Question 1088408
Determine weather triangle QRS is a right triangle for the given vertices. Explain Q(18,13) R(17,-3) S(-18,12)

sketch triangle first:

{{{drawing( 600, 600, -25, 25, -25,25,
circle(18,13,.12),circle(17,-3,.12),circle(-18,12,.12),
locate(18,13,Q(18,13)),locate(17,-3,R(17,-3)),locate(-18,12,S(-18,12)),

green(line(18,13,17,-3)),green(line(17,-3,-18,12)),green(line(18,13,-18,12)),
 graph( 600, 600, -25, 25, -25,25, 0)) }}}

{{{RS}}} is hypotenuse if triangle QRS  is a right triangle

find the length: S(-18,12) to R(17,-3)

{{{RS=sqrt((-18-17)^2+(12-(-3))^2)}}}

{{{RS=sqrt((-35)^2+15^2)}}}

{{{RS=sqrt(1450)}}}

{{{RS=sqrt(25*58)}}}

{{{RS=5sqrt(58)}}}


legs should be: {{{QR}}} and {{{QS}}}

Q(18,13) to R(17,-3)
{{{QR=sqrt((18-17)^2+(13-(-3))^2)}}}
{{{QR=sqrt(1^2+(13+3)^2)}}}
{{{QR=sqrt(1+16^2)}}}
{{{QR=sqrt(257)}}}

Q(18,13) to S(-18,12)

{{{QS=sqrt((18-(-18))^2+(13-12)^2)}}}
{{{QS=sqrt((18+18)^2+(1)^2)}}}
{{{QS=sqrt(36^2+1)}}}
{{{QS=sqrt(1296+1)}}}
{{{QS=sqrt(1297)}}}

check if: {{{(SR)^2=(QR)^2+(QS)^2}}}

{{{(5sqrt(58))^2=(sqrt(257))^2+(sqrt(1297))^2}}}

{{{1450=257+1297}}}

{{{1450<>1554}}}=> proves that triangle QRS  is {{{not}}} a right triangle

because 
a.
{{{QR=sqrt(257)}}}
{{{QS=sqrt(1297)}}}
{{{RS=5sqrt(58)}}}
{{{(SR)^2<>(QR)^2+(QS)^2}}}