Find an equation of the line that satisfies the given conditions. Through 1/2, -2/7 perpindicular to the line passing through

Through (1/2, -2/7) perpendicular to the line 5x-10y=1

5x-10y=1

5x-1=10y

$$y=\frac{5x}{10}-\frac{1}{10}\\\\
y=\frac{1x}{2}-\frac{1}{10}\\\\$$

gradient is 1/2

Gradient of perpendicular is -2/1=-2

now you just have to solve

$$\dfrac{y-\frac{-2}{7}}{x-\frac{1}{2}}=-2$$

I am in a bit of a hurry so i hope i haven't made any mistakes.

You are correct Melody; though it would probably be better to say "rearrange" your last equation, rather than "solve" it.  Doing this you get

$$y=-2x+\frac{5}{7}$$

or, if you prefer, by multiplying through by 7 and rearranging:

$$14x+7y=5$$

Yes, rearrange would have been a better choice of word.

Thanks Alan.

$$\mbox{Find an equation of the line that satisfies the given conditions.}\\
\mbox{Through (1/2, -2/7) perpendicular to the line 5x-10y=1}$$

$$\\5x-10y=1 \quad \Rightarrow \quad g(x) = 0.5x-0.1\\
f(x)=ax^2+bx+c\\
\boxed{\mbox{1. }f(x)=g(x)}\quad and \quad \boxed{\mbox{2. }f'(x)*g'(x)=-1}$$

$$f(x)=ax^2+bx+c\\
2=a*1^2+b*1+c\\
7=a*(-2)^2+b(-2)+c\\
(2)\quad 2=a+b+c\\
(1)\quad 7=4a-2b+c\\
(1)-(2)\quad 5=3a-3b \quad \Rightarrow \boxed{b=a-\frac{5}{3}\mbox{ and }c=\frac{11}{3}-2a}$$

$$f(x)=g(x)\\
ax^2+bx+c=0.5x-0.1\\
ax^2+(a-\frac{5}{3})x+\frac{11}{3}-2a=0.5x-0.1 \\
\Rightarrow \boxed{x_{1,2}=\frac{13-6a}{12a}\pm\sqrt{\frac{(6a-13)^2}{(12a)^2}-\frac{(113-60a)}{30a}}}$$

$$f'(x)*g'(x)=-1\\
f'(x)=2ax+b\\
g'(x)=0.5\\
(2ax+b)*0.5=-1 \quad b=a-\frac{5}{3}\\
\Rightarrow 2ax+b=-2\\
\Rightarrow 2ax+(a-\frac{5}{3})=-2\\
\boxed{2ax+a=-\frac{1}{3}}$$

$$2a\left[ \frac{13-6a}{12a}\pm\sqrt{\frac{(6a-13)^2}{(12a)^2}-\frac{113-60a}{30a}
}\right]+a=-\frac{1}{3}\\
\Rightarrow 324a^2-542.4a-56=0\\
\Rightarrow \boxed{a_{1,2}=\frac{22.6\pm\sqrt{636.76}}{27}
}$$

$$\mbox{a must be negative: }
\boxed{a=\frac{22.6-\sqrt{636.76}}{27}}$$

$${\mathtt{a}} = {\frac{\left({\mathtt{22.6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{636.76}}}}\right)}{{\mathtt{27}}}} \Rightarrow {\mathtt{a}} = -{\mathtt{0.097\: \!559\: \!403\: \!970\: \!835\: \!4}}$$

$${\mathtt{b}} = {\frac{\left({\mathtt{22.6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{636.76}}}}\right)}{{\mathtt{27}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{5}}}{{\mathtt{3}}}} \Rightarrow {\mathtt{b}} = -{\mathtt{1.764\: \!226\: \!070\: \!637\: \!502\: \!1}}$$

$${\mathtt{c}} = {\frac{{\mathtt{11}}}{{\mathtt{3}}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}\left({\frac{\left({\mathtt{22.6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{636.76}}}}\right)}{{\mathtt{27}}}}\right) \Rightarrow {\mathtt{c}} = {\mathtt{3.861\: \!785\: \!474\: \!608\: \!337\: \!5}}$$

The eqaution is:

$$\\f(x)=ax^2+bx+c\\
\boxed{f(x)=-0.0975594039708354x^2-1.7642260706375021x+3.8617854746083375
}$$

Nice one heureka!  Now do it for cubic, quartic, exponential,  ... etc. functions!

Life is complicated enough - let's keep it simple!!

You can have a point from me for lateral thinking anyway!

Hi Alan,

thank you.

Please proof your equation.

Many Greetings

heureka

Rearranging 5x -10y =1, we have  10y = 5x -1....dividing by 10 on both sides, we get

y = (5/10)x - 1/10

Since we want to write an euation of a line perpendicular to this one, the slope of our line has the negative reciprocal slope of  (5/10) = -(10/5) = -2

So, using point - slope form, we have

y - (-2/7) = -2(x - 1/2)

y + 2/7 = -2x +1         subtract 2/7 from both sides

y = -2x + 5/7             I actually like this form because I can see the slope and y intercept right away, but putting it into Ax + By = C form, we have   (by multiplying both sides by 7)

7y = -14x + 5        

14x + 7y = 5

And there you go.......

Sorry

I am not able well enough in English

heureka