5. Find the equation of the straight line that contains the POI of 3x+2y-12=0 and 5x-y-7=0 and that:
c) is perpendicular to the line y=5
I can do it doing simultaneous equations but thats beta shit and takes to long
How do i do it using k's
3x + 2y - 12 + k(5x - y - 7) = 0
(3+5k)x + (2-k)y + (-12-7k) = 0
WTF DO I DO NOW
Yeah i know the line has to be x = something because it has to be a straight line up. But im confused because with a straight line up the gradient is infinity so wtf do i do
HSC 2018: [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design]
Goal: ≥92.00 for B Advanced Mathematics (Hons) / B Engineering (Hons) (Computer) at UNSW
Do you do (2-k) = 0??? therefore k = 2?
someone plss a nibba got a math exam tmrw
fan96 liked your answer, meaning it's correct.
Here is a quick explanation. The line required only has an x-value since it's perpendicular to y=5. It also has to go through the point of intersection between the two lines mentioned in the question, so the first thing you need to think about is finding that point of intersection I(x,y) BUT since the y-value is irrelevant here (since the equation of our line is x='something'), we only need to find the x value;
Two ways you can go about it ( which are essentially the same); either find a value of k such that y is eliminated which you have done correctly or eliminate y using the elimination method of simultaneous equations, which is basically the same as the first method.
Last edited by KAIO7; 12 Jun 2018 at 12:28 AM.
Bsc (Advanced Mathematics)/(Neuroscience)/ TSP
Thank u so much ur amazing
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