Question

A(8, −6), B(−4, 2) and C(0, −10) are vertices of a triangle ABC. If P is the mid-point of AB and Q is the mid-point of AC, use co-ordinate geometry to show that PQ is parallel to BC. Give a special name to quadrilateral PBCQ.

Answer 1

P is the mid-point of AB.

So, the co-ordinate of point P are

`((8-4)/2 ,(-6 +2) /2)`

= `(4/2, (-4)/2)`

= (2, –2)

Q is the mid-point of AC.

So, the co-ordinate of point Q are

`((8 + 0)/2, (-6 - 10)/2)`

= `(8/2, (-16)/2)`

= (4, –8)

Slope of PQ =`(-8 + 2)/(4 - 2) = (-6)/2 = -3`

 Slope of BC =`(-10 - 2)/(0 + 4) = (-12)/4 = -3`

Since, slope of PQ = Slope of BC,

∴ PQ || BC

Also, we have :

Slope of PB = `(-2 - 2)/(2 + 4) = (-2)/3`

Slope of QC = `(-8 + 10)/(4 - 10) = 1/2`

Thus, PB is not parallel to QC.

Hence, PBCQ is a trapezium.