If the orthocenter of the triangle formed by the lines y = x + 1, y = 4x – 8 and y = mx + c is at (3, –1), then m – c is: [2025]

Question

If the orthocenter of the triangle formed by the lines y = x + 1, y = 4x – 8 and y = mx + c is at (3, –1), then m – c is: [2025]

Smart Achievers · Accepted Answer

(2)

Using lines PQ and QR, we get

Point Q = $(\frac{1 - c}{m - 1}, \frac{1 - c}{m - 1} + 1)$

$m_{Q H} = \frac{\frac{1 - c}{m - 1} + 2}{\frac{1 - c}{m - 1} - 3} = \frac{1 - c + 2 m - 2}{1 - c - 3 m + 3} = - \frac{1}{4} [∵ Q H ⊥ P R]$ ... (i)

$∵ m_{P H} = \frac{5}{0} \to \infty \Rightarrow Slope of line Q R (m) = 0$

Put value of m in equation (i), we get

$\frac{1 - c - 2}{1 - c + 3} = - \frac{1}{4} \Rightarrow c = 0$

$\Rightarrow m - c = 0$ .

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