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3 years ago

Find the equation of the line that is perpendicular to y=5 through the point (-4,-6)

Mathematics

1 answer:

Nikolay [14]3 years ago

4 0

Answer:

$x=-4$

Step-by-step explanation:

we know that

The given equation y=5 is a horizontal line (is parallel to the x-axis)

The slope of the given line is equal to zero

A perpendicular line to the given line is a vertical line (parallel to the y-axis)

The equation of a vertical line is equal to the x-coordinate of the point that passes through it

The point that passes through it is (-4,-6)

therefore

The equation of the perpendicular line is

$x=-4$

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$x^2+y^2-8x-6y+24=0\quad|-24\\\\(x^2-8x)+(y^2-6y)=-24$

Complete the square:

$x^2-8x+a^2\implies a=\dfrac{8x}{2x}=4\implies x^2-8x+4^2\\\\\\y^2-6y+b^2\implies b=\dfrac{6y}{2y}=3\implies y^2-6y+3^2$

And we have:

$(x^2-8x)+(y^2-6y)=-24\\\\(x^2-8x+4^2)-4^2+(y^2-6y+3^2)-3^2=-24\\\\
(x-4)^2+(y-3)^2-16-9=-24\\\\(x-4)^2+(y-3)^2-25=-24\quad|+25\\\\\boxed{(x-4)^2+(y-3)^2=1}$

So the answer is C (the same left side of equation).

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