JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Straight Lines

Let the three sides of a triangle are on the lines

4 x - 7 y + 10 = 0, x + y = 5

and

7 x + 4 y = 15

. Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines

x = 0, y = 0

and

x + y = 1

(A)5
(B) $5$ ✓ Correct
(C) $20$
(D)20

Answer: (B)

Solution

The lines

4 x - 7 y + 10 = 0

and

7 x + 4 y = 15

have slopes

\frac{4}{7}

and

- \frac{7}{4}

, whose product is

- 1

, so the triangle is right-angled and its orthocentre is the right-angle vertex. Solving these two lines together gives

x = 1, y = 2

, so that orthocentre is

(1, 2)

. The triangle on

x = 0, y = 0, x + y = 1

is right-angled at the origin, so its orthocentre is

(0, 0)

. The distance between

(1, 2)

and

(0, 0)

1 + 4 = 5

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Single correct option

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