JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Straight Lines

Let

C (α, β)

be the circumcenter of the triangle formed by the lines

4 x + 3 y = 69

4 y - 3 x = 17

and

x + 7 y = 61

Then

(α - β)^{2} + α + β

is equal to

Answer: (B)

Solution

The vertices are the pairwise intersections of the lines: solving

4 x + 3 y = 69

with

4 y - 3 x = 17

gives

(9, 11)

;

4 x + 3 y = 69

with

x + 7 y = 61

gives

(12, 7)

; and

4 y - 3 x = 17

with

x + 7 y = 61

gives

(5, 8)

.

The lines

4 x + 3 y = 69

and

4 y - 3 x = 17

have normals

(4, 3)

and

(- 3, 4)

with

(4) (- 3) + (3) (4) = 0

, so they are perpendicular and the triangle is right-angled at

(9, 11)

. For a right triangle the circumcentre is the midpoint of the hypotenuse, which joins

(12, 7)

and

(5, 8)

C (α, β) = (\frac{12 + 5}{2}, \frac{7 + 8}{2}) = (\frac{17}{2}, \frac{15}{2}) .

Therefore

(α - β)^{2} + α + β = (\frac{17}{2} - \frac{15}{2})^{2} + \frac{17}{2} + \frac{15}{2} = 1 + 16 = 17.

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Single correct option