Question

The coordinates of the orthocenter of the triangle whose sides are represented by the lines \(4x-7y+10 = 0\), \(x+y = 5\) and \(7x+4y = 15\) are \(\ldots\)

• A. \((1,2)\)
• B. \((1,-2)\)
• C. \((-1,2)\)
• D. \((-1,-2)\)

Hint:

Check whether two sides are perpendicular. If so, the vertex between them is the orthocentre.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The orthocentre is the point where the three altitudes meet. In a right-angled triangle, it is the vertex at the right angle.

Step 2: Key Formula or Approach:
Two lines are perpendicular if the product of their slopes is -1.

Step 3: Detailed Explanation:
Slope of \(4x - 7y + 10 = 0\): \(m_1 = \dfrac{4}{7}\).
Slope of \(7x + 4y = 15\): \(m_3 = -\dfrac{7}{4}\).
\[ m_1 m_3 = \frac{4}{7}\times\left(-\frac{7}{4}\right) = -1 \]
So these two sides are perpendicular and the triangle is right-angled at their intersection. Solve them together:
\[ 4x - 7y = -10, \qquad 7x + 4y = 15 \]
Multiply the first by 4 and the second by 7:
\[ 16x - 28y = -40, \qquad 49x + 28y = 105 \]
Add the two: \(65x = 65\), so \(x = 1\). Then \(4 - 7y = -10\) gives \(y = 2\).
The orthocentre is \((1, 2)\). It also lies on neither of the other options' coordinates, such as \((1,-2)\).

Final Answer:
The orthocentre is \((1, 2)\), option (A). \[ \boxed{(1,2) \text{ (A)}} \]