1.what is the length of the segment joining 3,6 and -2,-6 : 13 units
2.what is the center of the circle (x+6)^2+(y-8)^2=144 => (-6,8)
3.what is the slope of the line 3y+2x-6=0=> -2/3
Step-by-step explanation:
1.what is the length of the segment joining (3,6) and (-2,-6)?
Let
(x1,y1) = (3,6)
(x2,y2) = (-2,-6)
The length of a segment is given by:
[tex]d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\Putting\ values\\d = \sqrt{(-2-3)^2+(-6-6)^2}\\d = \sqrt{(-5)^2+(-12)^2}\\= \sqrt{25+144}\\= \sqrt{169}\\=13\ units[/tex]
2.what is the center of the circle (x+6)^2+(y-8)^2=144
The equation of circle is given by:
[tex](x-h)^2+(y-k)^2 = r^2[/tex]
Here, h and k are the coordinates of centre of circle
x - h = x+6
-h = 6
h = -6
y - 8 = y - k
-8 = - k
k = 8
So,
The center of circle is: (-6,8)
3.what is the slope of the line 3y+2x-6=0
We have to convert the equation in slope-intercept form to find the slope
Slope-intercept form is:
y = mx+b
Now,
[tex]3y+2x-6=0\\3y+2x = 6\\3y = -2x+6[/tex]
Dividing both sides by 3
[tex]\frac{3y}{3} = -\frac{2}{3}x+\frac{6}{3}\\y = -\frac{2}{3}x + 2[/tex]
In slope-intercept form, the co-efficient of x is the slope of the line so
m = -2/3
Keywords: Coordinate geometry, Slope
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