Respuesta :
The parallel line has the same slope, which is 4. y=4x+c and we find c by plugging in the point:
32=4×4+c, c=32-16=16 so y=4x+16, answer D.
32=4×4+c, c=32-16=16 so y=4x+16, answer D.
The equation of line that is parallel to the line [tex]y-1=4\left({x+3}\right)[/tex] and passes through the point [tex]\left({4,32}\right)[/tex] is [tex]\boxed{{\mathbf{y=4x+16}}}[/tex] and it matches with [tex]\boxed{{\mathbf{OPTION}}\,{\mathbf{D}}}[/tex] .
Further explanation:
It is given that the equation of line is [tex]y-1=4\left({x+3}\right)[/tex] and passes through point [tex]\left({4,32}\right)[/tex] .
Rewrite the given equation [tex]y-1=4\left({x+3}\right)[/tex] as follows:
[tex]\begin{aligned}y-1&=4\left({x+3}\right)\\y-1&=4x+12\\y&=4x+12+1\\y&=4x+13\\\end{aligned}[/tex]
Now, compare the obtained equation of line [tex]y=4x+13[/tex] with the standard equation of line [tex]y=mx+b[/tex] .
[tex]\begin{aligned}m&=4\\b&=13\\\end{aligned}[/tex]
Therefore, the slope is [tex]4[/tex] .
It is given that both lines are parallel to each other so the product of slope must be equal.
[tex]{m_1}={m_2}[/tex] …… (1)
Substitute [tex]4[/tex] for [tex]{m_1}[/tex] in equation (1) to obtain the value of slope [tex]{m_2}[/tex] .
[tex]{m_2}=4[/tex]
Therefore, the slope is [tex]4[/tex] .
It is given that the line passes through point [tex]\left({4,32}\right)[/tex] .
The point-slope form of the equation of a line with slope [tex]m[/tex] passes through point [tex]\left({{x_1},{y_1}}\right)[/tex] is represented as follows:
[tex]y-{y_1}=m\left({x-{x_1}}\right)[/tex] …… (2)
Substitute [tex]4[/tex] for [tex]{x_1}[/tex] , [tex]32[/tex] for [tex]{y_1}[/tex] and [tex]4[/tex] for [tex]m[/tex] in equation (2) to obtain the equation of line.
[tex]\begin{aligned}y-32&=4\left({x-4}\right)\\y-32&=4x-16\\y&=4x-16+32\\y&=4x+16\\\end{aligned}[/tex]
Therefore, the equation of line is [tex]y=4x+16[/tex] .
Now, the four options are given below.
[tex]\begin{gathered}{\text{OPTION A}}\to y=-x+33\hfill\\{\text{OPTION B}}\to y=-x+36\hfill\\{\text{OPTION C}}\to y=4x-16\hfill\\{\text{OPTION D}}\to y=4x+16\hfill\\\end{gathered}[/tex]
Since OPTION D matches the solution that is [tex]y=4x+16[/tex] .
Thus, the equation of line that is parallel to the line [tex]y-1=4\left({x+3}\right)[/tex] and passes through the point [tex]\left({4,32}\right)[/tex] is [tex]\boxed{{\mathbf{y=4x+16}}}[/tex] and it matches with [tex]\boxed{{\mathbf{OPTION}}\,{\mathbf{D}}}[/tex] .
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Answer Details:
Grade: Junior High School
Subject: Mathematics
Chapter: Coordinate Geometry
Keywords: Coordinate Geometry, linear equation, system of linear equations in two variables, variables, mathematics, equation of line, line, passes through point.
