If the orthocenter of the triangle formed by the lines y = x + 1, y = 4x - 8 and y = mx + …

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 :

Accepted Answer

Solve line PQ \& QR

$$\begin{aligned}
& \text { Point } \mathrm{Q}\left(\frac{1-\mathrm{c}}{\mathrm{~m}-1}, \frac{1-\mathrm{c}}{\mathrm{~m}-1}+1\right) \
& \mathrm{m}_{2 \mathrm{H}}=\frac{\frac{1-\mathrm{c}}{\mathrm{~m}-1}+2}{\frac{1-\mathrm{c}}{\mathrm{~m}-1}-3}=\frac{1-\mathrm{c}+2 \mathrm{~m}-2}{1-\mathrm{c}-3 \mathrm{~m}+3}=-\frac{1}{4} \quad\text{..... (1)}\
& \because \mathrm{~m}_{\mathrm{PH}}=\frac{5}{0} \rightarrow \infty \
& \Rightarrow \text { Slope of line } \mathrm{QR}(\mathrm{~m})=0
\end{aligned}$$

Put value of $m$ in equation (1)

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

so $\mathrm{m}-\mathrm{c}=0$ Ans.

Solutions · 1