Here all you need to know to solve this exercise:
1) When you write the equation of a line in the form
[tex]y=mx+q[/tex]
the slope is the coefficient m.
2) When you know that the line passes through a point [tex](x_0,y_0)[/tex] and has slope m, the equation of the line is given by
[tex]y-y_0=m(x-x_0)[/tex]
3) Let [tex]m_1,m_2[/tex] be the slopes of two lines. If the lines are parallel, then [tex]m_1=m_2[/tex]. If the lines are perpendicular, then [tex]m_1m_2=-1[/tex].
In the first exercise, using point 1, you can see that the slope of the given line is 2. We want a parallel line passing through the given point. So, our line has the same slope, and its equation is (see point 2)
[tex]y-2=2(x+1) \iff y = 2x+4[/tex]
Similarly, in the second exercise, the original slope is 1/3, so the perpendicular slope is -3. The equation will be, imposing the passage through the given point,
[tex]y-6 = -3(x-0) \iff y = -3x+6[/tex]
Answer:
see explanation
Step-by-step explanation:
19
The equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c yje y- intercept )
y = 2x - 3 is in this form with slope m = 2
• Parallel lines have equal slopes, hence
y = 2x + c ← partial equation of parallel line
To find c substitute (- 1, 2) into the partial equation
2 = - 2 + c ⇒ c = 2 + 2 = 4
y = 2x + 4 ← equation of parallel line
20
Rearrange x - 3y = 5 into slope- intercept form
Subtract x from both sides
- 3y = - x + 5 ( divide all terms by - 3 )
y = [tex]\frac{1}{3}[/tex] x - [tex]\frac{5}{3}[/tex] ← in slope- intercept form
with slope m = [tex]\frac{1}{3}[/tex]
Given a line with slope m then the slope of a line perpendicular to it is
[tex]m_{perpendicular}[/tex] = - [tex]\frac{1}{m}[/tex] = - [tex]\frac{1}{\frac{1}{3} }[/tex] = - 3, so
y = - 3x + c
The line passes through (0, 6) ⇒ c = 6
y = - 3x + 6 ← equation of perpendicular line