Question 4
Difficulty: Ill
The line $PQ$ 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) $5y+2x=0$
2) $2y-5x+21=0$
$35+2x-20=0$
$42y-5x-29=0$

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 $\boxed{{\displaystyle 2y-5x+21=0}}$