# Verify if the triangle PBC has a side on the line d:x+3y-4=0? P(2,3),B(-2,2),C(2,2/3)

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We have the triangle PBC and the line x+3y-4=0

if one of the sides of the triangle is on the line, then the line should verify 2 points from the triangle:

Let us verify:

The point P(2,3)

2+ 3(3) -4= 2+9 -4 = 6

Then P is not on line d.

The point B(-2,2):

-2+3(2)-4= -2 + 6-4= 0

Then B is on line d:

The point C(2, 2/3)

2+ 3(2/3) -4= 2+ 2-4 =0

Then the point C in on line d

Then there is 2 points B and C located on the line d, then one of the triangle sides is on the line d.

We'll state that a side of the triangle is on the line d, if and only if, the coordinates of the side verify the equation of the line.

For, the beginning, let's verify if the point P is on the line d:

xP + 3yP - 4 = 0

2 + 3*3 - 4 = 0

2+9-4=0

7=0 impossible, so P does not belong to the line d.

Let's verify if the point B is on the line d:

xB + 3yB - 4 = 0

-2 + 3*(2) - 4 = 0

4-4 = 0

0=0 true

The point B is on the line d.

Let's check if the point C is on the line d.

xC + 3yC - 4 = 0

2 + 3*2/3 - 4 = 0

2+2-4=0

0=0

The point C is on the line d.

Because the points B and C are on the line d, that means that the side BC is overlapping the line d.

We see whivh of the coordinates of the points P, B and C satisfy the equation of the line d: x+3y-4 = 0.

P(2,3): 2+3(3)-4 = 11-4 = 7 . P is not satisfying.

B(-2,2): -2+3(2) -4 = 4-4 = 0. B satisfies.satisfies.

C(2.2/3): 2+3(2/3)-2 = 2+2-4 = 0. C satisfies.

Therefore the side BC of the triangle PBC is the line x+3y-4 = 0.