Question

I challenge you.
From the point P(-2,4), if PQ is drawn perpendicular to the line 7x-24y+10=0, find the equation of the line PQ. Also determine the length of PQ.​

Answer 1

The perpendicular line to 7x-24y+10=0 is

24x+7y+k=0

The line passes point P

Plug in x=-2, y=4

-48+28+k=0

k=20

The line is

24x+7y+20=0

To find the point of intersection Q

Solve the system equation of two graphs

$\left \{ {{7x-24y+10=0} \atop {24x+7y+20=0}} \right.$

The solution is $(x,y)=(-\frac{22}{25} ,\frac{4}{25} )$ and it is the intersection

∴$Q(-\frac{22}{25} ,\frac{4}{25} )$

To find the length of PQ

Use the distance formula

(Distance) = $\sqrt{(-2-\frac{22}{25} )^2+(4-\frac{4}{25} )^2}$

$=\sqrt{(-\frac{72}{25} )^2+(\frac{96}{25} )^2}$

$=\frac{1}{25} \times \sqrt{(72)^2+(96)^2}$

$=\frac{1}{25} \times \sqrt{(2^3\times 3^2)^2+(2^5\times 3)^2}$

$=\frac{1}{25} \times \sqrt{2^6\times 3^4+2^{10}\times 3^2}$

$=\frac{1}{25} \times \sqrt{2^6\times 3^2\times (3^2+2^4)}$

$=\frac{1}{25} \times 2^3\times 3\times \sqrt{2^4+3^2}$

$=\frac{24}{25} \times \sqrt{25}$

$=\frac{24}{5}$

ANSWER

The line is determined as 24x+7y+20=0

The distance is $\frac{24}{5}$ or 4.8