KCET · Maths · Straight Lines
The orthocentre of the triangle with vertices \(A(0,0), B\left(0, \frac{3}{2}\right), C(-5,0)\) is
- A \(\left(\frac{5}{2}, \frac{3}{4}\right)\)
- B \(\left(\frac{-5}{2}, \frac{3}{4}\right)\)
- C \(\left(-5, \frac{3}{2}\right)\)
- D \((0,0)\)
Answer & Solution
Correct Answer
(D) \((0,0)\)
Step-by-step Solution
Detailed explanation
Let \(\triangle A O B\) is the given triangle.
Slope of \(A B=\frac{\frac{3}{2}-0}{0+5}=\frac{3}{10}\)
Slope of \(B O=\frac{0-0}{0+5}=0\)
The equation of line passing through \(A\) and perpendicular to \(B O\) is \(y-0=-0\left(x-\frac{3}{2}\right)\)

\(\Rightarrow \quad y=0\)
and equation of line passing through 0 and perpendicular to \(A B\) is \(y-0=-\frac{10}{3}(x-0)\) \(\Rightarrow \quad y=-\frac{10}{3} x\)
The intersection point of Eqs. (i) and (ii) is \((0,0)\), which is the required orthocentre.
Slope of \(A B=\frac{\frac{3}{2}-0}{0+5}=\frac{3}{10}\)
Slope of \(B O=\frac{0-0}{0+5}=0\)
The equation of line passing through \(A\) and perpendicular to \(B O\) is \(y-0=-0\left(x-\frac{3}{2}\right)\)

\(\Rightarrow \quad y=0\)
and equation of line passing through 0 and perpendicular to \(A B\) is \(y-0=-\frac{10}{3}(x-0)\) \(\Rightarrow \quad y=-\frac{10}{3} x\)
The intersection point of Eqs. (i) and (ii) is \((0,0)\), which is the required orthocentre.
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