# Coordinate Geometry Problems with Solutions

Given below are the Class 10 Maths Problems with Solutions for Coordinate Geometry

## Question 1 Prove that the points (a, b + c), (b, c + a) and (c, a + b) are collinear. Answer If the three point are collinear, then Area of the triangle will be zero Area of triangle is given by $A= \frac {1}{2} [x_1(y_2 - y_3) + x_2(y_3 -y_1) + x_3(y_1 - y_2)]$ $A= \frac {1}{2} [a(c+a -a -b) + b(a+b -b -c) + c (b+c - c-a)]$=0$Question 2 For what value of x will the points (x, -1), (2, 1) and (4, 5) lie on a line? Answer If the three point are collinear, then Area of the triangle will be zero Area of triangle is given by$A= \frac {1}{2} [x_1(y_2 - y_3) + x_2(y_3 -y_1) + x_3(y_1 - y_2)]0=\frac {1}{2} [x(1 -5) + 2(5+1) +4(-1-1)]0=-4x +12 -8$x=2 Question 3. Check whether the points (4, 5), (7, 6) and (6, 3) are collinear Answer If the point are collinear, then Area will be zero$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. Answer 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 ## Long Asnwer type Question 1. Find a point on the y – axis which is equidistant from the point A (6, 5) and B (-4, 3). Answer 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}}$36+y2+25-10y=16+y2+9-6y 4y=36 y=9 So point (0,9) Question 2. Show that the points (p, p), (-p, -p) and (-p3, p3) are the vertices of an equilateral triangle. Also find its area. Answer Let the point be A(p,p) ,B(-p,-p) and C(-p3, p3) Then AB=$\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}$=2 3p2 sq. units Question 3. Show that four points (0, -1), (6, 7), (-2, 3) and (8, 3) are the vertices of a rectangle. Also find its area. Answer Let the point be A(0,-1) ,B(6,7) , C(-2, 3) and D(8,3)3) Then AB=$\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}}$=√20 So Opposite sides are equal BC=AB AC=BD Diagonals are equal AB= CD So these are coordinates of the rectanges Now Area = Length X breath =2√20 X √20=40 Question 4. If two vertices of an equilateral triangle be (0,0), (3, 3), find the third vertex. Answer Two vertices of an equilateral triangle are (0, 0) and (3, √3). Let the third vertex of the equilaterla triangle be (x, y) Distance between (0, 0) and (x, y) = Distance between (0, 0) and (3, √3) = Distance between (x, y) and (3, √3) √(x2+ y2) = √(32); + 3) = √[(x - 3)2 + (y - √3)2] x2 + y2 = 12 --(A) √(x2 + y2)=√[(x - 3)2 + (y - √3)2] x2 + 9 - 6x + y2 + 3 - 2√3y = x2 + y2 12 - 6x - 2√3y = 0 3x + √3y = 6 x = (6 - √3y) / 3 -(B) Substituting the values of x in (A) [(6 - √3y)/3]2 + y2 = 12 (36 + 3y2 - 12√3y) / 9 + y2 = 12 36 + 3y2 - 12√3y + 9y2 = 108 - 12√3y + 12y2 - 72 = 0 -√3y + y2 - 6 = 0 (y - 2√3)(y + √3) = 0 y = 2√3 or - √3 If y = 2√3, x = (6 - 6) / 3 = 0 If y = -√3, x = (6 + 3) / 3 = 3 So, the third vertex of the equilateral triangle = (0, 2√3) or (3, -√3). Question 5. If P (2, -1), Q (3, 4), R (-2, 3) and S (-3, -2) be four points is a plane, show that PQRS is a rhombus but not a square. Find the area of the rhombus. Answer The given points are P(2,-1), Q(3,4), R(-2,3) and S(-3,-2).We have$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 2QS= \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$=24sq.units Question 6. Show that the points (-4, -1), (-2, -4), (4, 0) and (2, 3) are the vertices points of a rectangle. Answer let A(-4,-1), B(-2,4), C(4,0) and D(2,3) be the given points$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}$So diagonal are equal Hence, ABCD is a rectangle Question 7. Show that the points A (1, -2), B (3, 6), C (5, 10) and D (3, 2) are the vertices of a parallelogram. Answer$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}$Therefore Oppossite sides are equal AD=BC and AB=CD This proves that ABCD is a parallelogram Question 8. Prove that the points A (1, 7), B (4, 2), C (-1, -1) and D (-4, 4) are the vertices of a square. Answer$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}$Since All sides are equal and diagonal are equal to each other This is a square Question 9. Prove that the points (3, 0), (6, 4) and (-1, 3) are vertices of a right angled isosceles triangle. Answer Let A(3, 0), B(6, 4) and C(-1, 3)$AB=\sqrt {(3-6)^2 +(0-4)^2}=5BC=\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$, So it is right angled isosceles triangle. Question 10. Find the circumcenter of the triangle whose vertices are (-2, -3), (-1, 0), (7, -6) Answer Let the points be A(-2, -3), B(-1, 0), C(7, -6) Let the coordinates of the circumcentre of the triangle be O(x, y). Hence$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}$Squaring both the side and cutting coming terms 4x -3y=21 --(B) Solving (A) and (B), we get x=3 and y=-3 Therefore (3,-3) is the coordinate of circumcenter of the triangle Question 11. If two opposite vertices of a square are (5, 4) and (1, -6), find the co- ordinates of its remaining two vertices. Answer Let ABCD is the square and A (5,4) and C(1,-6). Lets us assume the coordinate of B(x,y) Now Sides are equal in square,Therefore AB=BC$\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 12. Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square. Question 13 If the points (p, q) (m, n) and (p - m, q -n) are collinear, show that pn = qm. Answer If the three point are collinear, then Area of the triangle will be zero Area of triangle is given by$A= \frac {1}{2} [x_1(y_2 - y_3) + x_2(y_3 -y_1) + x_3(y_1 - y_2)]0=\frac {1}{2} [p(n -q+n)) + m(q-n -q) +(p-m)(q-n)]0=-pq + 2pm -mn + pq-pn-pm +mn$pn = qm Question 14 Find k so tht the point P (-4, 6) lies on the line segment joining A (k, 10) and B (3, -8). Also, find the ratio in which P divides AB. Question 15 Find the area of the quadrilaterals, the co- ordinates of whose vertices are (i)(-3, 2), (5, 4), (7, -6) and (-5, -4) (ii)(1, 2), (6, 2), (5, 3) and (3, 4) (iii) (-4, -2), (-3, -5), (3, -2), (2, 3) Answer (i) let A(-3, 2), B(5, 4), C(7, -6) and D(-5, -4) Area of quadrilateral =Area of triangle ABC + Area of triangle ACD$=|\frac {1}{2} [-3(4 +6) + 5(-6 -2) +7(2-4)]| + |\frac {1}{2}[-3(-6 +4) + 7(-4 -2) -5(2+6)]|=|-42| + |-38|= 80 $(ii)Similarly A= 5.5 units Question 16 Show that the following sets of points are collinear (i) (2, 5), (4, 6) and (8, 8) (ii) (1, -1), (2, 1) and (4, 5) Answer If the three point are collinear, then Area of the triangle will be zero Area of triangle is given by$A= \frac {1}{2} [x_1(y_2 - y_3) + x_2(y_3 -y_1) + x_3(y_1 - y_2)]$(i)$A=\frac {1}{2} [2(6 -8) + 4(8 -5) +8(5-6)]$=0 (ii)$A=\frac {1}{2} [1(1 -5) + 2(5 +1) +4(-1-1)]$=0 Question 17. Find the value of x such that PQ = QR where the co- ordinates of P, Q and R are (6, -1), (1, 3) and (x, 8) respectively. Answer PQ=QR$\sqrt {(6-1)^2 + (-1-3)^2 } = \sqrt {(1-x)^2 + (3-8)^2}$Squaring both the sides$25 + 16 = 1 + x^2 -2x + 25x^2 -2x -15=0$x = 5 and -3 Also Read Go back to Class 10 Main Page using below links ### Practice Question Question 1 What is$1 - \sqrt {3}\$ ?
A) Non terminating repeating
B) Non terminating non repeating
C) Terminating
D) None of the above
Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
A) 19.4 cm3
B) 12 cm3
C) 78.6 cm3
D) 58.2 cm3
Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
A) 2 ,21,11
B) 1,10,19
C) -1 ,8,17
D) 2 ,11,20