Question

an isosceles triangle ABC where AB=BC.the line BD is perpendicular to the line AC.given that the gradient of line AC=-2/3.find coordinate of point A

Answer 1

To find the coordinate of A you can first determine the coordinate of D and apply the midpoint rule.

And to find the coordinates of D you must solve equations of AC and BD simultaneously.

AC has equation


`y=-(2)/(3) x+ (20)/(3)`

Or


`3y+2x=20 --(1)`

BD has equation


`y= (3)/(2)x -2`

Or


`2y-3x=-4 --(2)`

Multiply equation (1) by 2 and multiply equation (2) by 3 to obtain,


`6y+4x=40 --(3)`


`6y-9x=-12 --(4)`

Equation (3) minus equation (4)

gives


`13x=52`

`\therefore x=4`

Putting the value of x=4 in equation (4) gives


`6y-9(4)=-12`

This implies that


`6y-36=-12`


`6y=36-12`


`6y=24`

`\therefore y=4`






Solving the equations of AC and BD simultaneously give the coordinates of D to be;


`x= 4,y=4`

Let


`A(m,n)` be the coordinates.

Since D is the midpoint of AC


`(m+10)/(2)=4`

and


`(n+0)/(2)=4`

Solving for m and n, we have


`m=4×2-10=-2`

And


`n=4×2=8`

Therefore A has coordinates

[tex] (-2,8)/tex]