Free Coordinate Geometry 01 Practice Test - 10th Grade 

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

$\therefore PQ=\sqrt{(3-x{)}^2+(-6-2{)}^2}=10(3-x{)}^2+(8{)}^2=100(3-x{)}^2+64=100(3-x{)}^2=363-x=\pm 6x=-3\text{or}9$

If a point $P(x,y)$  divides a line segment joining $(x_1,y_1)\text{and}(x_2,y_2)$ in the ratio $m:n$, then the coordinates of $P$ are given by:

$x=\frac{m{x}_2+n{x}_1}{m+n},y=\frac{m{y}_2+n{y}_1}{m+n}$

Let $P(x,y)$ divide the line segment $AB$ joining $A(-1,7)$ and $B(4,-3)$ in the ratio $2:3$. Then by using section formula, the co-ordinates of $P$ are given by:

$(\frac{2\times 4+3\times (-1)}{2+3},\frac{2\times (-3)+3\times 7}{2+3})$
$=(\frac{8-3}{5},\frac{-6+21}{5})$
$=(\frac{5}{5},\frac{15}{5})$

Thus, the coordinates of $P$ is $(1,3)$

Midpoint of the line segment joining $(x_1,y_1)$ and $(x_2,y_2)$ is given by $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
$\therefore$ Mid-point of the line-segment joining the points (– 5, 4) and (9, – 8) = $(\frac{-5+9}{2},\frac{4-8}{2})$ = (2,-2)

For 3 points to be collinear the area of the triangle formed by these three points should be zero.

Area of a triangle formed by $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by
$\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$

$\therefore \frac{1}{2}\left[a\right(b-1)+0(1-0)+1(0-b\left)\right]=0\frac{1}{2}\left[a\right(b-1)+1(0-b\left)\right]=0ab=a+b\frac{1}{a}+\frac{1}{b}=1$

$\text{Any point on the y-axis has coordinates}\text{of the form}(0,y).\text{Since the point is equidistant from both}\text{points}(-3,4)\text{and}(7,6)\text{by distance formula,}{3}^2+{(y-4)}^2=7^2+{(y-6)}^{2}16–8y=40+36–12y4y=60y=15\therefore \text{the point is}(0,15).$

Area of a triangle with vertices $(x_1,y_1),(x_2,y_2)and(x_3,y_3)$ is given by
$Area=\frac{1}{2}\times |\left(x_1\right)(y_2-y_3)+\left(x_2\right)(y_3-y_1)+\left(x_3\right)(y_1-y_2)|$

Thus, $\text{Area of the given triangle}$
$=\frac{1}{2}|a(c+a-a-b)+b(a+b-b-c)+c(b+c-c-a)|$
$=\frac{1}{2}|ac-ab+ab-bc+bc-ac|=0$

From section formula, if a point $(x,y)$ divides the line segment joining the points ($x_1,y_1$) and ($x_2,y_2$) internally in the ratio m:n,  then $x=\frac{m{x}_2+n{x}_1}{m+n}$ and $y=\frac{m{y}_2+n{y}_1}{m+n}$.

So, $2=\frac{4m+n(-1)}{m+n}$

$⟹4m-n=2m+2n$

i.e., $2m=3n$

Thus,  $\frac{m}{n}$ = $\frac{3}{2}$  or $m:n=3:2$

The mid-point of the coordinates $(x_1,y_1),(x_2,y_2)$ is given by $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
$\therefore$ The mid-points of AB, BC and CA are $(\frac{-4+2}{2}$,  $\frac{3+3}{2})$, $(\frac{4+2}{2}$,  $\frac{3+5}{2})$ and $(\frac{-4+4}{2}$,  $\frac{5+3}{2})$ respectively. Let us denote these by D, E and F.
i.e.,  D(-1, 3), E(3, 4) and F(0, 4).

Area of the triangle formed by $(x_1,y_1),(x_2,y_2)(x_3,y_3)$ is given by
$\text{Area}=\frac{1}{2}$|$x_1$($y_2$-$y_3$)+$x_2$($y_3$-$y_1$)+$x_3$($y_1$-$y_2$)|

Now, let $x_1$ = -1,  $y_1$ = 3, $x_2$ = 3,  $y_2$ = 4 $x_3$ = 0 and  $y_3$ = 4

$\Rightarrow \text{Required area}=\frac{1}{2}$| -1(4 - 4) + 3(4 - 3) + 0(3 - 4)|  = 1.5 square units

$\therefore$ Area of the triangle = 1.5 square units

Consider the points 
$A(-4,-2),B(-3,-5),C(3,-2)$ and $D(2,3).A\left(△\right)=\frac{1}{2}\times |x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

$\therefore$ Area of  $△ABC$ 

$=\frac{1}{2}|(-4)(-5+2)-3(-2+2)+3(-2+5)|=\frac{1}{2}|20-8-6+15|=\frac{21}{2}A\left(△ABC\right)=10.5\text{sq. units}$

Similarly, area of   $△ACD$

$=\frac{1}{2}|(-4)(-2-3)+3(3+2)+2(-2+2)|=\frac{1}{2}|20+15|=\frac{35}{2}A\left(△ACD\right)=17.5\text{sq. units}$

$\text{Area of quadrilateral}ABCD=A\left(△ABC\right)+A\left(△ACD\right)=(10.5+17.5)=28\text{sq. units}$

We have to find a point on the y-axis. Therefore, its x-coordinate will be 0.
Let the point on y-axis be P (0,y)

Distance PA = Distance between $(0,y)$ and $(3,-4)=\sqrt{(0-3{)}^2+(y-(-4){)}^2}=\sqrt{(-3{)}^2+(y+4{)}^2}$

Distance PB = Distance between $(0,y)$ and $(-5,9)=\sqrt{(0-(-5){)}^2+(y-9{)}^2}=\sqrt{5^2+(y-9{)}^2}$

By the given condition, PA = PB, thus
$\sqrt{(-3{)}^2+(y+4{)}^2}=\sqrt{5^2+(y-9{)}^2}$

Squaring on both sides of the equation

$\therefore$ $(y+4{)}^2+9=(y-9{)}^2+25$

$y^2+16+8y+9=y^2-18y+81+25$

$26y=106-25$

$26y=81$

$y=\frac{81}{26}$

$\therefore$ The required point is $(0,\frac{81}{26})$.