What is the equation of a straight line passing through two points?

jahlin
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Homework Statement



whats the equation of a straight line passing through (1,4) and (5,1)?

Homework Equations


y=mx+c


The Attempt at a Solution



i found the equation to be y=-(3/4)x is it right?
Thanks
 
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No it's not right. How did you get y=(-3/4)x? The slope is right. The c part isn't.
 
well it doesn't intercept with the y-axis so i assumed it to be zero..
 
hi
actually the straight line intercepts with y-axis. you can substitute the x value and y value into the equation you found. Obviously they cannot fit.
so the gradient you get is -3/4, which is the correct one.
y = (-3/4)x + c;
one way to find c is by substituting (1, 4) into equation stated above.
then the correct equation of the straight line passing through two points can be found.

Cheers,
weesiang_loke
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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