Co-ordinate Geometry Help Please

**Brownhash** · July 18th 2008, 03:50 PM

Is anyone able to help me with these questions ? they are most likely easy but im in a hurry as it is due tomorrow.
Cheers

1)
Find the equation of the line in the form ax + by + c = 0 that
a) passes through (-2,5) and has a gradient of 2/3
b) passes through (-2,6) and (-3,-7)

2)
Find the equation of the line that passes through (4,-7) and is parallel to y=3x-2

3) Find the equation of the line that passes through (-3,8) and is perpindicular to the line 3x-4y+2=0

Any help will be highly appreciated thanks.

**Chris L T521** · July 18th 2008, 04:20 PM

Originally Posted by Brownhash

Is anyone able to help me with these questions ? they are most likely easy but im in a hurry as it is due tomorrow.
Cheers

1)
Find the equation of the line in the form ax + by + c = 0 that
a) passes through (-2,5) and has a gradient of 2/3 When you say gradient, are you referring to the slope?

b) passes through (-2,6) and (-3,-7)

2)
Find the equation of the line that passes through (4,-7) and is parallel to y=3x-2

3) Find the equation of the line that passes through (-3,8) and is perpindicular to the line 3x-4y+2=0

Any help will be highly appreciated thanks.

1b) If the line passes through 2 points, we can determine the slope of the line. Our two points are $(-2,6) \text{ and }(-3,-7)$

Thus $m=\frac{y_2-y_1}{x_2-x_1}=\frac{-7-6}{-3-(-2)}=\frac{-13}{-1}=13$

Thus, the line will have the form $y=13x+b$ .

To solve for $b$ , pick one of the points that the line passes through. I'll choose $(-2,6)$

$6=13(-2)+b\implies b=32$

Thus our line is $y=13x+32$ . Our line could also be written in the form $\color{red}\boxed{13x-y+32=0}$

2)

This time, we are to find a line that is $\parallel$ to $y=3x-2$ . It also passes through the point (4,-7).

If lines are parallel, they must have the same slope.

Thus our parallel line has the form $y=3x+b$ . To find b, plug in the point:

$-7=3(4)+b\implies b=-19$

Thus, our parallel line is ${\color{red}\boxed{y=3x-19}}=\color{red}\boxed{3x-y-19=0}$

3)

The line we are looking for is $\bot$ to $3x-4y+2=0$ and it passes through the point $(-3,8)$

First, rewrite the line in slope intercept form:

$3x-4y+2=0\implies y=\frac{3}{4}x+\frac{1}{2}$ .

The slope of the perpendicular line is the negative reciprocal of this line : $\therefore m=-\frac{4}{3}$

Thus the perpendicular line has the form of $y=-\frac{4}{3}x+b$

Plug in the point to find b:

$8=-\frac{4}{3}(-3)+b\implies b=4$

Thus, the line is ${\color{red}\boxed{y=-\frac{4}{3}x+4}}=\color{red}\boxed{4x+3y-12=0}$

Does this make sense?

--Chris

**Brownhash** · July 18th 2008, 04:39 PM

Yes that helps alot thanks for the help

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