Maths Book back answers and solution for Exercise questions - Mathematics : Coordinate Geometry: Area of a Triangle and Area of a Quadrilateral: Exercise Problem Questions with Answer
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(1,-1),\ (-4,6),\ (-3,-5) $$
(-10,-4),\ (-8,-1),\ (-3,-5) $$
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\left(-\frac12,3\right),\ (-5,6),\ (-8,8) $$
(a,b+c),\ (b,c+a),\ (c,a+b) $$
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(2,3),\ (4,a),\ (6,-3) $$
(a,2-2a),\ (-a+1,2a),\ (-4-a,6-2a) $$
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(-9,-2),\ (-8,-4),\ (2,2),\ (1,-3) $$
(-9,0),\ (-8,6),\ (-1,-2),\ (-6,-3) $$
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$$ (-4,-2),\ (-3,k),\ (3,-2),\ (2,3) $$
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$$ A(-3,9),\ B(a,b),\ C(4,-5) $$
are collinear and if
$$ a+b=1 $$
then find $a$ and $b$.
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$$ P(11,7),\ Q(13.5,4),\ R(9.5,4) $$
be the mid-points of the sides $AB, BC, AC$ respectively of $\triangle ABC$.
Find the coordinates of the vertices $A,B,C$.
Hence find the area of $\triangle ABC$ and compare this with area of $\triangle PQR$.
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$$ A(-5,-4),\ B(1,6),\ C(7,-4) $$
has to be painted.
If one bucket of paint covers $6$ square feet, how many buckets of paint will be required to paint the whole glass, if only one coat of paint is applied.
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# Answers
$$ 24 \text{ sq.units} $$
11.5 \text{ sq.units} $$
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Collinear
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$$ 44 $$
13 $$
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$$ a=0 $$
a=\frac12 \text{ or } -1 $$
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$$ 35 \text{ sq.units} $$
34 \text{ sq.units} $$
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k=-5 $$
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a=2,\quad b=-1 $$
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\text{Area}(\triangle ABC)=24 \text{ sq.units} $$
$$ \text{Area}(\triangle ABC)=4\times \text{Area}(\triangle PQR) $$
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38 \text{ sq.units} $$
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10 \text{ cans} $$
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$$ 3.75 \text{ sq.units} $$
3 \text{ sq.units} $$
13.88 \text{ sq.units} $$
# UNIT 5 : Coordinate Geometry
Maths Book back answers and solution for Exercise questions - Mathematics : Coordinate Geometry: Inclination of a line: Exercise Problem Questions with Answer
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90^\circ $$
0^\circ $$
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0 $$
1 $$
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(5,\sqrt5) $$
with the origin.
(\sin\theta,-\cos\theta) $$
and
$$ (-\sin\theta,\cos\theta) $$
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$$ (4,2)\text{ and }(-6,4) $$
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$$ (-3,-4),\ (7,2),\ (12,5) $$
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$$ (3,-1),\ (a,3),\ (1,-3) $$
are collinear, find the value of $a$.
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$$ (-2,a)\text{ and }(9,3) $$
has slope
$$ -\frac12 $$
Find the value of $a$.
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$$ (-2,6)\text{ and }(4,8) $$
is perpendicular to the line through the points
$$ (8,12)\text{ and }(x,24) $$
Find the value of $x$.
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A(1,-4),\ B(2,-3),\ C(4,-7) $$
L(0,5),\ M(9,12),\ N(3,14) $$
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$$ A(2.5,3.5),\ B(10,-4),\ C(2.5,-2.5),\ D(-5,5) $$
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$$ A(2,2),\ B(-2,-3),\ C(1,-3)\text{ and }D(x,y) $$
form a parallelogram then find the values of $x$ and $y$.
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$$ A(3,-4),\ B(9,-4),\ C(5,-7),\ D(7,-7) $$
Show that $ABCD$ is a trapezium.
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$$ A(-4,-2),\ B(5,-1),\ C(6,5),\ D(-7,6) $$
Show that the midpoints of its sides form a parallelogram.
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$$ QS=2PR $$
If the coordinates of $S$ and $M$ are
$$ (1,1)\text{ and }(2,-1) $$
respectively, find the coordinates of $P$.
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# Answers
$$ \text{Undefined} $$
0 $$
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$$ 0^\circ $$
45^\circ $$
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$$ \frac1{\sqrt5} $$
-\cot\theta $$
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3 $$
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a=7 $$
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a=\frac{17}{2} $$
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x=4 $$
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Yes
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x=5,\quad y=2 $$
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\left(1,-\frac32\right) $$
or
$$ \left(3,-\frac12\right) $$
# UNIT 5 : Coordinate Geometry
Maths Book back answers and solution for Exercise questions - Mathematics : Coordinate Geometry: Straight line: Exercise Problem Questions with Answer
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$$ (1,-5),\ (4,2) $$
and parallel to
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$$ 2(x-y)+5=0 $$
Find its
- slope - inclination - intercept on the $Y$-axis.
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$$ \sqrt3x+(1-\sqrt3)y=3 $$
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$$ (-2,3)\text{ and }(8,5) $$
is perpendicular to
$$ y=ax+2 $$
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$$ (19,3) $$
The inclination of the hill to the ground is $45^\circ$.
Find the equation of the hill joining the foot and top.
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\left(2,\frac23\right)\text{ and }\left(-\frac12,-2\right) $$
(2,3)\text{ and }(-7,-1) $$
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$$ (-6,-4) $$
in the xy-plane.
A bottle of milk is kept at
$$ (5,11) $$
The cat wishes to consume the milk travelling through the shortest possible distance.
Find the equation of the path it needs to take.
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$$ A(6,2),\ B(-5,-1),\ C(1,9) $$
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$$ -\frac54 $$
and passing through the point
$$ (-1,2) $$
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The percent $y$ (in decimal form) of megabytes remaining to get downloaded in $x$ seconds is given by
$$ y=-0.1x+1 $$
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4,\ -6 $$
-5,\ \frac34 $$
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3x-2y-6=0 $$
4x+3y+12=0 $$
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$$ (1,-4) $$
and has intercepts which are in the ratio $2:5$.
$$ (-8,4) $$
and making equal intercepts on the coordinate axes.
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# Answers
$$ (1,-5)\text{ and }(4,2) $$
is
$$ \left(\frac52,-\frac32\right) $$
$$ y=-\frac32 $$
$$ x=\frac52 $$
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$$ 2(x-y)+5=0 $$
$$ 2x-2y+5=0 $$
$$ y=x+\frac52 $$
### Slope
$$ m=1 $$
### Inclination
$$ 45^\circ $$
### Y-intercept
$$ \frac52 $$
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m=\tan30^\circ=\frac1{\sqrt3} $$
Intercept on y-axis:
$$ c=-3 $$
Equation:
$$ y=\frac1{\sqrt3}x-3 $$
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$$ \sqrt3x+(1-\sqrt3)y=3 $$
$$ y=-\frac{\sqrt3}{1-\sqrt3}x+\frac3{1-\sqrt3} $$
### Slope
$$ m=-\frac{\sqrt3}{1-\sqrt3} $$
### Y-intercept
$$ \frac3{1-\sqrt3} $$
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$$ m=\frac{5-3}{8-(-2)} $$
$$ =\frac2{10}=\frac15 $$
Perpendicular slope:
$$ a=-5 $$
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$$ 45^\circ $$
So slope:
$$ m=1 $$
Passing through
$$ (19,3) $$
Equation:
$$ y-3=x-19 $$
$$ y=x-16 $$
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Equation:
$$ 8x-5y-12=0 $$
$$ 4x-9y+19=0 $$
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$$ m=\frac{11-(-4)}{5-(-6)} $$
$$ =\frac{15}{11} $$
Equation:
$$ y+4=\frac{15}{11}(x+6) $$
$$ 15x-11y+46=0 $$
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Midpoint of $BC$:
$$ \left(-2,4\right) $$
Equation:
$$ x+4y-14=0 $$
### Altitude through A
Slope of $BC$:
$$ \frac{9-(-1)}{1-(-5)}=\frac{10}{6}=\frac53 $$
Perpendicular slope:
$$ -\frac35 $$
Equation:
$$ 3x+5y-28=0 $$
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$$ y-2=-\frac54(x+1) $$
$$ 5y-10=-5x-5 $$
$$ 5x+5y-5=0 $$
or
$$ x+y-1=0 $$
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Total MB of song:
$$ 1 $$
$$ 25\%\text{ remaining}=0.25 $$
$$ 0.25=-0.1x+1 $$
$$ x=7.5 $$
seconds.
$$ y=0 $$
$$ 0=-0.1x+1 $$
$$ x=10 $$
seconds.
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Intercept form:
$$ \frac{x}{4}+\frac{y}{-6}=1 $$
$$ 3x-2y=12 $$
\frac{x}{-5}+\frac{y}{3/4}=1 $$
$$ -3x+20y=15 $$
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$$ 3x-2y-6=0 $$
x-intercept:
$$ 2 $$
y-intercept:
$$ -3 $$
4x+3y+12=0 $$
x-intercept:
$$ -3 $$
y-intercept:
$$ -4 $$
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Equation:
$$ 5x+2y+3=0 $$
$$ \frac{x}{a}+\frac{y}{a}=1 $$
$$ x+y=a $$
Passing through
$$ (-8,4) $$
$$ -8+4=a $$
$$ a=-4 $$
Equation:
$$ x+y+4=0 $$
# UNIT 5 : Coordinate Geometry
Maths Book back answers and solution for Exercise questions - Mathematics : Coordinate Geometry: General Form of a Straight Line: Exercise Problem Questions with Answer
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5y-3=0 $$
7x-\frac{3}{17}=0 $$
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$$ y=0.7x-11 $$
$$ x=-11 $$
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\frac{x}{3}+\frac{y}{4}+\frac17=0 $$
and
$$ \frac{2x}{3}+\frac{y}{2}+\frac1{10}=0 $$
5x+23y+14=0 $$
and
$$ 23x-5y+9=0 $$
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$$ 12y=-(p+3)x+12 $$
and
$$ 12x-7y=16 $$
are perpendicular then find $p$.
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$$ P(-5,2) $$
and parallel to the line joining the points
$$ Q(3,-2)\text{ and }R(-5,4) $$
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$$ (6,-2) $$
and perpendicular to the line joining the points
$$ (6,7)\text{ and }(2,-3) $$
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Find the equation of the altitude through $A$ and $B$.
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$$ A(-4,2)\text{ and }B(6,-4) $$
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$$ 7x+3y=10 $$
$$ 5x-4y=1 $$
and parallel to the line
$$ 13x+5y+12=0 $$
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$$ 5x-6y=2 $$
$$ 3x+2y=10 $$
and perpendicular to the line
$$ 4x-7y+13=0 $$
---
$$ 3x+y+2=0 $$
and
$$ x-2y-4=0 $$
to the point of intersection of
$$ 7x-3y=-12 $$
and
$$ 2y=x+3 $$
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$$ 8x+3y=18 $$
$$ 4x+5y=9 $$
and bisecting the line segment joining the points
$$ (5,-4)\text{ and }(-7,6) $$
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# Answers
$$ 0 $$
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$$ 0.7 $$
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Parallel
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p=4 $$
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3x+4y+7=0 $$
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2x+5y-2=0 $$
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$$ 2x+5y+6=0 $$
Altitude through $B$:
$$ 5x+y-48=0 $$
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5x-3y-8=0 $$
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13x+5y-18=0 $$
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49x+28y-156=0 $$
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31x+15y+30=0 $$
---
4x+13y-9=0 $$
# UNIT 5 : Coordinate Geometry # Multiple Choice Questions
Mathematics : Coordinate Geometry: Multiple choice questions with answers / choose the correct answer with answers - Maths Book back 1 mark questions and answers with solution for Exercise Problems
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# Multiple Choice Questions
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Base $=10$
Height $=5$
$$ \text{Area}=\frac12\times10\times5 $$
$$ =25 $$
### Answer
$$ \boxed{(2)\ 25\text{ sq.units}} $$
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Distance from Y-axis is constant.
Hence:
$$ x=10 $$
### Answer
$$ \boxed{(1)\ x=10} $$
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$$ \boxed{(2)\ \text{parallel to Y-axis}} $$
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For collinear points:
$$ \frac{p-7}{3-5}=\frac{6-7}{6-5} $$
$$ \frac{p-7}{-2}=-1 $$
$$ p-7=2 $$
$$ p=9 $$
### Answer
$$ \boxed{(3)\ 9} $$
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Add equations:
$$ 4x=12 $$
$$ x=3 $$
Then:
$$ y=5 $$
### Answer
$$ \boxed{(3)\ (3,5)} $$
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$$ \frac{a-3}{4-12}=\frac18 $$
$$ \frac{a-3}{-8}=\frac18 $$
$$ a-3=-1 $$
$$ a=2 $$
### Answer
$$ \boxed{(4)\ 2} $$
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Slope of given line:
$$ \frac{8-0}{-8-0}=-1 $$
Perpendicular slope:
$$ 1 $$
### Answer
$$ \boxed{(2)\ 1} $$
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Perpendicular slope:
$$ -\frac1{1/\sqrt3} $$
$$ =-\sqrt3 $$
### Answer
$$ \boxed{(2)\ -\sqrt3} $$
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Intercept form:
$$ \frac{x}{5}+\frac{y}{8}=1 $$
$$ 8x+5y=40 $$
### Answer
$$ \boxed{(1)\ 8x+5y=40} $$
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Given slope:
$$ \frac73 $$
Perpendicular slope:
$$ -\frac37 $$
Passing through origin:
$$ y=-\frac37x $$
$$ 3x+7y=0 $$
### Answer
$$ \boxed{(3)\ 3x+7y=0} $$
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Slopes:
$$ m_1=\frac43 $$
$$ m_2=\frac34 $$
$$ m_3=-\frac34 $$
$$ m_4=-\frac43 $$
Since:
$$ m_2\times m_4=-1 $$
they are perpendicular.
### Answer
$$ \boxed{(3)\ l_2 \text{ and } l_4 \text{ are perpendicular}} $$
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$$ y=\frac12x+\frac{21}{8} $$
Slope:
$$ 0.5 $$
Intercept:
$$ 2.625\approx2.6 $$
### Answer
$$ \boxed{(1)} $$
---
$$ \boxed{(2)} $$
---
$$ \boxed{(2)} $$
---
Check option (2):
$$ 2+1=3 $$
$$ 3(2)+1=7 $$
Both satisfy.
### Answer
$$ \boxed{(2)} $$
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# Answer Key
| Q.No | Answer | |---|---| | 1 | 2 | | 2 | 1 | | 3 | 2 | | 4 | 3 | | 5 | 3 | | 6 | 4 | | 7 | 2 | | 8 | 2 | | 9 | 1 | | 10 | 3 | | 11 | 3 | | 12 | 1 | | 13 | 2 | | 14 | 2 | | 15 | 2 | # UNIT 5 : Coordinate Geometry
Rhombus
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$$ \left(\frac72,\frac{13}2\right) $$
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$$ 0\text{ sq.units} $$
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$$ k=-5 $$
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Find slopes:
Slope of line joining
$$ (-2,-1)\text{ and }(4,0) $$
$$ =\frac{0+1}{4+2} =\frac16 $$
Slope of line joining
$$ (3,3)\text{ and }(-3,2) $$
$$ =\frac{2-3}{-3-3} =\frac{-1}{-6} =\frac16 $$
Hence opposite sides are parallel.
Similarly,
Slope of line joining
$$ (4,0)\text{ and }(3,3) $$
$$ =\frac{3-0}{3-4} =-3 $$
Slope of line joining
$$ (-2,-1)\text{ and }(-3,2) $$
$$ =\frac{2+1}{-3+2} =-3 $$
Hence both pairs of opposite sides are parallel.
Therefore the quadrilateral is a parallelogram.
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$$ 2x-3y-6=0 $$
$$ 3x-2y+6=0 $$
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$$ 1340\text{ litres} $$
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$$ (-1,-4) $$
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$$ 13x+13y-6=0 $$
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$$ 119x+102y-125=0 $$
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# Answer Key
| Q.No | Answer | |---|---| | 1 | Rhombus | | 2 | $(7/2,\ 13/2)$ | | 3 | $0$ sq.units | | 4 | $-5$ | | 5 | Parallelogram | | 6 | $2x-3y-6=0,\ 3x-2y+6=0$ | | 7 | $1340$ litres | | 8 | $(-1,-4)$ | | 9 | $13x+13y-6=0$ | | 10 | $119x+102y-125=0$ |
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