Mathematics · Straight lines

JEE Main 2025 — 4 April, Morning Shift — Question 23

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

$x+y=1$ is

Answer: B

$A$ is orthocentre of above $\Delta$ .

$O$ is orthocentre of above $\Delta$ .

Analyzing the First Triangle Let the lines be:

L_1: 4x - 7y + 10 &= 0 \implies m_1 = \frac{4}{7} \\ L_2: x + y &= 5 \implies m_2 = -1 \\ L_3: 7x + 4y &= 15 \implies m_3 = -\frac{7}{4} \end{aligned}$$ Since $m_1 \cdot m_3 = \frac{4}{7} \cdot (-\frac{7}{4}) = -1$, $L_1 \perp L_3$. In a right-angled triangle, the orthocenter is the vertex containing the right angle. Solving $L_1$ and $L_3$:

4x - 7y = -10 \quad \dots (i)

7x + 4y = 15 \quad \dots (ii)

Solving these simultaneously gives $x = 1$ and $y = 2$. Thus, $H_1 = (1, 2)$. Analyzing the Second Triangle The lines $x=0$, $y=0$, and $x+y=1$ form a right-angled triangle at the origin. Therefore, the orthocenter $H_2 = (0, 0)$. Distance Calculation The distance $d$ between $H_1(1, 2)$ and $H_2(0, 0)$ is:

d = \sqrt{(1-0)^2 + (2-0)^2} = \sqrt{1 + 4} = \sqrt{5}

Answer key and solution verified before publishing.

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.