Coordinate Geometry (1 Viewer)

[FDownes]

Having trouble with this question, can anyone show me the best way to tackle it? It asks;

On a number plane, the points P, Q and R have coordinates of (3, 0), (0, 2) and (6, 4) respectively. Find the shortest distance from the point R to the line PQ.

Do I simply find the equation of the line perpendicular to PQ and solve simultaneously, or is there a better way?

 

[Aerath]

Find equation of line PQ, and find perpendicular distance from R to PQ.

I think I've done it right (well, I hope so).
Edit: Edits in blue.

 

[conics2008]

FIND EQN OF LINE PQ ...

Then USE P.D Distance frompoint R to the line PQ.

 

[FDownes]

Yes, that's right. Thanks. Time for another;

In the diagram, the line l₁ has equation 2x - 3y + 5 = 0 and the line l₂ passes through the points P(0, 6) and Q (4, 0). Find the area of the triangle RSQ.

 

[Aerath]

Find perpendicular distance of point S to the x axis. Then use Area = 0.5xbxh, where b = RQ and h = SN (perpendicular distance).

 

[tacogym27101990]

find RQ
and find the perp distance from s to the line x-axis (rq) equation is y=0

then use A=0.5bh

 

[conics2008]

just find point of intersection between those two lines. and see that point you have

get the height of the triangle by the y coordinate..

hence 1/2 * base whichi s RQ times the height.

good day.