JEE Mathematics 2025 Question with Solution

Given: The sides of the first triangle lie on $4x-7y+10=0$, $x+y=5$, and $7x+4y=15$.

Find: The distance between its orthocenter and the orthocenter of the triangle formed by $x=0$, $y=0$, and $x+y=1$.

The orthocenter of a triangle is the intersection of its altitudes.

For the first triangle, check whether any two sides are perpendicular.

The line $4x-7y+10=0$ has slope $m_1=\frac{4}{7}$. The line $7x+4y=15$ has slope $m_3=-\frac{7}{4}$. Since

$$m_1m_3=\frac{4}{7}\cdot\left(-\frac{7}{4}\right)=-1$$

these two lines are perpendicular. Therefore, the triangle is right-angled at their point of intersection, so its orthocenter is that intersection point.

Solve

$$\begin{aligned} 4x-7y&=-10 \\ 7x+4y&=15 \end{aligned}$$

which gives $x=1$ and $y=2$. Hence the orthocenter of the first triangle is $H=(1,2)$.

Now consider the triangle formed by $x=0$, $y=0$, and $x+y=1$. This is a right triangle at the origin, so its orthocenter is $H_0=(0,0)$.

Therefore, the required distance is

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

So the correct option is B.

Given: The triangle is formed by the lines $4x-7y+10=0$, $x+y=5$, and $7x+4y=15$.

Find: The distance from its orthocenter to the orthocenter of the triangle formed by $x=0$, $y=0$, and $x+y=1$.

Principle used: In a right triangle, the orthocenter lies at the vertex of the right angle.

For $4x-7y+10=0$,

$$7y=4x+10$$

so its slope is $\frac{4}{7}$.

For $7x+4y=15$,

$$4y=-7x+15$$

so its slope is $-\frac{7}{4}$.

Their product is

$$\frac{4}{7}\cdot\left(-\frac{7}{4}\right)=-1$$

Hence these two sides are perpendicular.

So the orthocenter of the first triangle is the intersection of

$$4x-7y+10=0$$

and

$$7x+4y=15$$

Using

$$\begin{aligned} 4x-7y&=-10 \\ 7x+4y&=15 \end{aligned}$$

we obtain $H=(1,2)$.

The second triangle is bounded by the coordinate axes and the line $x+y=1$, so it is right-angled at $O=(0,0)$. Therefore its orthocenter is $H_0=(0,0)$.

Now compute the distance:

$$\begin{aligned} HH_0&=\sqrt{(1-0)^2+(2-0)^2} \\ &=\sqrt{1+4} \\ &=\sqrt{5} \end{aligned}$$

Therefore, the distance is $\sqrt{5}$, so the correct option is B.