Write an equation of the line that is perpendicular to 3x + 9y = 7 and passes through the point (6, 4)
2 answers:
The perpendicular line will have the x and y coefficients swapped and one of them negated. We can write the desired line as
9(x -6) -3(y -4) = 0
where the coordinates of the point of interest are (6, 4).
Dividing by 3, this is
3(x -6) -(y -4) = 0
3x -y = 14
An equation for the line of interest is ...
3x -y = 14
9(x -6) -3(y -4) = 0
where the coordinates of the point of interest are (6, 4).
Dividing by 3, this is
3(x -6) -(y -4) = 0
3x -y = 14
An equation for the line of interest is ...
3x -y = 14
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Answer:
y= -2x²+4x.
Step-by-step explanation:
the common equation of the parabola is y=ax²+bx+c, where a, b, c - numbers.
1) the coordinates of the vertex are:
2) if according to the condition x₀=1, then b= -2a.
If according to the condition y₀=2, then
it means that c=a+2.
3) if b= -2a; c=a+2 and the point (3;-6) belongs to the given parabola, then it is possible to substitute them into the common equation of the parabola:
-6=3²*a-6a+a+2; ⇔ a= -2.
4) if a= -2, then b=-2a=4, and c=a+2=0.
5) if a=-2, b=4 and c=0, then the required equation of the parabola is:
y= -2x²+4x.