Question about 'showing' something is a chord (1 Viewer)

[asianese]

I think someone posted a similar question but I wanted to clarify:

'If

I have subbed in P, and shown that the LHS=RHS, and then for Q also, LHS=RHS

I tried doing it from y-y1=(m)(x-x1) but it got way too messy. Is this a valid 'show' that the chord PQ has that equation - just subbing it in and then showing that both satisfy?

Thanks

 

[lolcakes52]

if it was a one or two marker then subbing in the points of p and q would be fine. Otherwise you would have to derive the equation. I would use y-y1 = m(x-x1) then solve using by resolving tan and sec in terms of sin and cos then using sin(A-B) and stuff to simplify the gradient.

 

[asianese]

Damn...it was 4 marks so I guess i'd have to derive it. Erghhh ok thanks

 

[Nooblet94]

I think someone posted a similar question but I wanted to clarify:

'If

I have subbed in P, and shown that the LHS=RHS, and then for Q also, LHS=RHS

I tried doing it from y-y1=(m)(x-x1) but it got way too messy. Is this a valid 'show' that the chord PQ has that equation - just subbing it in and then showing that both satisfy?

Thanks

I just did it and it's not messy at all - quite the opposite in fact. When you find the expression for the gradient, simplify it before you sub it into the point-gradient form.

 

[asianese]

Lols i probably did something wrong then. No biggie.