Mathematics, 16.10.2021 22:20 ahrozycki

Write the equation of the line with the given characteristics Perpendicular to x = - 5 passes through (- 8, 13)

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This is a complicated problem, so I hope you read through my explanation and find it helpful.

Given x = -5, and the point (-8, 13):

x = -5 is a vertical line.

Vertical lines are symbolically represented by the equation, x = a where a is the x-intercept.

Vertical lines have undefined slope. Since any two points on a vertical line have the same x-coordinate (x = -5), the slope of the equation cannot be computed as a finite number according to the slope formula, m = (y2 - y1)/(x2 - x1), because division by zero is an undefined operation.

Perpendicular lines have slopes that are negative reciprocals of one another (meaning, multiplying the slopes of both lines will have a product of -1).

However, because x = -5 is a vertical line, then it means that the other line perpendicular to it must be a HORIZONTAL LINE, so that they intersect at a 90° angle and qualify as perpendicular lines.

Horizontal lines are symbolically represented by the equation, y = b where b is the y-intercept. Therefore, to come up with a horizontal line, you must use the y-coordinate of point, (-8, 13), to come up with the equation, y = 13.

Therefore, the equation of the perpendicular line is y = 13.

As a proof, I’m including a screenshot of the graphed equations, which also contains the given point, (-8, 13).

Please mark my answers as the Brainliest, if you find this helpful :).
Write the equation of the line with the given characteristics

Perpendicular to x = - 5 passes thro

Write the equation of the line with the given characteristics

Perpendicular to x = - 5 passes thro

Answer from: Quest

center (1,3) with radius 6

step-by-step explanation:

we must complete the square to find the center and radius of the circle.

first make sure the x and y squared terms have 1 as their coefficients. we also make sure x and y terms together.

$x^2-2x+y^2-6y=26$

we now create space between the x and y terms with parenthesis.

$(x^2-2x)+(y^2-6y)=26$

we complete the square by taking the middle terms -2x and the -6y - divide each and square them.

$\frac{-2}{2} =(-1)^{2} =1$

$\frac{-6}{2} =(-3)^{2} =9$

we add the squares to both sides.

$(x^2-2x+1)+(y^2-6y+9)=26+1+9$

simplify.

$(x^2-2x+1)+(y^2-6y+9)=36$

and write the quadratics in factored form.

$(x-1)^{2} +(y-3)^{2} =36$

the center is (h,k) or (1,3). the radius is the square root of 36 which is 6.

Answer from: Quest

Ican't read this very blury
Ineed an equation to the word problem above

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