Question

Let the three sides of a triangle are on the lines \( 4x - 7y + 10 = 0,\quad x + y = 5,\quad 7x + 4y = 15 \).
Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines \( x = 0,\quad y = 0,\quad x + y = 1 \) is

• A. 5
• B. $\sqrt{5}$
• C. $\sqrt{20}$
• D. 20

Hint:

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

Answer 1

The Correct Option is B

Lines of the first triangle: \(4x-7y+10=0,\; x+y=5,\; 7x+4y=15\).

Concept Used:

In a right triangle, the orthocenter is the vertex where the two perpendicular sides meet.

Step-by-Step Solution:

Step 1: Check perpendicular sides.

Slopes: \(m_1=\dfrac{4}{7}\) for \(4x-7y+10=0\), and \(m_3=-\dfrac{7}{4}\) for \(7x+4y=15\). Since \(m_1 m_3=-1\), these two lines are perpendicular. Hence the triangle is right-angled at their intersection.

Step 2: Orthocenter \(H\) of the first triangle is their intersection.

\[ \begin{cases} 4x-7y+10=0\\ 7x+4y-15=0 \end{cases} \Rightarrow \begin{aligned} 4x-7y&=-10\\ 7x+4y&=15 \end{aligned} \Rightarrow x=1,\; y=2. \] So \(H=(1,2)\).

 

Step 3: Orthocenter of the triangle formed by \(x=0,\; y=0,\; x+y=1\).

It’s a right triangle at the origin, so orthocenter \(H_0=(0,0)\).

Final Computation & Result

\[ \text{Distance } = \sqrt{(1-0)^2+(2-0)^2}=\sqrt{1+4}=\boxed{\sqrt{5}}. \]