Class 10 Maths Important Questions for Coordinate Geometry
Given below are the Class 10 Maths Important Questions for Coordinate Geometry
(a) Concepts questions
(b) Calculation problems
(c) Multiple choice questions
(d) Long answer questions
(e) Fill in the blank's
Short answer type
Question 1
Calculate the Following
Distance between the point (1,3) and ( 2,4)
Mid-point of line segment AB where A(2,5) and B( -5,5)
Area of the triangle formed by joining the line segments (0,0) ,( 2,0) and (3,0)
Distance of point (5,0) from Origin
Distance of point (5,-5) from Origin
Coordinate of the point M which divided the line segment A(2,3) and B( 5,6) in the ratio 2:3
Quadrant of the Mid-point of the line segment A(2,3) and B( 5,6)
the coordinates of a point A, where AB is the diameter of circle whose center is (2,−3) and B is (1, 4)
Solution
(a) $D=\sqrt{(1-2)^{2}+(3-4)^{2}}=\sqrt{2}$
(b)Mid-point is given by (2-5)/2,(5+5)/2 or (-3/2, 5)
(c) $A=\frac{1}{2}[0(0-0)+2(0-0)+3(0-0)]=0$
Since the three points are collinear, the area is zero
(d) $D=\sqrt{5^{2}+0^{2}}=5$
(e)$D=\sqrt{(-5)^{2}+0^{2}}=5$
(f) Coordinates of point M is given by
$x=\frac{2X3+3X2}{2+3}=\frac{12}{5}$
$y=\frac{2X6+3X5}{2+3}=\frac{27}{5}$
(g) Mid point is given by (7/2, 9/2) which lies in First quadrant
(h) We know that center is mid point of AB, So
$2=\frac{1+x}{2}$
$-3=\frac{4+y}{2}$
Solving these, we get (3,-10)
True or False statement
Question 2
(a) Point A( 0,0) B( 0,3) ,C( 0,7) and D( 2,0) formed a quadrilateral
(b) The point P (-2, 4) lies on a circle of radius 6 and center C (3, 5)
(c) Triangle PQR with vertices P (-2, 0), Q (2, 0) and R (0, 2) is similar to Δ XYZ with
Vertices X (-4, 0) Y (4, 0) and Z (0, 4).
(d) Point X (2, 2) Y (0, 0) and Z (3, 0) are not collinear
(e) The triangle formed by joining the point A( -3,0) , B( 0,0) and C( 0,2) is a right angle triangle
(f) A circle has its center at the origin and a point A (5, 0) lies on it. The point B (6, 8) lies inside the circle
(g) The points A (-1, -2), B (4, 3), C (2, 5) and D (-3, 0) in that order form a rectangle Solution
False, As three point are A,B and C are collinear, So no quadrilateral can be formed
False, As the distance between the point P and C is $\sqrt{26}$ which is less than 6.So point lies inside the circle
True. Both the triangle are equilateral triangle with side 4 and 8 respectively
True. As the Area formed by the triangle XYZ is not zero
True, If we plot the point on the Coordinate system, it becomes clear that it is right angle at origin
False. The radius of the circle is 5 and distance of the point B is more than 5,So it lies outside the circle
True. If we calculate the distance between two points, it becomes clear that opposite side are equal, also the diagonal are equal. So it is a rectangle
Multiple choice Questions
Question 3
Find the centroid of the triangle XYZ whose vertices are X (3, - 5) Y (- 3, 4) and Z (9, - 2).
(a) (0, 0)
(b) (3, 1)
(c) (2, 3)
(d) (3,-1) Solution
(d)
Centroid of the triangle is given by
$x=\frac{x_{1}+x_{2}+x_{3}}{3}=\frac{3-3+9}{3}=3$
$y=\frac{y_{1}+y_{2}+y_{3}}{3}=\frac{-5+4-2}{3}=-1 $
Question 4
The area of the triangle ABC with coordinates as A (1, 2) B (2, 5) and C (- 2, - 5)
(a)-1
(b) .4
(c)2
(d) 1 Solution
(d)
$A=\frac{1}{2}[1(5+5)+2(-5-2)-2(2-5)]=1$
Question 5
Find the value of p for which these point are collinear (7,-2) , (5,1) ,(3,p)?
(a) 2
(b) 4
(c) 3
(d) None of these Solution
a
For these points to be collinear
A=0
Or
$\frac{1}{2}[7(1-p)+5(p+2)+3(-2-1)]=0$
7-7p+5p+10-9=0
p=2
Question 6
Determine the ratio in which the line 2x + y - 4 = 0 divides the line segment joining the points A (2, - 2) and B (3, 7).
(a) 2:9
(b) 1:9
(c)1:2
(d) 2:3 Solution
(a)
Let the ratio be m: n
Now
Coordinate of the intersection
$x=\frac{3m+3n}{m+n}$
$y=\frac{7m-2n}{m+n}$
Now these points should lie of the line, So
$2(\frac{3m+2n}{m+n})+(\frac{7m-n}{m+n})-4=0$
m:n=2:9
Question 7
If the mid-point of the line segment joining the points A (3, 4) and B (a, 4) is P (x, y) and x + y - 20 = 0,then find the value of a
(a) 0
(b) 1
(c) 40
(d)45 Solution (d)
id point (3+a)/2, 4
Now
(3+a)/2 -4 -20=0
3+a=48
A=45
Long answer type
Question 8
Prove that the points (a, b + c), (b, c + a) and (c, a + b) are collinear. Solution
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 9
For what value of x will the points (x, -1), (2, 1) and (4, 5) lie on a line? Solution
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 10
If the points (p, q) (m, n) and (p - m, q -n) are collinear, show that pn = qm. Solution
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 11
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 12
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) Solution
(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 13
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) Solution
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 14.
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. Solution
PQ=QR
$\sqrt {(6-1)^2 + (-1-3)^2 } = \sqrt {(1-x)^2 + (3-8)^2}$
Squaring both the sides
$25 + 16 = 1 + x^2 -2x + 25$
$x^2 -2x -15=0$
x = 5 and -3
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