Question

.1. If (1,x) is at a distance of √10 units from (0,0) , find the value of x.
2. Find the value of k such that AB=BC, where A= (6,-1), B= (1,3), C= (k,8) respectively.
3. Find the point on the y- axis equidistant from (7,6) and (-3,4).
4. Show that (-2,2), (8,-2) and (-4,-3) are the vertices of a right triangle.
5. Show that A= (3,4) , B=(4,2), C=(5,-4), D=(4,-10) are the vertices of a rhombus.
6.Show that (-3,-4) , (12,5),(14,12) and (-1,3) are the vertices of a parallelogram.
7. Show that (-1,-8),(4,-6),(2,-1),(-3,-3) are the vertices of a square.
8. If A= (-3,-2), B=(3,2) , C = (2√3, 3√3), show that triangle ABC is equilateral.
9.If a circle passes through (1,2) with its center at origin, find its radius.
10. Find the ratio in which  the line segment joining the points  (-2,3) and (5,-4) is divided by i) x-axis ii)y-axis
11. the centroid of a triangle is (2,7). if two of its vertices are (4,8) and (-2,6). find the third vertex.
12.Find the fourth vertex of the parallelogram ABCD , if A=(-2,-1) , B= (1,0) and C=(4,3)
13.The slope of the line perpendicular to the line 2x-3y+6=0 is 
14. the line passing through the points (2,6) and (-2,6) is 
     a) parallel to x-axis.    b) parallel to y-axis

Answer 1

1)
$\sqrt{10} = \sqrt{(1-0)^2+(x-0)^2}\\ \\10 = 1 + x^2\\x = +3,\ or\ -3$

2)
$AB^2=(6-1)^2+(-1-3)^2=25+16=41\\ \\BC^2=(k-1)^2+(8-3)^2=(k-1)^2+25\\ \\(k-1)^2=41-25=16\\ \\k-1=+4,\ or\ -4\\ \\k = -3,\ or\ 5$

3)
$(x-7)^2+(y-6)^2=(x+3)^2+(y-4)^2\\ \\x^2-14x+49+y^2+36-12y=x^2+6x+9+y^2-8y+16\\ \\20x+4y-60=0$
All the points on the straight line 5x+y-12 =0 will be equidistant from given points. In fact this is the perpendicular bisector for the line joining given points.

4)
slope of AB = (-2-2)/(8-(-2))= -4/10 = -2/5
Slope of BC = (-3-(-2))/(-4-8) = -1/-12 = 1/12
Slope of CA = (-3-2)/(-4-(-2))=-5/-2 = 5/2 
Product of slopes of AB and CA is 1. So they are perpendicular.

5)
coordinates of A or C are typed wrong.   A = (3,-4)

slope of AC = (-4-(-4)/(5-3) = 0  ie., parallel to x axis
Slope of BD = ( -10-2) / (4-4 ) = infinity  ie., parallel to y axis.

AB = √(6²+1²) = √37   BC = √6²+1² = √37  
CD = √(6²+1²) = √37    DA also.   so Rhombus

6)
slope of AB = (5+4)/(12+3) = 3/5         Slope of CD = (12-3)/(14-(-1))=3/5        
slope of BC = 7/2         slope of AD = -7/-2 
So opposite sides are parallel.

7)
AB² = 5^2+2^2            BC² = (4-2)²+(-6+1)²         CD²= (-3-2)²+(-3+1)²
DA² = 5^2+(-3+1)^2 
Slope of AB = 2/5      slope of BC = 5/-2            product = -1  so perpendicular
slope of CD is also perpendicular to DA

8)
Coordinates of C are  wrongly mentioned :  2√3 ,  -3√3

AB² = 6²+4²    BC²=(-3√3-2)²+(3-2√3)² = 31+12√3+21-12√3 =52
CA² = (2√3+3)²+(-3√3+2)² = 31+21=52     so equilateral  Δ

9)
radius = distance from origin of (1,2) = √(1-0)²+(2-0)² = √5

10)
the point of intersection of y axis and given line is (0, y1)

let the ratio = r  
  x coordinate = 0 = (-2 * 1 + 5 * r) / (1 + r)    => r = 2/5        y axis divides in 2/5

11)
2 = (4-2+x)/3  x = 4        7 = (8+8+y)/3       y = 5
         so (4,5)

12)
slope of AB = 1/3   = (y-3)/(x-4)          3y-9=x-4      x = 3y-5
slope of BC = 3/3 = 1                   ( y+1 )/(x+2 ) = 1    x = y -1
y = 4/2 = 2    x = 1

13)
  slope of given line = 2/3         slope of line perpendicular to it is  -3/2

14)
Since the y coordinate is same , it is parallel to x axis. perpendicular to y axis.

Answer 2

Hi this is answered by Prmkulk1978 :  brainly.in/question/2479570

Answer:

The third vertex of triangle is :(3,1)

Step-by-step explanation:

Let the third vertex be (x,y)  and other two vertices are (-3,1) and (0,-2)

Coordinates of  centroid  of triangle  =(0,0)

∴ [-3+0+x/3 , 1-2+y/3] = (0,0)

-3+x/3 = 0

x-3=0

x=3

and 1-2+y/3 =0

y-1/3=0

y-1=0

y=1

∴ x=3 and y=1