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Coordinate geometry (1 Viewer)

tommykins

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回复: Re: Coordinate geometry

A line that's perpindicular to the x axis is the line x = a where a is any integer, and vice versa.

I made a mistake on my working, I apologise.
 

Fortian09

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Re: 回复: Re: Coordinate geometry

Hey i got a qn?


If i'm trying to find the distance between two parallel lines, i can find the intercept for one of them and use perpendicular distance formula correct?

And... problems...


1.If the distance from the point (2,1) to the line 2x-y+k-0 is 2rt5, find the value of k.

2. The equations of the parallel sides of a trapezium are 2x+y-5=0 and 4x+2y-7=0.
Find the length of its altitude.
 

tommykins

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Re: 回复: Re: Coordinate geometry

Just pick x = 0, find the y value for one line and use that point in your perpindicular distance formula.

1. 2sqrt5 = 2(2) - 1(1) + k/sqrt[2²+1²]
2*5 = 4 - 1 + k
10 = 3 + k
k = 7
 
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lyounamu

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Re: 回复: Re: Coordinate geometry

Fortian09 said:
Hey i got a qn?


If i'm trying to find the distance between two parallel lines, i can find the intercept for one of them and use perpendicular distance formula correct?

And... problems...


1.If the distance from the point (2,1) to the line 2x-y+k-0 is 2rt5, find the value of k.

2. The equations of the parallel sides of a trapezium are 2x+y-5=0 and 4x+2y-7=0.
Find the length of its altitude.
1. Hint: use the perpendicular distance formula where (2,1) = (x1,x2) and
a=2, b=-1 and c=k

The formula is: AV(ax1+by1+c)/SR(a^2+b^2)

2. Hint: there are two equations. For one of the equation, find the x and y value, e.g. when x=1, y = 3 for the 1st equation: 2x+y-5=0.

Then use the perpendicular distance formula where (1,3) = (x1,x2) and a=4, b=2 and c=-4 and this value comes from the 2nd equation: 4x+2y-7=0

Good luck, I just think that the act of gemilit chasidim is more important than the act of tzedekah.

Damn, the purpose of my act of gemilit chasidim is totally crushed due to tommy...
 

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