Question

How to do this last coordinate geometry question? Thanks in advance!

Answer 1

`C=(6,3),B=(8,9)" and "D=(4,2)`

Explanation:

`"the coordinates can be found using "color(blue)"simultaneous equations"`

`"point C lies on both the equations"`

`y=1/2xto(1)`

`y=3x-15to(2)`

`"substitute "y=1/2x" into equation "(2)`

`rArr3x-15=1/2x`

`rArr5/2x=15rArrx=6`

`"substitute "x=6" in "(1)`

`rArry=1/2xx6=3`

`rArr"coordinates of C"=(6,3)`
`color(blue)"--------------------------------------------"`

`"obtain the equation of AB with m"=1/2`

`y=1/2x+blarrcolor(blue)"is partial equation"`

`"to find b substitute "(2,6)" into the partial equation"`

`6=1+brArrb=5`

`"equation of AB is "y=1/2x+5`

`"point B lies on both the equations"`

`y=1/2x+5to(1)`

`y=3x-15to(2)`

`rArr3x-15=1/2x+5`

`rArr5/2x=20rArrx=8`

`"substitute "x=8" in "(1)`

`rArry=(1/2xx8)+5=9`

`rArr"coordinates of B "=(8,9)`
`color(blue)"--------------------------------------"`

`m_(AD)=-1/(1/2)=-2`

`y=-2x+blarr"using point "(2,6)`

`6=-4+brArrb=10`

`"equation of AD is "y=-2x+10`

`"point D lies on both the equations"`

`y=1/2xto(1)`

`y=-2x+10to(2)`

`rArr1/2x=-2x+10`

`rArr5/2x=10rArrx=4`

`"substitute "x=4" in "(1)`

`rArry=1/2xx4=2`

`rArr"coordinates of D "=(4,2)`
`color(blue)"----------------------------------"`

Answer 2

See below.

Explanation:

The line passing through points A and B, doesn't have equation `y=1/2x`. The equation has an y intercept. we can find this since we know a point on the line, namely A( 2 , 6 ).

`6=1/2(2)+b=>b=5` `:.` `y=1/2x+5`

This equation and the equation of CB `y=3x-15` intersect at point B. So:

`1/2x+5=3x-15=>x=8`

`:.`

`y=1/2(8)+5=>y=9`

Coordinates of B`->color(blue)( ( 8 , 9 )`

The line OC intersects the lineCB at point C:

`:.`

`1/2x=3x-15=>x=6`

`y=1/2(6)=>y=3`

Coordinates of C`-> color(blue)(( 6 , 3 )`

The line AD is perpendicular to the line OC, so will have a gradient:

`-2` ( the product of the gradients of perpendicular line is `-1`

i.e.

`-2*1/2=-1`

This line passes through A `( 2, 6 )`

`:.`

`y-6=-2(x-2)`

`y=-2x+10`

This intersects with line OC at point D

`:.`

`-2x+10=1/2x=>x=4`

`y=1/2(4)=>y=2`

Coordinates of point D`->color(blue)( ( 4 , 2 )`

Answer 3

`B(8,9)`.

Explanation:

I would like to place my comment on finding the co-ords. of the

point `B` in an another way.

Recall that the diagonals `OB and AC` bisect each other, meaning

that their mid-points are the same.

So, if `B=B(x,y),` then,

`"Mid-pt. of "OB"=the mid-pt. "of AC`,

`rArr ((0+x)/2,(0+y)/2)=((2+6)/2,(6+3)/2)`,

`rArr x=8,y=9," giving, "B(x,y)=B(8,9)`.