Answer:
17. Equation : [tex]y=-x-1[/tex]
18: Equation: [tex]y=\frac{1}{2}x+8[/tex]
19: Equation: [tex]y=3x-6[/tex]
20: Equation: [tex]y=\frac{1}{3}x-2[/tex]
Step-by-step explanation:
Using the Slope-Intercept Form for the following questions:
Equation of line: [tex]y=mx+b[/tex] where m is the slope and b is the y-intercept.
17. A(3, -4), [tex]y=-x+8[/tex]
First substitute the slope from the original line [tex]y=-x+8[/tex] (i.e. -1 in this case) into the equation of line.
[tex]y=(-1)x+b[/tex] .....(1)
Now, substitute the given point A(3,-4) in the equation (1), we have;
[tex]-4=(-1)\cdot (3)+b[/tex]
[tex]-4=-3+b[/tex]
[tex]-4+3=b[/tex]
∴[tex]b=-1[/tex]
Substitute the value of b=-1 in the Slope-Intercept form i.e, [tex]y=-x+(-1)[/tex]
[tex]y=-x-1[/tex]
Therefore, the equation of the line passing through the given point that is parallel to the given line is, [tex]y=-x-1[/tex].
18. A(-6, 5) , [tex]y=\frac{1}{2}x -7[/tex]
First substitute the slope from the original line [tex]y=\frac{1}{2}x -7[/tex] (i.e. [tex]\frac{1}{2}[/tex] in this case) into the equation of line.
[tex]y=\frac{1}{2} x+b[/tex] .....(1)
Now, substitute the given point A(-6, 5) in the equation (1), we have;
[tex]5=\frac{1}{2} \cdot (-6)+b[/tex]
[tex]5=-3+b[/tex]
[tex]5+3=b[/tex]
∴[tex]b=8[/tex]
Substitute the value of b=8 in the Slope-Intercept form i.e, [tex]y=\frac{1}{2}x+8[/tex]
[tex]y=\frac{1}{2}x+8[/tex]
Therefore, the equation of the line passing through the given point that is parallel to the given line is, [tex]y=\frac{1}{2}x+8[/tex].
19. A(2,0), [tex]y=3x-5[/tex]
First substitute the slope from the original line [tex]y=3x-5[/tex] (i.e. 3 in this case) into the equation of line.
[tex]y=3x+b[/tex] .....(1)
Now, substitute the given point A(2, 0) in the equation (1), we have;
[tex]0=3\cdot (2)+b[/tex]
[tex]0=6+b[/tex]
∴[tex]b=-6[/tex]
Substitute the value of b=-6 in the Slope-Intercept form i.e, [tex]y=3x+(-6)[/tex]
[tex]y=3x-6[/tex]
Therefore, the equation of the line passing through the given point that is parallel to the given line is, [tex]y=3x-6[/tex]
20. A(3, -1), [tex]y=\frac{1}{3}x+10[/tex]
First substitute the slope from the original line [tex]y=\frac{1}{3}x +10[/tex] (i.e. [tex]\frac{1}{3}[/tex] in this case) into the equation of line.
[tex]y=\frac{1}{3} x+b[/tex] .....(1)
Now, substitute the given point A(3, -1) in the equation (1), we have;
[tex]-1=\frac{1}{3} \cdot (3)+b[/tex]
[tex]-1=1+b[/tex]
[tex]-1-1=b[/tex]
∴[tex]b=-2[/tex]
Substitute the value of b=-2 in the Slope-Intercept form i.e, [tex]y=\frac{1}{3}x+(-2)[/tex]
[tex]y=\frac{1}{3}x-2[/tex]
Therefore, the equation of the line passing through the given point that is parallel to the given line is, [tex]y=\frac{1}{3}x-2[/tex].
