Find the equation of the straight line

[chwala]

Homework Statement

See attached

Relevant Equations

straight line equations

Find the question here ( this one is a pretty easy question).

I have attempted this in the past using varied approach, this is in reference to part(b) of the question... i have previously used pythagoras theorem to finding co-ordinates of $B$ and $D$...

Anyway find my current approach on this,

The gradient of the line $AC=\frac{-1}{3}$, it follows that the gradient of line $BD=3$, with mid-point of $BD$=$(4,3)$.
For part (a),
The equation of line $BD$, is given by
$y=mx+c$
$3=12+c$
$c=-9$,
Therefore, $y=3x-9$

For part (b),
Let the co-ordinates of $B$=$(x_1,y_1)$
$D$=$(x_2,y_2)$
then it follows that (using gradient),
$$\frac {3}{1}=\frac {3-y_2}{4-x_2}$$
$$⇒D(x_2,y_2)=(3,0)$$
Also,
$$\frac {3}{1}=\frac {y_1-3}{x_1-4}$$
$$⇒B(x_1,y_1)=(5,6)$$

Any other way for part (2) only. Cheers

 

[anuttarasammyak]

AB is perpendicular to BC.

C and D are on the circle centered at (4,3) with radius $\sqrt{10}$.

 

[chwala]

$$(x-4)^2+(y-3)^2=10$$
$$(x-4)^2+(3x-12)^2=10$$
$$x^2-8x+15=0$$
$x_1=5,⇒y_1=15-9=6$
$x_2=3⇒y_2=9-9=0$

Bingo! anutta...

 

[anuttarasammyak]

The equation of the former of #2 is
$$(x-7)(x-1)+(y-2)(y-4)=0$$
which coincides with the equation of the circle.

 

[chwala]

The equation of the former of #2 is
$$(x-7)(x-1)+(y-2)(y-4)=0$$
which coincides with the equation of the circle.

Agreed, i just used your approach in post $3$ unless you want me to show all steps. Its clear mate.