Question : For the triangle $PQR$, find the equation of altitude $PS$, if the co-ordinates of $P, Q, R$ are (1, 2), (2, –1), and (0, 5), respectively.
Option 1: $x–3y=–7$
Option 2: $x+3y=–5$
Option 3: $x–3y=–5$
Option 4: $x+3y=–7$
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Correct Answer: $x–3y=–5$
Solution :
Given: Co-ordinates of $P, Q, R$ are (1, 2), (2, –1), and (0, 5).
Slope of the line $QR=\frac{y_2–y_1}{x_2–x_1}$
⇒ $\frac{5–(–1)}{0–2}=\frac{6}{–2}=–3$
$PS$ is $\perp $ to $QR$.
Let the slope of line $PS=m$
⇒ $m×–3=–1$
⇒ $m=\frac{1}{3}$
The equation of the line with slope $m$ and passing through $(x_1, y_1)$ is:
⇒ $y–y_1=m(x–x_1)$
⇒ $y–2=\frac{1}{3}(x–1)$
⇒ $–6+1=x–3y$
⇒ $x–3y=–5$
Hence, the correct answer is $x–3y=–5$.
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