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

$y=-\frac{2}{3} x+ \frac{20}{3}$

Or

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

BD has equation

$y= \frac{3}{2}x -2$

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

$\frac{m+10}{2}=4$

and

$\frac{n+0}{2}=4$

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]