10th Grade > Mathematics
COORDINATE GEOMETRY MCQs
Coordinate Geometry
:
B
Area of a triangle having vertices (x1,y1),(x2,y2) and (x3,y3)
=12[x1(y2−y3)+x2(y3−y1)+x3(y1−(y2)]
Given, (x1,y1) = (-2, 1)
(x2,y2) = (3, 3)
(x3,y3) = (1, -2)
=12[−2(3+2)+3(−2−1)+1(1−3)]=−10.5
Since area is not a negative quantity, take the absolute value.
Thus, area of the given triangle = 10.5 sq. units.
:
A
Since A, B and C are the vertices of a triangle, it follows that sum of two sides is greater than the third side. So (i) is correct.
And since the area of a triangle with vertices (x1, y1), (x2 , y2 ) and (x3 , y3 ) is
12[ x1(y2 - y3) + x2( y3 - y1) + x3(y1 - y2)]
= 12 [a (d - f) + c (f - b) + e (b - d)] (by putting x1 = a, y1 = b, x2 = c, y2 = d, x3 = e, y3 = f)
Since the above area is not the same as that given in (ii), (ii) is incorrect.
:
A
Let, (x,y) be coordinates of the mid-point.
From the section formula if (x,y) are the midpoints of the line segment joining (x1,y1) and (x2,y2),
then x=x1+x22 and y=y1+y22
x=2+12 and y=−3+12
So, (x,y)=(32,−1)
:
B
False. Any point that lies on the coordinate axes doesn't belong to any of the quadrants. The point (2,0) lies on the coordinate axes.
:
C and D
Given, area of triangle = 5 sq. units.
Let the third vertex be (x,y).
Area of a triangle formed by (x1,y1), (x2,y2) & (x3,y3)=12|x1(y2−y3)+x2(y3−y1)+x3(y1−y2)|
5=12|2(−2−y)+3(y−1)+3x|5=12|−4−2y+3y−3+3x|
3x+y−7=10 or 3x+y−7=−10
⇒3x+y=17 or 3x+y=−3
Also given that the third vertex lies on y=x+3. It means that the point of intersection of both the lines is the vertex.
Let us now solve y=x+3⇒x−y=−3 −−−(1) and 3x+y=17 −−−(2).
Multiplying (1) by 3, ⇒3x−3y=−9 −−−(3)
(2)−(3)⇒ 3x+y= 17−3x∓3y=∓9––––––––––––––––– 4y=26 y=132.
Substituting this value of y in (1).
x=y−3=132+3=72
∴x=72 and y=132
Now, to solve 3x−3y=−9−−−(3) and 3x+y=−3−−−(4)
(4)−(3)⇒ 3x+y =−3−3x∓3y=∓9––––––––––––––––– 4y=6 y=32.
Substituting this value of y in (1).
x=y−3=32−3=−32
∴x=−32 and y=32.
Vertex are (72,132) & (−32,32)
:
B and C
Distance between the points (x1,y1) and (x2,y2) is given by
√(x2−x1)2 + (y2−y1)2
∴PQ=√(3−x)2+(−6−2)2=10(3−x)2+(8)2=100(3−x)2+64=100(3−x)2=363−x=±6x=−3 or 9
:
A
If a point P(x,y) divides a line segment joining (x1,y1) and (x2,y2) in the ratio m:n, then the coordinates of P are given by:
x=mx2+nx1m+n, y=my2+ny1m+n
Let P(x,y) divide the line segment AB joining A(−1,7) and B(4,−3) in the ratio 2:3. Then by using section formula, the co-ordinates of P are given by:
(2×4+3×(−1)2+3,2×(−3)+3×72+3)
=(8−35,−6+215)
=(55,155)
Thus, the coordinates of P is (1,3)
:
B
Midpoint of the line segment joining (x1,y1) and (x2,y2) is given by (x1+x22,y1+y22)
∴ Mid-point of the line-segment joining the points (– 5, 4) and (9, – 8) = (−5+92,4−82) = (2,-2)
:
B
For 3 points to be collinear the area of the triangle formed by these three points should be zero.
Area of a triangle formed by (x1,y1), (x2,y2), (x3,y3) is given by
12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]
∴ 12[a(b−1)+0(1−0)+1(0−b)]=012[a(b−1)+1(0−b)]=0ab=a+b1a+1b = 1