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
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
Show Hint
- 5
- $\sqrt{5}$
- $\sqrt{20}$
- 20
The Correct Option is B
Solution and Explanation
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
\[ \text{Distance } = \sqrt{(1-0)^2+(2-0)^2}=\sqrt{1+4}=\boxed{\sqrt{5}}. \]
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