hey,
write down the equation of the straight line that:
a) passes through (0,-10) and is parrallel to the line with equation y=6x+3
b) passes through the point (0,-1) and is perpendicular to the line with equation 3x-2y+5=0
thanks
hey,
write down the equation of the straight line that:
a) passes through (0,-10) and is parrallel to the line with equation y=6x+3
b) passes through the point (0,-1) and is perpendicular to the line with equation 3x-2y+5=0
thanks
Hello,
If two lines are parallel, then the slopes are equal.
The slope of y=6x+3 is 6.
Therefore, the equation of the line we want to find is y=6x+a, with a to determine.
Because it passes through (0,-10), we have -10=6*0+a. So a=-10.
Thus the equatio of the line is $\displaystyle y=6x-10$
y=ax+bb) passes through the point (0,-1) and is perpendicular to the line with equation 3x-2y+5=0
It passes through the point 0,-1. Therefore, -1=a*0+b --> b=-1
ax-y-1=0.
A normal vector to it is $\displaystyle (a,-1)$.
A normal vector to 3x-2y+5=0 is $\displaystyle (3,-2)$.
Because the two lines are perpendicular, their normal vector are perpendicular too. In particular, the scalar product is 0.
$\displaystyle 0=(a,-1) \dot (3,-2)=3a+2 \implies a=-\frac 23$
You have the equation of the line now;..
i hope i didn't make any mistake...
@Moo: You're mistake-free.
@fresh_: Yes, of course.
Here I listed the steps for you:
Given: $\displaystyle y= -\frac{2x}{3} - 1$
Move 1 to the other side: $\displaystyle y + 1 = -\frac{2x}{3}$
Cross multiply: $\displaystyle 3y + 3 = -2x$
Switch sides with 2x and 3: $\displaystyle 3y + 2x = -3$
OR you can simply do this:
$\displaystyle \left(y + \frac{2x}{3} = - 1\right) \cdot 3$
$\displaystyle 3y + 2x = -3$