Answer :
Answer:
[tex]\textsf{a)}\quad y=\dfrac{4}{3}x-21[/tex]
[tex]\textsf{b)} \quad y=-\dfrac{3}{4}x+4[/tex]
Step-by-step explanation:
To write equations that are parallel or perpendicular to the line 4x - 3y = 0, we first need to rearrange the equation into slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept.
[tex]4x-3y=0\\\\\\3y=4x\\\\\\y=\dfrac{4}{3}x[/tex]
Therefore, the slope of the line is 4/3.
Part (a)
Parallel lines have the same slope. Therefore, the slope of the line that is parallel to 4x - 3y = 0 is:
[tex]m=\dfrac{4}{3}[/tex]
To write an equation of the line that passes through point (12, -5) and is parallel to the line 4x - 3y = 0, substitute the slope m = 4/3 and the point (12, -5) into the point-slope form of a linear equation:
[tex]y-y_1=m(x-x_1)\\\\\\y-(-5)=\dfrac{4}{3}(x-12)\\\\\\y+5=\dfrac{4}{3}x-16\\\\\\y=\dfrac{4}{3}x-21[/tex]
Part (b)
The slopes of perpendicular lines are negative reciprocals of each other. Therefore, the slope of the line that is perpendicular to the line 4x - 3y = 0 is:
[tex]m = -\dfrac{3}{4}[/tex]
To write an equation of the line that passes through point (12, -5) and is perpendicular to the line 4x - 3y = 0, substitute the slope m = -3/4 and the point (12, -5) into the point-slope form of a linear equation:
[tex]y-y_1=m(x-x_1)\\\\\\y-(-5)=-\dfrac{3}{4}(x-12)\\\\\\y+5=-\dfrac{3}{4}x+9\\\\\\y=-\dfrac{3}{4}x+4[/tex]
Answer:
A) [tex]y = \dfrac{4}{3}x - 21[/tex]
B) [tex]y = -\dfrac{3}{4}x + 4 [/tex]
Step-by-step explanation:
To solve these problems, we'll use the concept of slopes and the equations of lines.
a) Parallel Line:
To find the equation of a line parallel to [tex]4x - 3y = 0[/tex], we first need to find the slope of the given line.
Rewrite the equation in slope-intercept form:
[tex]y = mx + b[/tex]
where
[tex]m[/tex] is the slope.
Given line:
[tex]4x - 3y = 0[/tex]
[tex]3y = 4x[/tex]
[tex]y = \dfrac{4}{3}x[/tex]
The slope [tex]m[/tex] of the given line is [tex] \dfrac{4}{3} [/tex].
Since parallel lines have the same slope, a parallel line passing through the point [tex](12, -5)[/tex] will also have the slope [tex] \dfrac{4}{3} [/tex].
Now, use the point-slope form of the equation of a line:
[tex]y - y_1 = m(x - x_1)[/tex]
where
- [tex](x_1, y_1)[/tex] is the given point [tex](12, -5)[/tex] and
- [tex]m = \dfrac{4}{3}[/tex].
Plug in the values:
[tex]y - (-5) = \dfrac{4}{3}(x - 12)[/tex]
[tex]y + 5 = \dfrac{4}{3}(x - 12)[/tex]
Now, solve for [tex]y[/tex]:
[tex]y + 5 = \dfrac{4}{3}x - 16[/tex]
[tex]y = \dfrac{4}{3}x - 16 - 5[/tex]
[tex]y = \dfrac{4}{3}x - 21[/tex]
Therefore, the equation of the line that passes through the point [tex](12, -5)[/tex] and is parallel to [tex]4x - 3y = 0[/tex] is:
[tex] \boxed{y = \dfrac{4}{3}x - 21} [/tex]
b) Perpendicular Line:
To find the equation of a line perpendicular to [tex]4x - 3y = 0[/tex], we first need to find the slope of the given line. Again, rewrite the equation in slope-intercept form.
Given line:
[tex]4x - 3y = 0[/tex]
[tex]3y = 4x[/tex]
[tex]y = \dfrac{4}{3}x[/tex]
The slope [tex]m[/tex] of the given line is [tex] \dfrac{4}{3} [/tex].
The slope of a line perpendicular to another line is the negative reciprocal of its slope.
Therefore, the slope of the perpendicular line will be [tex] -\dfrac{3}{4} [/tex] (negative reciprocal of [tex] \dfrac{4}{3} [/tex]).
Now, use the point-slope form of the equation of a line with the given point [tex](12, -5)[/tex] and slope [tex] -\dfrac{3}{4}[/tex]:
[tex]y - (-5) = -\dfrac{3}{4}(x - 12)[/tex]
[tex]y + 5 = -\dfrac{3}{4}(x - 12)[/tex]
Now, solve for [tex]y[/tex]:
[tex]y + 5 = -\dfrac{3}{4}x + 9[/tex]
[tex]y = -\dfrac{3}{4}x + 9 - 5[/tex]
[tex]y = -\dfrac{3}{4}x + 4[/tex]
Therefore, the equation of the line that passes through the point [tex](12, -5)[/tex] and is perpendicular to [tex]4x - 3y = 0[/tex] is:
[tex] \boxed{y = -\dfrac{3}{4}x + 4} [/tex]
