# Thread: Determine point on the line 5x-4y-2=0

1. ## Determine point on the line 5x-4y-2=0

The problem

Determine the point on the line 5x-4y-2=0 where the normal that passes through the point (7,-2) intersects the line.

Attempt at a solution

The normal to y= mx+ b has slope -1/m.
5x- 4y- 2= 0 is the same as 4y= 5x- 2 or y= (5/4)x- 1/2. Its slope is 5/4 so the slope of any normal to it is -4/5.

What is the equation of the line through (7, -2) with slope -4/5?

r = (7,-2) + t (-4,5)

But where do I go from here? Can anyone point me in the right direction?

Sorry if I'm being a little frustrating, I had pneumonia and missed nearly all the lessons, and I'm really struggling to catch back up.

2. What is the equation of the line through (7, -2) with slope -4/5?
Equation of a line passing through (a,b) having slope "M" is

$(y-b)= M(x-a)$

Use it you are going the correct way

3. Originally Posted by Random-Hero-
The problem
Determine the point on the line 5x-4y-2=0 where the normal that passes through the point (7,-2) intersects the line.
The normal is $4x+5y-18=0$.

4. So what do I do to find the actual point?

5. Originally Posted by Random-Hero-
So what do I do to find the actual point?
-First you got that equation of normal

-this intersects with the initial line

-Solve these two equations to get the point of intersection

6. After solving the two I got the point (-2,-3)

Is this correct?

7. Scratch that, I got (2,2)

8. The normal is .

The line is $5x- 4y- 2= 0$

you need to find the point of intersection

which is done as

$5x-4y-2=0$

$5x = 4y+2$

$x= \frac{4y+2}{5}
$

Put it in equation of normal

4(\frac{4y+2}{5} + 5y -18 = 0

\frac{16y+8}{5} = 18-5y

Multiply both sides by 5

$16y+8 = 90-25y$

$41y= 82$

$y=2$

Hence $x= \frac{8+2}{5}$

$x= 2$

Hence point is (2,2)

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9. Thanks all! Really helped me out This is a great community!