InfyTQ Geometry Quiz 3

The distance of the point (5,-5) from the origin is determined by the following way:-

\sqrt{(5-0)^{2}+(-5-0)^{2}}

\sqrt{(5)^{2}+(-5)^{2}}

\sqrt{50}

√50 = 7.07 units

 

The line segment joining the points A(2,3) and B(5,6).

Let point P divides the line segment AB internally in the ratio 2:3.

The section formula is

P = \left ( \frac{x_{2}m +x_{1}n}{m + n}, \frac{y_{2}m + y_{1}n}{m+n} \right )

P = \left ( \frac {5(2)+ 2(3)}{2 + 3}, \frac{6(2) + 3(3)}{2+3} \right )

P = \left ( \frac {10+6}{2 + 3}, \frac{12 + 9}{2+3} \right )

P = \left ( \frac {16}{5}, \frac{21}{5} \right )

Solution

The point lies in the first quadrant

solution:Centre is the midpoint of PS, hence,
2 = (1+x)/2 = 3
-3 = (4+y)/2 = -10
= (3, -10)

Solution:

Area enclosed by the lines must be zero for all these points to be collinear
Area = 1/2 [7 (1-p) + 5 (p + 2) + 3 (-2-1)]
= 7 – 7p + 5p +10 -9
P= 2

Solution:
Formula to calculate area =

\frac{1}{2}\left [ x_{1}(y_{2}-y_{3}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2}) \right ]

\frac{1}{2}\left [ 1(5-(-5) + 2((-5) - 2) + (-2)(2 - 5) \right ] = 1

Solution:
Let us assume the ratio is m: n
Coordinate of the intersection becomes
X= \frac{3(m) + 2(n)}{m+n}
Y= \frac{7m + (-2n)}{m+n}
Substituting the values in the given equation, we get
2\left [\frac{3(m) + 2(n)}{m+n}+\frac{7m + (-2n)}{m+n} \right ]-4 = 0
Therefore, m: n= 2: 9

Solution:
The midpoint can be calculated as,
\frac{(3 +a)}{2}, 4
Substituting these values in the equation, we get
\frac{(3 +a)}{2} -4 -20 =0
3+a =48
A= 45

Solution:
Centroid of the triangle = \frac{x_{1}+x_{2}+x_{3}}{3} , \frac{y_{1}+y_{2}+y_{3}}{3}
= \frac{3-7+10}{3}, \frac{-5+4-2}{3}
= (2, -1)

Solution:
Considering the points to be A, B, C, and D,
The midpoint of AC = Midpoint of BD
\frac{x+1}{2} , \frac{6+2}{2} = \frac{4+3}{2}, \frac{y+5}{2}
⇒ \frac{x+1}{2} = \frac{7}{2} and \frac{6+2}{2} = \frac{y+5}{2}
X+1 = 7 and 8 = y+5
X= 6, and y = 3