Free Coordinate Geometry 03 Practice Test - 10th Grade
The co-ordinates of the points which divides the line joining (– 2, – 2) and (– 5, 7) in the ratio 2 : 1 is _____.
(4, – 4)
(– 3, 1)
(– 4, 4)
(1, – 3)
Solution : C
If a point P(x,y) divides the line joining (x1,y1) and (x2,y2) in the ratio m:n
then x = mx2+nx1m+n and y = my2+ny1m+n
∴x = mx2+nx1m+n = 2(−5)+1(−2)2+1 = -4
and y = my2+ny1m+n = 2(7)+1(−2)2+1 = 4
∴ the required point is (-4,4).
The distance of any point (x,y) from the origin is =
Solution : C
The distance of a point (x,y) from the origin (0,0) is √(x−0)2+(y−0)2=√x2+y2
What is the x coordinate of point C?
Point C lies on the line 'l' which is parallel to Y-axis. So the x-coordinate of any point on this line will be constant.
Therefore its x coordinate is 4
The ratio in which the line segment joining the points (1, – 7) and (6, 4) is divided by x-axis is _____.
Solution : D
Let C(x,0) divide AB in the ratio k:1.
By section formula, the coordinates of C are given by: (6k+1k+1,4k−7k+1)
But C(x,0)=(6k+1k+1,4k−7k+1) ⇒4k−7k+1=0
⇒4k−7 = 0 ⇒k = 74
i.e., the x-axis divides AB in the ratio 7:4.
The co-ordinates of the vertices of Triangle ABC are A(4,1),B(–3,2) and C(0,k). Given that the area of △ABC is 12 sq. units. Find the value(s) of k.
Solution : A and B
Area of a triangle formed by (x1,y1), (x2,y2) (x3,y3) = 12|x1(y2 - y3) + x2(y3 - y1)+ x3(y1 - y2)|
Thus, the area of ΔABC formed by the given points A(4, 1), B(-3, 2) and C(0, k) is
But given that the area of ΔABC=12 sq. unit
±(11−7k)=24 ⇒11−7k=24 or −(11−7k)=24
If11−7k=24, then −7k=24−11=13 ⇒k=−137
Hence the values of k are : 5, −137
The value of p for which the points (–1, 3), (2, p), (5, –1) are collinear is
Solution : B
The given points A(-1, 3), B(2, p), C(5, -1) are collinear.
⇒ Area ΔABC formed by these points should be zero.
⇒ The area of ΔABC = 0
⇒ 12( x1( y2- y3) + x2( y3- y1) + x3( y1 - y2)) = 0
⇒ -1(p + 1) + 2 (-1 - 3) + 5(3 -p) = 0
⇒ -p - 1 - 8 + 15 - 5p = 0
⇒ -6p +15 - 9 = 0 ⇒ -6p = -6 ⇒ p = 1
Hence the value of p is 1.
The distance between the points (a + b, b + c) and (a – b, c – b) is :
Solution : C
Let, A be (a+b, b+c) and B be (a−b, c−b)
∵ Distance between the points (x1,y1) and (x2,y2) is
√(x2−x1)2 + (y2−y1)2
∴AB=√(a−b−a−b)2+(c−b−b−c)2 =√(−2b)2+(−2b)2=√8b2AB=2√2 b
∴ Distance between the points = 2√2 b
Points P and Q are two points in the coordinate plane. The x-coordinate of the point equidistant from the points P and Q and lying on the line y = -1 is
The required point is equidistant from the points P and Q, and also lies on the line y = -1.
Let the point be A, it lies on y = -1
Therefore consider the point A(x_a, -1)
Distance AP = Distance AQ
√(xa−1)2+(−1−3)2) = √(xa+5)2+(−1+5)2
(xa−1)2 + (−4)2 = (xa+5)2 + (4)2
-2xa + 1 = 10xa + 25
-12xa = 24
xa = -2
The coordinates of the point A is (-2, -1).
In the figure below, is any side of ΔABC parallel to X axis? If so, the side is
Any line whose y-coordinate is a constant will be parallel to the X axis. Observe that B
and C have the same y coordinate, c, and hence this side BC is parallel to the X-axis.
Similarly, A and B have the same x – coordinates, -a, hence this side is parallel to the Y axis.
A, B, C are the three vertices of a triangle such that they are equidistant from the origin. A and B lie on the positive X and Y axes respectively and C lies on the line y + x = a. If AB = √2a, find the area of the triangle. (C lies in the first quadrant)
Solution : D
A and B are equidistant from origin and AB is √2a.
If we assume the points A and B to be (x,0) and (0,x) respectively,
AB= √x2+x2 = √2x = √2a
So A(0,a) and B(a,0) are the vertices.
Observe that A,B both lie on the line x+y=a and since C also lies on the line y+x = a, they are collinear.
Hence area of the triangle formed by the three points is 0.