P,Q,R and S have coordinates (13,5) (3,7) and (5,-3) respectively and are three vertices of a rhombus PQRS.

Find the equations of PS and QS

Hence find the oordinates of the fourth vertex S

Thanks for any help.

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# Straight Line Geometry Problem

Started by gary, Dec 21 2004 07:18 PM

2 replies to this topic

### #1

Posted 21 December 2004 - 07:18 PM

### #2

Posted 21 December 2004 - 09:16 PM

could you check you have copied the question down properly because i think the bit asking for equations may be wrong

If i am not here i am somewhere else

### #3

Posted 21 December 2004 - 09:42 PM

actually maybe its not

equation of QS

gradient = that of PR so use gradient equation y2-y2/x2-x1 = 1

using y-b = m(x-a) sub in Q for a and b and you get

y=x + 4

equation of PS

gradient is found using mQR * mPs = -1

find gradient of QR (same as above) to get -5

sub in and you get -5 * 1/5 = -1 => gradient = 1/5

use y-b=m(x-1) again to get equation which is 5y = x +12

to find S:

we know 2 equations: y = x + 4 and 5y = x + 12

at point s the y co-ordinates will equal the same in each equation ie x+4=.2x + 2.4

we can use this to find x

x+4=.2x + 2.4

0.8X + 4 = 2.4

0.8x= -1.6

x = -2

sub in x in equation 1

y = -2 + 4

y = 2

S(-2,2)

equation of QS

gradient = that of PR so use gradient equation y2-y2/x2-x1 = 1

using y-b = m(x-a) sub in Q for a and b and you get

y=x + 4

equation of PS

gradient is found using mQR * mPs = -1

find gradient of QR (same as above) to get -5

sub in and you get -5 * 1/5 = -1 => gradient = 1/5

use y-b=m(x-1) again to get equation which is 5y = x +12

to find S:

we know 2 equations: y = x + 4 and 5y = x + 12

at point s the y co-ordinates will equal the same in each equation ie x+4=.2x + 2.4

we can use this to find x

x+4=.2x + 2.4

0.8X + 4 = 2.4

0.8x= -1.6

x = -2

sub in x in equation 1

y = -2 + 4

y = 2

S(-2,2)

If i am not here i am somewhere else

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