Given below are the Class 10 Maths Problems with Solutions for Coordinate Geometry
(a) Concepts questions
(b) Calculation problems
(c) Long answer questions

Question 1.
Find a point on the y – axis which is equidistant from the point A (6, 5) and B (-4, 3). Solution

Let the point of Y-axis is (0,y)
Then
$\sqrt{(0-6)^{2}+(y-5)^{2}}=\sqrt{(0+4)^{2}+(y-3)^{2}}$
Squaring both the sides
${(0-6)^{2}+(y-5)^{2}}={(0+4)^{2}+(y-3)^{2}}$
36+y^{2}+25-10y=16+y^{2}+9-6y
4y=36
y=9
So point (0,9)

Question 2.
Show that the points (p, p), (-p, -p) and (-p√3, p√3) are the vertices of an equilateral triangle. Also find its area. Solution

Let the point be A(p,p) ,B(-p,-p) and C(-p√3, p√3)
Then
AB=$\sqrt{(-p-p)^{2}+(-p-p)^{2}}$=2p√2
BC=$\sqrt{(-p\sqrt{3}+p)^{2}+(p\sqrt{3}+p)^{2}}$=2p√2
AC=$\sqrt{(-p\sqrt{3}-p)^{2}+(p\sqrt{3}-p)^{2}}$=2p√2
All Sides equal,It is an equilateral triangle
Area of triangle
=$\frac{(side)^2\sqrt {3}}{4}$
=2 √3p^{2} sq. units

Question 3.
Check whether the points (4, 5), (7, 6) and (6, 3) are collinear Solution

If the point are collinear, then Area will be zero
$A=\frac{1}{2}[4(6-3)+7(3-5)+6(5-6)]=-8$
So they are not collinear

Question 4.
Find the value of q for which the points (7, -2), (5, 1), and (3, q) are collinear. Solution

If the point are collinear, then Area must be zero
$0=\frac{1}{2}[7(1-q)+5(q+2)+3(-2-1)]$
q=4

Question 5.
Show that four points (0, -1), (6, 7), (-2, 3) and (8, 3) are the vertices of a rectangle. Also find its area. Solution

Let the point be A(0,-1) ,B(6,7) , C(-2, 3) and D(8,3)3)
Then
AB=$\sqrt{(0-6)^{2}+(-1-7)^{2}}$=10
BC=$\sqrt{(6+2)^{2}+(7-3)^{2}}$=2√20
CD=$\sqrt{(-2-8)^{2}+(3-3)^{2}}$=10
AD=$\sqrt{(0-8)^{2}+(3+1)^{2}}$=2√20
AC=$\sqrt{(0+2)^{2}+(-1-3)^{2}}$=√20
BD=$\sqrt{(6-8)^{2}+(7-3)^{2}}$=√20

So Opposite sides are equal
BC=AB
AC=BD
Diagonals are equal
AB= CD
So these are coordinates of the rectanges

Now Area = Length X breath
=2√20 X √20=40

Question 6.
If two vertices of an equilateral triangle be (0,0), (3, √3), find the third vertex. Solution

Two vertices of an equilateral triangle are (0, 0) and (3, √3).

Let the third vertex of the equilaterla triangle be (x, y)

Distance between (0, 0) and (x, y) = Distance between (0, 0) and (3, √3) = Distance between (x, y) and (3, √3)

If y = 2√3, x = (6 - 6) / 3 = 0
If y = -√3, x = (6 + 3) / 3 = 3

So, the third vertex of the equilateral triangle = (0, 2√3) or (3, -√3).

Question 7. If P (2, -1), Q (3, 4), R (-2, 3) and S (-3, -2) be four points is a plane, show that PQRS is a rhombus but not a square. Find the area of the rhombus. Solution

The given points are P(2,-1), Q(3,4), R(-2,3) and S(-3,-2).We have
$PQ= \sqrt {(3-2)^2 +(4+1)^2}=\sqrt {26}$
$QR= \sqrt {(-2-3)^2 +(3-4)^2} =\sqrt {26}$
$RS= \sqrt {(-3+2)^2 +(-2-3)^2} =\sqrt {26}$
$SP= \sqrt {(-3-2)^2 +(-2-3)^2} =\sqrt {26}$
$PR= \sqrt {(-2-2)^2 +(3+1)^2} = 4 \sqrt 2$
$QS= \sqrt {(-3-3)^2 +(-2-4)^2}=6 \sqrt 2$

Therefore Sides are equal
PQ=QR=RS=SP= $\sqrt {26}$
and Diagonal are not equal
$PR \ne QS$
Hence PQRS is a rhombus but not a square.
Now Area of rhombus PQRS is given by
$A= \frac {1}{2} D_1 \times D_2$
Where $D_1$ and $D_2$ are diagonals
So
$A=\frac {1}{2} 4 \sqrt 2 \times 6 \sqrt 2$
=24sq.units

Question 8.
Show that the points (-4, -1), (-2, -4), (4, 0) and (2, 3) are the vertices points of a rectangle. Solution

let A(-4,-1), B(-2,4), C(4,0) and D(2,3) be the given points
$AB= \sqrt {(-2+4)^2 +(-4+1)^2}=\sqrt {13}$
$BC= \sqrt {(4+2)^2 +(0+4)^2}= \sqrt {52}$
$CD= \sqrt {(2-4)^2 +(3-0)^2} = \sqrt {13}$
$AD= \sqrt {(2+4)^2 +(3+1)^2}= \sqrt {52}$
Therefore Oppossite sides are equal
AD=BC and AB=CD
This proves that ABCD is a parallelogram
Now lets look at the diagonals
$AC= \sqrt {(4+4)^2 +(0+1)^2} = \sqrt {65}$
?$BD= \sqrt {(2+2)^2 +(3+4)^2} = \sqrt {65}$
So diagonal are equal
Hence, ABCD is a rectangle

Question 9.
Show that the points A (1, -2), B (3, 6), C (5, 10) and D (3, 2) are the vertices of a parallelogram. Solution

$AB= \sqrt {(1-3)^2 +(-2-6)^2}=\sqrt {68}$
$BC= \sqrt {(3-5)^2 +(6-10)^2}= \sqrt {20}$
$CD= \sqrt {(5-3)^2 +(10-2)^2} = \sqrt {68}$
$AD= \sqrt {(1-3)^2 +(-2-2)^2}= \sqrt {20}$
Therefore Oppossite sides are equal
AD=BC and AB=CD
This proves that ABCD is a parallelogram

Question 10.
Prove that the points A (1, 7), B (4, 2), C (-1, -1) and D (-4, 4) are the vertices of a square. Solution

$AB= \sqrt {(1-4)^2 +(7-2)^2}=\sqrt {34}$
$BC= \sqrt {(4+1)^2 +(2+1)^2}= \sqrt {34}$
$CD= \sqrt {(-1+4)^2 +(-1-4)^2} = \sqrt {34}$
$AD= \sqrt {(1+4)^2 +(7-4)^2}= \sqrt {34}$
$AC= \sqrt {(1+1)^2 +(7+1)^2}=\sqrt {68}$
$BD= \sqrt {(4+4)^2 +(2-4)^2}=\sqrt {68}$
Since All sides are equal and diagonal are equal to each other
This is a square

Question 11.
Prove that the points (3, 0), (6, 4) and (-1, 3) are vertices of a right angled isosceles triangle. Solution

Let A(3, 0), B(6, 4) and C(-1, 3)
$AB=\sqrt {(3-6)^2 +(0-4)^2}=5$
$BC=\sqrt {(6+1)^2 +(4-3)^2}=\sqrt {50}$
$AC=\sqrt {(3+1)^2 +(0-3)^2}=5$
Two sides are equal,so it is isosceles triangle
Also
$BC^2 = AB^2 + AC^2$, So it is right angled isosceles triangle.

Question 12.
Find the circumcenter of the triangle whose vertices are (-2, -3), (-1, 0), (7, -6) Solution

Let the points be A(-2, -3), B(-1, 0), C(7, -6)
Let the coordinates of the circumcentre of the triangle be O(x, y).
Hence
$OA= \sqrt {(x+2)^2 +(y+3)^2}$
$OB= \sqrt {(x+1)^2 +(y)^2}$
$OC= \sqrt {(x-7)^2 +(y+6)^2}$
Now
(x, y) will the equidistant from the vertices of the triangle as it is the circumcenter
OA=OB=OC
Taking OA=OB
$\sqrt {(x+2)^2 +(y+3)^2}=\sqrt {(x+1)^2 +(y)^2}$
Squaring both the side and cutting coming terms
x + 3y = -6 --(A)
Taking OB=OC
$\sqrt {(x+1)^2 +(y)^2}=\sqrt {(x-7)^2 +(y+6)^2}$
Squaring both the side and cutting coming terms
4x -3y=21 --(B)
Solving (A) and (B), we get
x=3 and y=-3
Therefore (3,-3) is the coordinate of circumcenter of the triangle

Question 13.
If two opposite vertices of a square are (5, 4) and (1, -6), find the co- ordinates of its remaining two vertices. Solution

Let ABCD is the square and A (5,4) and C(1,-6). Lets us assume the coordinate of B(x,y)
Now Sides are equal in square,Therefore
AB=BC
$\sqrt {(x-5)^2 +(y-4)^2}=\sqrt {(x-1)^2 +(y+6)^2}$
solving this
2x + 5y=1
or
$x=\frac {1- 5y}{2}$ -(1)
Now
Diagonal= AC = $\sqrt {(5-1)^2 +(4+6)^2}= 2 \sqrt {29}$
Now $\text{Diagonal} = \text{side} \times \sqrt 2$
Therefore
$\text{side}= \sqrt {58}$
$\text{side}=AB= \sqrt {(x-5)^2 +(y-4)^2}$
or
$58 = (x-5)^2 +(y-4)^2$
Putting the value of x from (1), we get
$[(\frac {1- 5y}{2} -5)^2 +(y-4)^2]=58$
Solving it ,we get
$29y^2 + 58y -87=0$
Solving this quadratic equation ,we get
y=-3 or 1
Substituting the value of y in (1) ,we get
x=8 or -2
So coordinates of the other vertices are (8, -3) and (-2, 1)

Question 14.
Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square.

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