Question

A triangle ∆P QR has vertices P(3, 2, −4), Q(1, 0, −4), R(2, 1, 1). Use the distance formula to decide which one of the following properties the triangle has. 1. isoceles with |P Q| = |P R| 2. isoceles with |RP| = |RQ| 3. not isoceles 4. isoceles with |QP| = |QR|

Answer 1

Answer:

a. FALSE

b.TRUE

C. FALSE

Explanation:

The formula fot the distance between two points is given as

$d=\sqrt{(x_{2}-x_{1})^{2} +(y_{2}-y_{1})^{2} +(z_{2}-z_{1})^{2}}$

hence we determine the distances between all the points

a.P(3,2,-4), Q(1,0,-4), R(2,1,1)

$PQ=\sqrt{(1-3)^{2} +(0-2)^{2} +(-4-(-4))^{2}}\\PQ=\sqrt{4+4+0}\\ PQ=\sqrt{8}$

For point PR

we have

$PR=\sqrt{(2-3)^{2} +(1-2)^{2} +(1-(-4))^{2}}\\PR=\sqrt{1+1+9}\\ PR=\sqrt{11}$

$|PQ|\neq |PR|$

B. For point RP

$RP=\sqrt{(3-2)^{2} +(2-1)^{2} +(-4-1)^{2}}\\RP=\sqrt{1+1+25}\\ RP=\sqrt{27}$

for point RQ  we have

$RQ=\sqrt{(1-2)^{2} +(0-1)^{2} +(-4-1)^{2}}\\RQ=\sqrt{1+1+25}\\ RQ=\sqrt{27}$

|RP|=|R Q|

C.

$QP=\sqrt{(3-1)^{2} +(2-0)^{2} +(-4+4)^{2}}\\QP=\sqrt{4+4+0}\\ QP=\sqrt{8}$

For point Q R

$QR=\sqrt{(2-1)^{2} +(1-0)^{2} +(1-(-4))^{2}}\\QR=\sqrt{1+1+9}\\ QR=\sqrt{11}$

$QP\neq QR$