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Vlad1618 [11]
4 years ago
15

A line passes through (8,2) and (11,-13) write the equation of thr line in standerd form

Mathematics
1 answer:
docker41 [41]4 years ago
8 0

Answer:

Step-by-step explanation:

(8,2),(11,-13)

slope(m) = (-13 - 2) / (11 - 8) = -15/3 = -5

y = mx + b

slope(m) = -5

(8,2)...x = 8 and y = 2

now we sub and find b, the y int

2 = -5(8) + b

2 = -40 + b

2 + 40 = b

42 = b

so ur equation is : y = -5x + 42...now we need it in standard form

Ax + By = C

y = -5x  42

5x + y = 42 <====

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What is the equation of a circle whose center at (-5, -2) and radius is 2?​
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The equation of a circle whose centre (-5,-2) and radius 2 is x^{2}+y^{2}+10x+4y+25=0

What is equation of a circle?

A circle is a closed curve drawn from a fixed point called centre of the circle. The distance between the centre of the circle and the arc of the circle is called the radius of the circle.

The equation of a circle with centre (h, k) radius r is

(x-h)^{2} +(y-k)^{2}=r^{2}

Given center = (-5,-2)  and Radius = 2

Put the Given value in the equation:

= (x-(-5))^{2} +(y-(-2))^{2} = 2^{2}

= (x+5)^{2} +(y+2)^{2}=2^{2}

= (x^{2} +10x+25)+(y^{2}+4y+4)= 4

= x^{2} +y^{2}+10x+ 4y+ 29 = 4

= x^{2} +y^{2}+10x +4y+ 29-4=0

= x^{2} +y^{2}+10x+4y+25=0

So, the equation of the circle whose centre is (-5,-2) and Radius 2 is

x^{2}+y^{2}+10x+4y+25=0

To read more about the equation of a circle. you can follow:

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