# Thread: graph of a point that passes through a line

1. ## graph of a point that passes through a line

no clue what to do, help please

write an equation for a line tha tasses through (5,4) and is perpinducular to the graph of 2x - 3y = 1

thank you!!!1

2. Originally Posted by Ilovemath123 no clue what to do, help please

write an equation for a line tha tasses through (5,4) and is perpinducular to the graph of 2x - 3y = 1

thank you!!!1
You need to find two things: the gradient of the line and a point on the line.

Get those two things and substitute into the general equation

$\displaystyle y - y_1 = m(x - x_1)$

where m is the gradient and $\displaystyle \, (x_1, \, y_1)\,$ is a point on the line.

The point on the line is given on a platter. It's (5, 4).

As for the gradient, you know that it's perpendicular to the line $\displaystyle 2x - 3y = 1 \Rightarrow 2x - 1 = 3y \Rightarrow y = \frac{2}{3} x - \frac{1}{3}$.

Therefore: $\displaystyle m \times \frac{2}{3} = -1 \Rightarrow m = .....$

(You do know that when two lines are perpendicular, the product of their gradients is equal to -1, right?)

3. Originally Posted by Ilovemath123 no clue what to do, help please

write an equation for a line tha tasses through (5,4) and is perpinducular to the graph of 2x - 3y = 1

thank you!!!1
Two lines are perpendicular to another if their slopes are the negative reciprocals of each other.

y=mx+b where m is the slope, b is y-intercept.
y=(2/3)x-(1/3)
m=2/3

the line you want has slope -(3/2).
y=-(3/2)x+b

plug in (5,4) to solve for b.

4=-(3/2)5+b
b=11.5

your line is y=-(3/2)x+11.5

4. Originally Posted by Ilovemath123 write an equation for a line tha tasses through (5,4) and is perpinducular to the graph of 2x - 3y = 1
Unfortunately I have no chance against all those fast-typers...

So you get a drawing

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