Free Coordinate Geometry 02 Practice Test - 10th Grade 

Let's name the points as $A(a,a),B(-a,-a)$ and $C(–\sqrt{3}a,\sqrt{3}a)$.

By distance formula, distance between points $(x_1,y_1)$ and $(x_2,y_2)$

$=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

$BC=\sqrt{(-\sqrt{3}a+a{)}^2+(\sqrt{3}a+a{)}^2}=\sqrt{a^2+(\sqrt{3}a{)}^2-2\times a\times \sqrt{3}a+\sqrt{3}a{)}^2+a^2+2\times a\times \sqrt{3}a}BC=\sqrt{8a^2}=2\sqrt{2}a\text{units}\text{Similarly}AB=\sqrt{(a+a{)}^2+(a+a{)}^2}=\sqrt{8a^2}=2\sqrt{2}a\text{units}AC=\sqrt{(-\sqrt{3}a-a{)}^2+(\sqrt{3}a-a{)}^2}=\sqrt{8a^2}=2\sqrt{2}a\text{units}$

$\text{We get,}AB=BC=AC=2\sqrt{2}a\text{units}\therefore △\text{ABC is an equilateral triangle}.$

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 $C(-1,2)$ divide the line joining $A(2,5)$ and $B(x,y)$ in the ratio 3:4.

Then, 
$C(\frac{3x+8}{7},\frac{3y+20}{7})$ = C(-1, 2)

$\Rightarrow \frac{3x+8}{7}=-13x+8=-7\Rightarrow x=-5\Rightarrow \frac{3y+20}{7}=23y+20=14\Rightarrow y=-2$

Thus, the coordinates of B are $B(-5,-2).$

True. Midpoint of the line joining the points ($x_1,y_1$) and ($x_2,y_2$) is $(\frac{1\left(x_1\right)+1\left(x_2\right)}{1+1},\frac{1\left(y_1\right)+1\left(y_2\right)}{1+1})$ = $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$.

Point M (-2, y) is equidistant from the points A (5, 7) and B(3 , -4).

Distance AM = Distance BM
$\Rightarrow \sqrt{\left(\right(x_a-x_m{)}^2+(y_a-y_m{)}^2)})$ = $\sqrt{\left(\right(x_b-x_m{)}^2+(y_b-y_m{)}^2)})$
$\Rightarrow (5-(-2){)}^2$ + $(7-y{)}^2$ = $(3-(-2){)}^2$ + $(-4-y{)}^2$
$\Rightarrow 49+49-14y+y^2=25+16+8y+y^2$
$\Rightarrow 22y=57$
$\Rightarrow y=\frac{57}{22}$

Area of a triangle having vertices $(x_1,y_1),(x_2,y_2)\text{and}(x_3,y_3)$
$=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-\left(y_2\right)]$

Given, ($x_1,y_1$) = (-2, 1)

            ($x_2,y_2$) = (3, 3)

            ($x_3,y_3$) = (1, -2)

  $=\frac{1}{2}[-2(3+2)+3(-2-1)+1(1-3\left)\right]=-10.5$

Since area is not a negative quantity, take the absolute value.

Thus, area of the given triangle = 10.5 sq. units.

Since A, B and C are the vertices of a triangle, it follows that sum of two sides is greater than the third side. So (i) is correct.
And since the area of a triangle with vertices ($x_1$, $y_1$), ($x_2$ , $y_2$ ) and ($x_3$ , $y_3$ ) is
$\frac{1}{2}$[  $x_1$($y_2$  - $y_3$) +   $x_2$(  $y_3$ -   $y_1$) + $x_3$($y_1$ - $y_2$)]
= $\frac{1}{2}$ [a (d - f) + c (f - b) + e (b - d)] (by putting $x_1$  = a, $y_1$ = b, $x_2$ = c, $y_2$ = d, $x_3$ = e, $y_3$ = f)
Since the above area is not the same as that given in (ii), (ii) is incorrect.
 

Let, $(x,y)$ be coordinates of the mid-point.

From the section formula if $(x,y)$ are the midpoints of the line segment joining $(x_1,y_1)$ and $(x_2,y_2)$,

then $x=\frac{x_1+x_2}{2}$ and $y=\frac{y_1+y_2}{2}$

$x=\frac{2+1}{2}$ and $y=\frac{-3+1}{2}$

So, $(x,y)=(\frac{3}{2},-1)$

Distance formula for the points $(x_1,y_1)and(x_2,y_2)$ = $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
$\therefore$ Distance between the points (-1, 5) and (2, 1),
d = $\sqrt{{(2-(-1\left)\right)}^2+{(1-5)}^2}$
= $\sqrt{{(2+1)}^2+{(-4)}^2}$
= $\sqrt{9+16}$ = $\sqrt{25}$
= 5 units

Given, area of triangle = 5 sq. units.
Let the third vertex be $(x,y)$.
Area of a triangle formed by $(x_1,y_1),(x_2,y_2)\&(x_3,y_3)=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

$5=\frac{1}{2}|2(-2-y)+3(y-1)+3x|5=\frac{1}{2}|-4-2y+3y-3+3x|$

$3x+y-7=10\text{or}3x+y-7=-10$

$\Rightarrow 3x+y=17\text{or}3x+y=-3$

Also given that the third vertex lies on $y=x+3$. It means that the point of intersection of both the lines is the vertex.

Let us now solve $y=x+3\Rightarrow x-y=-3---\left(1\right)$ and $3x+y=17---\left(2\right)$.

Multiplying $\left(1\right)$ by 3, $\Rightarrow 3x-3y=-9---\left(3\right)$
$\left(2\right)-\left(3\right)\Rightarrow \underset{–}{3x+y=17{}^{}{}_{-}3x\mp 3y=\mp 9}4y=26y=\frac{13}{2}$.

Substituting this value of $y$ in $\left(1\right)$.
$x=y-3=\frac{13}{2}+3=\frac{7}{2}$
$\therefore x=\frac{7}{2}$ and $y=\frac{13}{2}$
 
Now, to solve  $3x-3y=-9---\left(3\right)$ and $3x+y=-3---\left(4\right)$

$\left(4\right)-\left(3\right)\Rightarrow \underset{–}{3x+y=-3{}^{}{}_{-}3x\mp 3y=\mp 9}4y=6y=\frac{3}{2}$.
Substituting this value of $y$ in $\left(1\right)$.
$x=y-3=\frac{3}{2}-3=-\frac{3}{2}$
$\therefore x=-\frac{3}{2}$ and $y=\frac{3}{2}$.


Vertex are $(\frac{7}{2},\frac{13}{2})\&(-\frac{3}{2},\frac{3}{2})$