Coordinate Geometry Questions for XAT 2022 – Download PDF
Download Coordinate Geometry Questions for XAT PDF – XAT Coordinate Geometry questions pdf by Cracku. Top 10 very Important Coordinate Geometry Questions for XAT based on asked questions in previous exam papers.
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Question 1: The vertices of a triangle are (0,0), (4,0) and (3,9). The area of the circle passing through these three points is
a) $\frac{14\pi}{3}$
b) $\frac{123\pi}{7}$
c) $\frac{12\pi}{5}$
d) $\frac{205\pi}{9}$
Question 2: The following 2 points (5,5) and (-5,-5) form 2 vertices of a square. Which of the following points does not represent a vertex of that square?
a) (15,-5)
b) (-5,5)
c) (5,15)
d) (-5,15)
Question 3: A triangle is formed by the lines $l_1 : 3x -2y +18 =0 $ , $l_2 : 2x-y+7 =0 $ and $l_3 : 5x-2y+6 =0 $. What is sum (in sq units) of the square of the length of the sides?
a) 56
b) 86
c) 122
d) 140
Question 4: A triangle is formed by line $x+y=4$ and coordinate-axis. What is the ratio of inradius to circumradius for the given triangle?
a) $\sqrt{\ 2}$
b) $\frac{\sqrt{\ 2}}{2-\sqrt{\ 2}}$
c) $\frac{\sqrt{\ 2}}{2+\sqrt{\ 2}}$
d) $\frac{1}{2}$
Question 5: Find the coordinates of the point of reflection of the point (3.4) along the line 3x+y=6.
a) (-6/5,13/5
b) (-8/5,-4/5)
c) (-8/5,14/5)
d) (-8/5,21/5)
e) e
Question 6: The reflection of point P around the X axis is P’. The reflection of P’ around Y=X is P’’. If the coordinates of P’’ are (2,5), the coordinates of P are?
a) (5,-2)
b) (5,2)
c) (2,-5)
d) (-2,-5)
Question 7: If the point of intersection of the lines bx+4y+1=0 and ax+8y+1=0 lies on the line cx+12y+1=0, then what is the relation between a,b and c ?
a) a+c = 2b
b) a+b = 2c
c) b+c = 2a
d) a+2c =b
Question 8: In what ratio does the point T(x,0) divide the segment joining the points S(-4,-1) and U(1,4)?
a) 1 : 4
b) 4 : 1
c) 1 : 2
d) 2 : 1
Question 9: Slope of the line AB is -2/3. Co-ordinates of points A and B are (x , -3) and (5 , 2) respectively. What is the value of x?
a) 4
b) -14
c) 12.5
d) -4
Question 10: What are the co-ordinates of the centroid of a triangle, whose vertices are A(2 , 5), B(-4 , 0) and C(5 , 4)?
a) (-1 , 3)
b) (1 , 3)
c) (1 , -3)
d) (-1 , -3)
Answers & Solutions:
1) Answer (D)
Equation of circle $x^2+y^2+2gx+2fy+c=0$
It passes through (0,0), (4,0) and (3,9). Substitute each point in the above equation:
=> On substituting the value (0,0) in the above equation, we obtain: $c=0$
=> On substituting the value (4,0) in the above equation, we obtain: $16+0+8g+0 = 0$ ; $g=-2$
=> On substituting the value (3,9) in the above equation, we obtain: $9+81-12+18f = 0$ ; $f= -13/3$
Radius of the circle r = $\sqrt{\ g^2+f^2-c}$ => $r^2=\frac{205}{9}$
Therefore, Area = $\pi\ r^2=\frac{205\pi\ }{9}$
2) Answer (C)
We have been given 2 vertices. It is possible that they are adjacent vertices or opposite vertices.
Case 1: The given points are on a diagonal. We come up with the following square
Hence (-5,5) satisfies the condition for a vertex of the square.
Case 2: The given points are on a side. We come up with the following 2 possible squares:
Hence (15,-5) satisfies the condition for a vertex of the square.
Hence (-5,15) satisfies the condition for a vertex of the square.
Hence, we can see all the mentioned points apart from (5,15) satisfies the condition for the vertex of the square.
3) Answer (C)
Let us take intersection of $l_1 : 3x -2y +18 =0 $ , $l_2 : 2x-y+7 =0 $.
$ l_1 -2l_2 : -x+4 = 0$ or $x=4$ which gives y=15. Thus (4,15) is the point of intersection
Let us take intersection of $l_2 : 2x-y+7 =0 $ and $l_3 : 5x-2y+6 =0 $
$ l_3 – 2l_2 : x-8 = 0$ or $x=8$ which gives $y=23$
Let us take intersection of $l_1 : 3x -2y +18 =0 $ and $l_3 : 5x-2y+6 =0 $
$l_3 – l_1 : 2x-12 = 0$ or $x=6$ which gives $y=18$
Thus point of intersection are (4,15), (8,23) and (6,18)
length of side 1 = $\sqrt{\left(8-4\right)^2+\left(23-15\right)^2}=\sqrt{\ 4^2+8^2}=4\sqrt{\ 5}$
length of side 2 = $\sqrt{\left(8-6\right)^2+\left(23-18\right)^2}=\sqrt{\ 2^2+5^2}=\sqrt{\ 29}$
length of side 3= $\sqrt{\left(6-4\right)^2+\left(18-15\right)^2}=\sqrt{\ 2^2+3^2}=\sqrt{\ 13}$
Sum of square of sides = 80+29+13 =122
4) Answer (C)
Since it is a right angle triangle, circumradius = $\frac{hypotenuse}{2}$ = $=\ \frac{\ \sqrt{\ 4^2+4^2}}{2}=\frac{4\sqrt{\ 2}}{2}=2\sqrt{\ 2}$
inradius = $\frac{area}{semi-perimeter}=\frac{\frac{1}{2}\left(4\right)\left(4\right)}{\frac{\left(4+4+4\sqrt{\ 2}\right)}{2}}\ =\frac{8}{4+2\sqrt{\ 2}}\ =\frac{4}{2+\sqrt{\ 2}}$
Ratio = $\frac{\left(\ \frac{4}{2+\sqrt{\ 2}}\right)}{2\sqrt{\ 2}}=\frac{\sqrt{\ 2}}{2+\sqrt{\ 2}}$
5) Answer (A)
x1’ = 2*d – x1
y1’ = 2*d*a – y1 + 2*c
and
d=(𝑥1+(𝑦1−𝑐)∗𝑎)/(1+𝑎2)
d=(3+(4-6)*(-3))/(1+9)
d=9/10
x1’ = 2*9/10-3
=-6/5
y1’ = 2*9/10*(-3)-4+12
=13/5
6) Answer (A)
if P” is (2,5) ; reflection P’ will be (5,2) which if reflected along x axis would be (5,-2)
7) Answer (C)
The point of intersection of lines bx+4y+1=0 and ax+8y+1=0 is $\left[\frac{1}{a-2b},\frac{1}{4}\cdot\frac{\left(b-a\right)}{a-2b}\right]$
Substituting these points in cx+12y+1 = 0 , we get $c\cdot\frac{1}{a-2b}+12\cdot\frac{1}{4}\cdot\frac{\left(b-a\right)}{a-2b}+1=0$
=> c + 3(b -a) + (a – 2b) = 0
=> b+c = 2a
8) Answer (A)
(x , y) = ($\frac{mx_{2}+nx_{1}}{m+n}$ , $\frac{my_{2}+ny_{1}}{m+n}$)
(x , 0) = ($\frac{m-4n}{m+n}$ , $\frac{4m+n}{m+n}$)
4m+n = 0
4m = -n
m/n = -1/4
So the answer is option A.
9) Answer (C)
slope = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{2+3}{5-x}=\frac{5}{5-x}=\frac{-2}{3}$
$\rightarrow 15 = -10 + 2x$
$\rightarrow x = \frac{25}{2} = 12.5$
so the answer is option C.
10) Answer (B)
centroid = $(\frac{x_{1}+x_{2}+x_{3}}{3} , \frac{y_{1}+y_{2}+y_{3}}{3})$ = $(\frac{2-4+5}{3} , \frac{5+0+4}{3})$ = $(1 , 3)$
so the answer is option B.
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