- #1

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- 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

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

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