Free Coordinate Geometry 03 Practice Test - 10th Grade 

If a point P(x,y) divides the line joining $(x_1,y_1)\text{and}(x_2,y_2)$ in the ratio m:n
then $x$ = $\frac{m{x}_2+n{x}_1}{m+n}$ and y = $\frac{m{y}_2+n{y}_1}{m+n}$

$\therefore x$ = $\frac{m{x}_2+n{x}_1}{m+n}$ = $\frac{2(-5)+1(-2)}{2+1}$ = -4

and $y$ = $\frac{m{y}_2+n{y}_1}{m+n}$ = $\frac{2\left(7\right)+1(-2)}{2+1}$ = 4

$\therefore$ the required point is (-4,4).

The distance of a point $(x,y)$ from the origin $(0,0)$ is $\sqrt{(x-0{)}^2+(y-0{)}^2}=\sqrt{x^2+y^2}$

Point C lies on the line 'l' which is parallel to Y-axis. So the x-coordinate of any point on this line will be constant.

Therefore its x coordinate is 4

Let $C(x,0)$ divide $AB$ in the ratio $k:1$.

By section formula, the coordinates of $C$ are given by: $(\frac{6k+1}{k+1},\frac{4k-7}{k+1})$

But $C(x,0)=(\frac{6k+1}{k+1},\frac{4k-7}{k+1})$ $\Rightarrow \frac{4k-7}{k+1}=0$

$\Rightarrow 4k-7$ = 0 $\Rightarrow k$ = $\frac{7}{4}$

i.e., the x-axis divides $AB$ in the ratio $7:4$.

Area of a triangle formed by $(x_1,y_1),(x_2,y_2)(x_3,y_3)$ = $\frac{1}{2}$|$x_1$($y_2$ - $y_3$) + $x_2$($y_3$ - $y_1$)+ $x_3$($y_1$ - $y_2$)| 
Thus, the area of  $\Delta ABC$ formed by the given points A(4, 1), B(-3, 2) and C(0, k) is

$=\frac{1}{2}|4(2-k)+(-3)(k-1)+0(1-2)|$

$=\frac{1}{2}|8-4k-3k+3|=\frac{1}{2}|11-7k|$

But given that the area of $\Delta ABC=12\text{sq. unit}$

 $\Rightarrow |11-7k|=24$

$\pm (11-7k)=24$            $\Rightarrow 11-7k=24$ or $-(11-7k)=24$

$If11-7k=24,then-7k=24-11=13$       $\Rightarrow k=\frac{-13}{7}$

$-(11-7k)=24⟹-11+7k=24$

  $\Rightarrow 7k=24+11=35$    $\Rightarrow k=\frac{35}{7}=5$

Hence the values of k are : 5,   $\frac{-13}{7}$

The given points A(-1, 3), B(2, p), C(5, -1) are collinear.

⇒ Area  ΔABC formed by these points should be zero.

⇒ The area of  ΔABC = 0

⇒  $\frac{1}{2}$( $x_1$( $y_2$- $y_3$) +  $x_2$( $y_3$- $y_1$) +  $x_3$( $y_1$ -  $y_2$)) = 0

⇒ -1(p + 1) + 2 (-1 - 3) + 5(3 -p) = 0 

⇒ -p - 1 - 8 + 15 - 5p = 0

⇒ -6p +15 - 9 = 0 ⇒ -6p = -6 ⇒ p = 1

Hence the value of p is 1.

Let, $A$ be $(a+b,b+c)$ and $B$ be $(a-b,c-b)$

$\because$ Distance between the points $(x_1,y_1)\text{and}(x_2,y_2)$ is 
$\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$\therefore AB=\sqrt{(a-b-a-b{)}^2+(c-b-b-c{)}^2}=\sqrt{(-2b{)}^2+(-2b{)}^2}=\sqrt{8b^2}AB=2\sqrt{2}b$

$\therefore$ Distance between the points = $2\sqrt{2}b$

The required point is equidistant from the points P and Q, and also lies on the line y = -1.

Let the point be A, it lies on y = -1

Therefore consider the point A(x_a, -1)

Distance AP = Distance AQ

$\sqrt{(x_a-1{)}^2+(-1-3{)}^2})$ = $\sqrt{(x_a+5{)}^2+(-1+5{)}^2}$
${(x_a-1)}^2$ + $(-4{)}^2$ = ${(x_a+5)}^2$ + $(4{)}^2$
-2$x_a$ + 1 = 10$x_a$ + 25
-12$x_a$ = 24
$x_a$ = -2
The coordinates of the point A is (-2, -1).

Any line whose y-coordinate is a constant will be parallel to the X axis. Observe that B
and C have the same y coordinate, c, and hence this side BC is parallel to the X-axis.

Similarly, A and B have the same x – coordinates, -a, hence this side is parallel to the Y axis.

A and B are equidistant from origin and AB is $\sqrt{2}$a.
If we assume the points A and B to be (x,0) and (0,x) respectively,
AB= $\sqrt{x^2+x^2}$ = $\sqrt{2}$x = $\sqrt{2}$a
x=a
So A(0,a) and B(a,0) are the vertices.
Observe that A,B both lie on the line x+y=a and since C also lies on the line y+x = a, they are collinear.
Hence area of the triangle formed by the three points is 0.