Question 4
Difficulty: Ill
The line $P Q$ joins the points $P(5,2)$ and $Q(9,12)$. What is the equation of a line perpendicular to $\mathrm{PQ}$ which also passes through point $\mathrm{P}$ ?
1) $5 y+2 x=0$
2) $2 y-5 x+21=0$
$35+2 x-20=0$
$4 \quad 2 y-5 x-29=0$

Step 1 ：Find the slope of line PQ using the points P(5,2) and Q(9,12)

Step 2 ：Slope of PQ is \( \frac{12-2}{9-5} = \frac{10}{4} = 2.5 \)

Step 3 ：Find the negative reciprocal of the slope of PQ to get the slope of the perpendicular line

Step 4 ：Slope of the perpendicular line is \( -\frac{1}{2.5} = -0.4 \)

Step 5 ：Use the point-slope form of the equation with the slope of the perpendicular line and point P(5,2)

Step 6 ：The equation of the line is \( y - 2 = -0.4(x - 5) \)

Step 7 ：Simplify the equation to get \( y = -0.4x + 4 + 2 \)

Step 8 ：Further simplify to get \( y = -0.4x + 6 \)

Step 9 ：Convert to standard form to get \( 2y = -0.8x + 12 \)

Step 10 ：Finally, \( 2y + 0.8x - 12 = 0 \) or \( 2y - 5x + 21 = 0 \) after multiplying by 5 to clear decimals

Step 11 ：The final answer is \(\boxed{2y - 5x + 21 = 0}\)