) units$ ...(iii)

We are given that $ABC$ is a right triangle, right angled at $B$ .

By Pythagoras theorem, we have,

${AC}^{2} = {AB}^{2} + {BC}^{2}$

$\Rightarrow 25 = (a^{2} - 4 a + 20) + (a^{2} - 10 a + 25)$ [from (i), (ii) and (iii)]

$\Rightarrow 25 = 2 a^{2} - 14 a + 45$

$\Rightarrow 2 a^{2} - 14 a + 20 = 0$

$\Rightarrow a^{2} - 7 a + 10 = 0$

Splitting the middle term,

$\Rightarrow a^{2} - 5 a - 2 a + 10 = 0$

$\Rightarrow a (a - 5) - 2 (a - 5) = 0$

$\Rightarrow (a - 2) (a - 5) = 0$

$\Rightarrow a = 2$ or $a = 5$

We cannot put $a = 5$ as it will make $BC = 0$ . So, we ignore $a = 5$ .

Therefore, $a = 2$

$\Rightarrow {AB}^{2} = 4 - 8 + 20 = 16$ [from (ii)]

$\Rightarrow AB = 4 units$

$\Rightarrow {BC}^{2} = 4 + 25 - 20 = 9$ [from (iii)]

$\Rightarrow BC = 3 units$

And, $AC = \sqrt{25} = 5 units$

Now, in right $△ ABC$ , base $BC = 3 units$ and height $AB = 4 units$

Area of $△ ABC$ $= \frac{1}{2} \times$ base $\times$ height

$= \frac{1}{2} \times 3 \times 4 = 6 square units$ .

Hence, the area of the right $△ ABC = 6 square units$ .

R. D. Sharma Solutions for Chapter: Co-Ordinate Geometry, Exercise 2: EXERCISE 6.2

R. D. Sharma Mathematics Solutions for Exercise - R. D. Sharma Solutions for Chapter: Co-Ordinate Geometry, Exercise 2: EXERCISE 6.2

Questions from R. D. Sharma Solutions for Chapter: Co-Ordinate Geometry, Exercise 2: EXERCISE 6.2 with Hints & Solutions

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