Quiz- 3

Let us assume that the x axis point coordinates are (x, 0)
Distance = (X2 – x1)2 + (y2 – y1)2
5= (x-6)2 + (0-(-3))2
25 = x2 – 12 x + 36 + 9
X = 2, 10
Therefore, the required points are (2, 0), and (10, 0)

2^nd and 4^th quadrant have different signs

Applying the formula,

(y2-y1)/(x2 -x1)
= (7-3)/(2-(-2))
= 4/4
= 1

solution:

The slope of the line = 2
C= -3
Substituting values in the equation, we get
Y = mx + c
We get
Y = 2x – 3

Solution:

Formula for internal division =
(mx2 + nx1) / (m + n)
Y = (my2 + ny1)/ (m + n)
Hence the point is (11/3, 17/3)

Formula for calculating area of triangle when vertices are given,
= 1/2[x1( y2 - y3 )+ x2 (y3 –y1)+ x3 (y1 –y2)]
= 1/2 [4 (12 + 2) + 10 (-2-5) + (-3) (5 - 12)]
= 3.5

Distance
= √(〖(1-2)〗^2+〖(3-4)〗^2 )
= √2

Area = 1/2 [0 (0 - 0) + 2 (0 -0) + 3 (0 - 0)]
= 0
Since all the three points are collinear, the area will be 0.

Using a formula to calculate the centroid of the triangle
= (9+2+1)/3, (0+8+4)/3
= (4, 4)

Distance = √(5^2+0^2 )
= 5