JEE Main 2025 (Online) 4th April Morning ShiftQuestion 1
Let the three sides of a triangle are on the lines $4x-7y+10=0$, $x+y=5$ and $7x+4y=15$. 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
Points and Straight Lines
Solution
Key fact. In a right-angled triangle the orthocentre is the vertex of the right angle (the two legs are themselves altitudes). Step 1 - First triangle. Slopes: $4x-7y+10=0$ has slope $\tfrac47$; $7x+4y=15$ has slope $-\tfrac74$. Their product is $-1$, so these two sides are perpendicular. The orthocentre is their intersection: $$16x-28y=-40,\quad 49x+28y=105\ \Rightarrow\ 65x=65\ \Rightarrow\ x=1,\ y=\tfrac{4+10}{7}=2 .$$ Orthocentre $H_1=(1,2)$. Step 2 - Second triangle. $x=0$ and $y=0$ meet at a right angle at the origin, so $H_2=(0,0)$. Step 3 - Distance. $$H_1H_2=\sqrt{1^2+2^2}=\sqrt5 .$$ Answer: (C)