∆ABC is a triangle whose vertices are A(0, 0), B(a, 5) and C(-5, 5). If the triangle is right-angled at A, then find the value of a.

Correct option is B

Given:

Vertices of triangle: A(0, 0), B(a, 5), C(-5, 5)

Triangle is right-angled at A

Formula Used:

The distance between two points $$(x_1, y_1)$$​ and $$(x_2, y_2)$$​ is given by:

Distance =$$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$​

Right angle is at A, the lengths of sides AB and AC must satisfy the Pythagorean theorem:

​$$AB^2 + AC^2 = BC^2$$​

Solution:

The distance between A(0, 0) and B(a, 5) is:

AB =$$\sqrt{(a - 0)^2 + (5 - 0)^2} = \sqrt{a^2 + 25}$$​

The distance between A(0, 0) and C(-5, 5) is:

AC =$$\sqrt{(-5 - 0)^2 + (5 - 0)^2} = \sqrt{25 + 25} = \sqrt{50}$$

The distance between B(a, 5) and C(-5, 5) is:

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

Since the triangle is right-angled at A, we have:

​$$AB^2 + AC^2 = BC^2$$​

​$$(a^2 + 25) + 50 = (a + 5)^2$$​

​$$a^2 + 75 = a^2 + 10a + 25$$​

75 = 10a + 25

50 = 10a

a = 5

Alternate Method:

Concept Used:
In a triangle right-angled at a point, the vectors from that point to the other two vertices are perpendicular.
Two vectors are perpendicular if their dot product is zero.

Solution:
Let$$\vec{AB} = (a - 0, 5 - 0) = (a, 5)$$​​
Let$$\vec{AC} = (-5 - 0, 5 - 0) = (-5, 5)$$​​

Dot product:

​$$\vec{AB} \cdot \vec{AC} = 0$$​

​$$a \cdot (-5) + 5 \cdot 5 = 0$$​

-5a + 25 = 0

5a = 25

a = 5