CBSE 10th Standard Maths Subject Coordinate Geometry Ncert Exemplar 2 Marks Questions 2021
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CBSE 10th Standard Maths Subject Coordinate Geometry Ncert Exemplar 2 Marks Questions 2021
10th Standard CBSE

Reg.No. :
Maths

Name the type of triangle formed by the points A(5, 6), B(4, 2) and C(7, 5).
(a) 
Find the value of m if the points (5, 1), (2, 3) and (8, 2m) are collinear.
(a) 
The points A(2, 9), B(a, 5) and C(5, 5) Are the vertices of \(\triangle ABC\) right angled at B. Find the value of a and hence the area of \(\triangle ABC\).
(a) 
Name the type of triangle PQR formed by the points \(P(\sqrt { 2 } ,\sqrt { 2 } ),Q(\sqrt { 2 } ,\sqrt { 2 } )\) and \(R(\sqrt { 6 } ,\sqrt { 6 } )\)
(a) 
Show that \(\Delta \)ABC with vertices A(2, 0), B(0, 2) and C(2,0) is similar to \(\Delta \)DFE with vertices D( 4, 0), E(4, 0) and F(0, 4).
(a) 
Find the area of a triangle with vertices (a, b +c), (b, c + a) and (c, a + b).
(a) 
Check, whether the points (4, 0), (4, 0) and (0, 3) are the vertices of an isosceles triangle or equilateral triangle.
(a) 
Show that the points A (6, 10), B(4, 6) and C(3, 8) are collinear, such that \(AB=\frac { 2 }{ 9 } AC\) .
(a) 
Find the angle subtended by these points \(P\left( \sqrt { 2 } ,\sqrt { 2 } \right) ,Q(\sqrt { 2 } ,\sqrt { 2 } )\quad and\quad R(\sqrt { 6 } ,\sqrt { 6 } ).\)
(a) 
Find the perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0).
[Hint The perimeter of a triangle is the sum of lengths of its three sides, so first find the length of three sides and then add them.](a)
2 Marks
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CBSE 10th Standard Maths Subject Coordinate Geometry Ncert Exemplar 2 Marks Questions 2021 Answer Keys

scalene triangle

\(m=\frac { 19 }{ 14 } \)

\(a=2,\ OR\ \left( \triangle ABC \right) =6\ sq.units\)

We have P(\(\sqrt{2}\) , \(\sqrt{2}\)) , Q(\(\sqrt{2}\),\(\sqrt{2}\)) and R(\(\sqrt{6}\),\(\sqrt{6}\))
\(\therefore\) PQ = \(\sqrt { \left( \sqrt { 2 } +\sqrt { 2 } \right) ^{ 2 }+\left( \sqrt { 2 } +\sqrt { 2 } \right) ^{ 2 } } \)
\(=\sqrt { \left( 2\sqrt { 2 } \right) ^{ 2 }+\left( 2\sqrt { 2 } \right) ^{ 2 } } \)
\(=\sqrt { 4\times 2+4\times 2 } =\sqrt { 8+8 } \)
\(=\sqrt { 16 } \)=4 units
PR = \(\sqrt { \left( \sqrt { 2 } +\sqrt { 6 } \right) ^{ 2 }+\left( \sqrt { 2 } +\sqrt { 6 } \right) ^{ 2 } } \)
\(=\sqrt { 2+6+2\sqrt { 2 } +2+62\sqrt { 2 } } \)
\(=\sqrt { 2+6+2+6 } \)
\(=\sqrt { 16 } \) = 4 units
RQ = \(\sqrt { [(\sqrt { 2 } )+\sqrt { 6 } ^{ 2 }+\left( \sqrt { 2 } \sqrt { 6 } \right) ^{ 2 } } \)
\(=\sqrt { 2+62\sqrt { 2 } +2+6+2\sqrt { 2 } } \)
\(=\sqrt { 2+6+2+6 } \)
\(=\sqrt { 16 } \) =4 units
Since PQ=PR=RQ = 4 units
∴ PQR is an equilateral triangle. 
Given, vertices of \(\Delta \)ABC are A(2, 0), B(0, 2) and C(2,0) and D( 4, 0), E(4, 0) and F(0, 4).
Now, AB = \(\sqrt { { (0+2) }^{ 2 }+{ (20) }^{ 2 } } =\sqrt { 4+4 } =2\sqrt { 2 } \) units [\(\because \) distance=\(\sqrt { { \left( { x }_{ 2 }{ x }_{ 1 } \right) }^{ 2 }{ \left( { y }_{ 2 }{ y }_{ 1 } \right) }^{ 2 } } \)]
BC = \(\sqrt { { (20) }^{ 2 }+{ (02) }^{ 2 } } \) \( =\sqrt { 4+4 } =2\sqrt { 2 } \) units
CA = \(\sqrt { { (22) }^{ 2 }+{ (00) }^{ 2 } } \) = \(\sqrt { { (4) }^{ 2 }+0 } \)= 4 units
FD = \(\sqrt { { (0+4) }^{ 2 }+{ (40) }^{ 2 } } \) = \(\sqrt { { (4) }^{ 2 }+{ (4) }^{ 2 } } \) = \(4\sqrt { 2 } \) units
FE = \(\sqrt { { (40) }^{ 2 }+{ (04) }^{ 2 } } \) = \(\sqrt { { (4) }^{ 2 }+{ (4) }^{ 2 } } \) = \(4\sqrt { 2 } \) units
and ED = \(\sqrt { { (44) }^{ 2 }+{ (00) }^{ 2 } } \) = \(\sqrt { { (8) }^{ 2 }} \) = \(\sqrt { {64} }\) = 8 units
Here, we see that sides of \(\Delta \) DEF are twice the sides of \(\Delta \)ABC.
Hence, both the triangle are similar.
Hence proved. 
Here, (x_{1}, y_{1}) = (a, b + c), (x_{2}, y_{2}) = (b,c + a) and (x_{3}, y_{3}) = (c, a + b)
Now, area of a triangle
=\(\frac { 1 }{ 2 } \)[x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{1}  x_{1}y_{3}  x_{2}y_{1}  x_{3}y_{2}]
=\(\frac { 1 }{ 2 } \)[a(c + a) + b(a + b) + c(b + c)a(a + b)  b(b + c)  c(c + a)]
=\(\frac { 1 }{ 2 } \)[ac + a^{2} + ba + b^{2} + cb + c^{2 } a^{2}  ab  b^{2}  bc  c^{2}  ca]
=\(\frac { 1 }{ 2 } \)(0) = 0
Hence, the area of a triangle with vertices (a,b + c), (b, c + a) and (c, a + b) is zero. 
Let A = (x_{1},y_{1}) = (4, 0), B = (x_{2},y_{2}) = (4, 0) and C = (x_{3}, y_{3}) = (0, 3)
Now, AB = \(\sqrt { { [4(4)] }^{ 2 }+{ (00) }^{ 2 } } \) [ using distance formula ]
\(=\sqrt { { (4+4) }^{ 2 } } =\sqrt { { 8 }^{ 2 } } =8\quad units\)
\(BC=\sqrt { (04{ ) }^{ 2 }+{ (30) }^{ 2 } } =\sqrt { { (4) }^{ 2 }+{ (3) }^{ 2 } } \)
\(=\sqrt { 16+9 } =\sqrt { 25 } =5\quad units\)
and AC \(=\sqrt { [0(4){ ] }^{ 2 }+(30{ ) }^{ 2 } } \)
\(=\sqrt { 16+9 } =\sqrt { 25 } =5\quad units\)
\(\because BC = AC\)
So, \(\triangle \) ABC is an isosceles triangle. 
Here, coordinates of \(A\equiv ({ x }_{ 1 },{ y }_{ 2 })\) = (6, 10),
Coordinates of B \(\equiv \) (x_{2}, y_{2}) = (4, 6) and
Coordinates of C \(\equiv \) (x_{3}, y_{3}) = (3, 8).
We know that,
Area of triangle = \(\frac { 1 }{ 2 } \left { x }_{ 1 }({ y }_{ 2 }{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }{ y }_{ 2 }) \right \)
\(\therefore\) Area of \(\Delta ABC\) = \(\frac { 1 }{ 2 } \left 6\{ 6(8)\} +(4)(810)+3(106) \right \)
\(=\frac { 1 }{ 2 } \left 6(14)+(4)(18)+2(4) \right \)
\(=\frac { 1 }{ 2 } \left 84+72+12 \right =0\)
Since, area of \(\Delta ABC\) is zero. So, points A, B and C are collinear.
Now, \(AB=\sqrt { { (4+6) }^{ 2 }+{ (610) }^{ 2 } } \)
[ using distance formula ]
\(=\sqrt { { 2 }^{ 2 }+{ (4) }^{ 2 } } =\sqrt { 4+16 } =\sqrt { 20 } \)
\(=2\sqrt { 5 } units\)
\(AC=\sqrt { ({ 3+6) }^{ 2 }+(810)^{ 2 } } \)
\(=\sqrt { { 9 }^{ 2 }+({ 18 })^{ 2 } } =\sqrt { 81+324 } \)
\(=\sqrt { 405 } =\sqrt { 81\times 5 } =9\sqrt { 5 } units\)
\(\therefore \quad AB=2\sqrt { 5 } \times \frac { 9 }{ 9 } =\frac { 2 }{ 9 } AC\)
Hence proved. 
First find the rype of triangle using di ranee formula and hence obtain the angles.
First find the rype of triangle using di ranee formula and hence obtain the angles.

It is clear from the figure that,
Perimeter of \(\Delta\)AOB
=Distance (AO) + Distance (OB) + Distance (AB)
\(=\sqrt { { (00) }^{ 2 }+{ (40) }^{ 2 } } +\sqrt { { (03) }^{ 2 }+{ (00) }^{ 2 } } +\sqrt { { (03) }^{ 2 }+{ (40) }^{ 2 } } \)
[ using distance formula]
\(=\sqrt { { ({ x }_{ 1 }{ x }_{ 2 }) }^{ 2 }+{ ({ y }_{ 1 }{ y }_{ 2 }) }^{ 2 } } \)
\(=\sqrt { { (4) }^{ 2 } } +\sqrt { { (3) }^{ 2 } } +\sqrt { 9+16 } \)
= 4+3+\(\sqrt { 25 } \) = 4 + 3 + 5 = 12 units
Hence, the perimeter of given triangle is 12 units.
2 Marks