Solving for a Cartesian Equation of a Line with Given Vector and Point

Jeann25
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Homework Statement



Find a Cartesian equation for the line p + tv, where p=<2,5> and v=<-3,1>

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The Attempt at a Solution



I know the vector form will look likse <2,5>+t<-3,1>. But I'm not sure how to use this information to find the equation of the line in y=mx+b form.
 
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Just substitute two points into the line y=mx+b to find m and b. Also, the slope m can be obtained immediately from the direction vector v.
 
Or, if you write the vector form in components, eliminate t from this system:

<br /> \left\{ \begin{array}{l}<br /> x = 2 - 3t \\ <br /> y = 5 + t \\ <br /> \end{array} \right.<br />
 
Now, solve one equation for t and (I recomment y= 5+ t) and substitute that into the other equation.
 
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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