Coordinate Geometry IV.

105

12
Coordinate Geometry

From the CAT point of view, Coordinate Geometry by itself The x coordinate of the point is also known as the abscissa
is not a very significant chapter. Basically, applied questions while the y coordinate is also known as the ordinate.
are asked in the form of tabular representation or regarding
1. Distance Formula If two points P and Q are such that
the shape of the structure formed. However, it is advised
to go through the basics and important formulae to have a they are represented by the points (x1, y1) and (x2, y2) on
feel-good effect as also to be prepared for surprises, if any, the x-y plane (cartesian plane), then the distance between
in the examination. Logical questions might be asked based the points P and Q = ( x1 - x2 ) 2 + ( y1 - y2 ) 2 .
on the formulae and concepts contained in this chapter.
Besides, the student will have an improved understanding
of the graphical representation of functions if he/she has Illustration
gone through coordinate geometry.
The students who face any problems in this chapter can Question 1: Find the distance between the points (5, 2)
stop after solving LOD II and can skip LOD III. and (3, 4).

Answer: Distance = (5 - 3) 2 + (2 - 4) 2
CARTESIAN COORDINATE SYSTEM
È Using the formula ( x - x ) 2 + ( y - y ) 2 ˘
Î 1 2 1 2 ˚
Rectangular Coordinate Axes
= 2 2 units
Let X¢OX and Y’OY be two mutually perpendicular lines
through any point O in the plane of the paper. Point O is
2. Section Formula If any point (x, y) divides the line
known as the origin. The line X¢OX is called the x-axis or segment joining the points (x1, y1) and (x2, y2) in the ratio
axis of x; the line Y¢OY is known as the y-axis or axis of m : n internally,
y; and the two lines taken together are called the coordinate then x = (mx2 + nx1)/(m + n)
axes or the axes of coordinates.
y = (my2 + ny1)/(m + n) (See figure)
y
II I
Quadrant 3 Quadrant
2 m n
(–, +) (+, +)
1 A P B
x¢ x (x 1, y 1 ) (x 2, y 2 )
–3 –2 –1 –11 2 3
III –2 IV Fig. 12.2
Quadrant –3 Quadrant
(–, –) y ¢ (+, –) If any point (x, y) divides the line segment joining the
Fig. 12.1 points (x1, y1) and (x2, y2) in the ratio m : n externally,

Any point can be represented on the plane described by then x = (mx2 – nx1)/(m – n)
the coordinate axes by specifying its x and y coordinates. y = (my2 – ny1)/(m – n)
IV.106 How to Prepare for Quantitative Aptitude for CAT

Illustration Illustration
Question 2: Find the point which divides the line seg- Question 4: Find the centroid of the triangle whose ver-
ment joining (2, 5) and (1, 2) in the ratio 2 : 1 internally. tices are (5, 3), (4, 6) and (8, 2).
Answers: X = (2.1 + 1.2) /(1 + 2) = 4/3 Answer: X coordinate = (5 + 4 + 8)/3 = 17/3
Y = (2.2 + 1.5)/(2 + 1) = 9/3 = 3
Y coordinate = (3 + 6 + 2)/3 = 11/3
3. Area of a Triangle The area of a triangle whose ver-
tices are A (x1, y1), B (x2, y2) and C (x3, y3) is given by 5. In-centre of a triangle The centre of the circle that
touches the sides of a triangle is called its In-centre. In
È{x1 ( y2 - y3 ) + x2 ( y3 - y1 ) + x3 ( y1 - y2 )} ˘ other words, if the three sides of the triangle are tangential
ÍÎ 2 ˙˚
to the circle then the centre of that circle represents the
A in-centre of the triangle.
(x 1, y 1 )
The in-centre is also the point of intersection of the
internal bisectors of the angles of the triangle. The distance
of the in-centre from the sides of the triangle is the same
and this distance is called the in-radius of the triangle.
B C
(x 2, y 2 ) (x 3, y 3 ) If A (x1, y1), B (x2, y2) and C (x3, y3) are the coordinates
of the vertices of a triangle, then the coordinates of its
Fig. 12.3 in-centre are
[Note: Since the area cannot be negative, we have to take
(ax1 + bx2 + cx3 ) (ay1 + by2 + cy3 )
the modulus value given by the above equation.] x= and y=
(a + b + c) (a + b + c)
Corollary: If one of the vertices of the triangle is at the
origin and the other two vertices are A (x1, y1), B (x2, y2), where BC = a, AB = c and AC = b.
( x1 y2 - x2 y1 ) A
then the area of triangle is .
2

Illustration I

Question 3: Find the area of the triangle (0, 4), (3, 6) and
B C
(–8, –2).
Answer: Area of triangle = | 1/2 {0 (6– (–2)) + 3 ((–2) ­–4) Fig. 12.5
+ (–8) (4–6)}|
= | 1/2 {(0) + 3 (–6) + (–8) (–2)}|
= | 1/2 (–2)| = |–1| = 1 square unit.
Illustration
4. Centre of gravity or centroid of a triangle The cen- Question 5: Find the in-centre of the right angled isos-
troid of a triangle is the point of intersection of its medians celes triangle having one vertex at the origin and having
(the line joining the vertex to the middle point of the opposite the other two vertices at (6, 0) and (0, 6).
side). Centroid divides the medians in the ratio 2 : 1. In other
words, the CG or the centroid can be viewed as a point at Answer: Obviously, the length of the two sides AB and
which the whole weight of the triangle is concentrated. BC of the triangle is 6 units and the length of the third
side is (62 + 62)1/2.
Formula: If A (x1, y1), B (x2, y2) and C (x3, y3) are the co-
ordinates of the vertices of a triangle, then the coordinates Hence a = c = 6, b = 6 2
of the centroid G of that triangle are
x = (x1 + x2 + x3)/3 and y = (y1 + y2 + y3)/3
A

B C

Fig. 12.4 Fig. 12.6

 Coordinate Geometry IV.107

In-centre will be at Solving (i) and (ii), we get x = 3, y = –3.

Hence, the coordinates of the circumcentre of D ABC
(6.0 + 6 2 .0 + 6.6) (6.6 + 6 2 .0 + 6.0) are (3, –3).
,
(6 + 6 + 6 2 ) (6 + 6 + 6 2 )
7. Orthocentre of a triangle The orthocentre of a trian-
36 36 gle is the point of intersection of the perpendiculars drawn
= , from the vertices to the opposite sides of the triangle.
12 + 6 2 12 + 6 2
A
6. Circumcentre of a triangle The point of intersection
of the perpendicular bisectors of the sides of a triangle is
called its circumcentre. It is equidistant from the vertices O
of the triangle. It is also known as the centre of the circle
which passes through the three vertices of a triangle (or the B C
centre of the circle that circumscribes the triangle.) Fig. 12.8
Let ABC be a triangle. If O is the circumcentre of the
triangle ABC, then OA = OB = OC and each of these three
represent the circum radius.
Illustration
A Question 7: Find the orthocentre of the triangle whose
sides have the equations y = 15, 3x = 4y, and 5x + 12y = 0.
O Answer: Let ABC be the triangle whose sides BC, CA
and AB have the equations y = 15, 3x = 4y, and 5x + 12y
B C = 0 respectively.
Solving these equations pairwise, we get coordinates of
A, B and C as (0, 0), (–36, 15) and (20, 15) respectively.
Fig. 12.7 AD is a line passing through A (0, 0) and perpendicular
to y = 15.
So, equation of AD is x = 0.
Illustration
The equation of any line perpendicular to 3x – 4y = 0
Question 6: What will be the circumcentre of a triangle is represented by 4x + 3y + k = 0.
whose sides are 3x – y + 3 = 0, 3x + 4y + 3 = 0 and This line will pass through (–36, 15) if –144 + 45 + k =
x + 3y + 11 = 0? 0 fi k = 99.
Answer: Let ABC be the triangle whose sides AB, BC So the equation of BE is 4x + 3y + 99 = 0.
and CA have the equations 3x – y + 3 = 0, 3x + 4y + 3 = 0 Solving the equations of AD and BE we get x = 0, y =
and x + 3y + 11 = 0 respectively. – 33.
Solving the equations, we get the points A, B and C as Hence the coordinates of the orthocentre are (0, –33).
(–2, –3), (–1, 0) and (7, – 6) respectively. 8. Collinearity of three points: Three given points A,
The equation of a line perpendicular to BC is 4x – 3y B and C are said to be collinear, that is, lie on the same
+ k = 0. straight line, if any of the following conditions occur:
[For students unaware of this formula, read the section
on straight lines later in the chapter.] (i) Area of triangle formed by these three points is zero.
This will pass through (3, –3), the mid-point of BC, if (ii) Slope of AB = Slope of AC.
12 + 9 + k = 0 fi k = –21 (iii) Any one of the three points (say C) lies on the straight
Putting k1 = –21 in 4x – 3y + k = 0, we get 4x – 3y line joining the other two points (here A and B).
– 21 = 0 (i)
as the equation of the perpendicular bisector of BC. A B C
Again, the equation of a line perpendicular to CA is Fig. 12.9
3x – y + k1 = 0.
This will pass through (5/2, –9/2), the mid-point of Illustration
AC if
15/2 + 9/2 + k1 = 0 fi k1 = –12 Question 8: Select the right option the points (–a, –b),
Putting k1 = –12 in 3x – y + k1 = 0, we get 3x – y (0, 0) and (a, b) are
– 12 = 0 (ii) (a) collinear (b) vertices of square
as the perpendicular bisector of AC. (c) vertices of a rectangle (d) None of these
IV.108 How to Prepare for Quantitative Aptitude for CAT

Answer: We can use either of the three methods to check 10. Different Forms of the Equations of a Straight Line
whether the points are collinear. (a) General Form The general form of the equation of a
But the most convenient one is (ii) in this case. straight line is ax + by + c = 0.
Let A, B, C are the points whose coordinates are (–a, (first degree equation in x and y). where a, b and c
–b), (0, 0) and (a, b) are real constants and a, b are not simultaneously equal
Slope of BC = b/a to zero.
Slope of AB = b/a -a
In this equation, slope of the line is given by .
So, the straight line made by points A, B and C is b
collinear. The general form is also given by y = mx + c; where m
Hence, (a) is the answer. is the slope and c is the intercept on y-axis.
[If you have not understood this here, you are requested to In this equation, slope of the line is given by m.
read the following section on straight lines and their slopes (b) Line Parallel to the X-axis The equation of a straight
and then re-read this solution] line parallel to the x-axis and at a distance b from it, is
Alternative: Draw the points on paper assuming the pa- given by y = b.
per to be a graph paper. This will give you an indication Obviously, the equation of the x-axis is y = 0
regarding the nature of points. In the above question, point
(a, b) is in first quadrant for a > 0, b > 0 and point (–a, –b) (c) Line Parallel to Y-axis The equation of a straight
line parallel to the y-axis and at a distance a from it, is
is directly opposite to the point (a, b) in the third quadrant
given by x = a.
with the third point (0, 0) in the middle of the straight line
Obviously, the equation of y-axis is x = 0
joining the points A and B.
You can check this by assuming any value for ‘a’ and ‘b’. (d) Slope Intercept Form The equation of a straight line
Also, you can use this method for solving any problem passing through the point A (x1, y1) and having a slope m
involving points and diagrams made by those points. is given by
However you should be fast enough to trace the points on ( y – y1) = m (x – x1)
paper. A little practice of tracing points might help you.
(e) Two Points Form The equation of a straight line
9. Slope of a line The slope of a line joining two points passing through two points A (x1, y1) and B (x2, y2) is
A (x1, y1) and B (x2, y2) is denoted by m and is given by m given by
= ( y2 – y1)/(x2 – x1) = tan q, where q is the angle that the ( y2 - y1 ) ( x - x1 )
line makes with the positive direction of x-axis. This angle ( y – y1) =
( x2 - x1 )
q is taken positive when it is measured in the anti-clockwise
direction from the positive direction of the axis of x. Ê y - y1 ˆ
Its slope = Á 2
Y Y Ë x2 - x1 ˜¯
B (f) Intercept Form The equation of a straight line mak-
B q ing intercepts a and b on the axes of x and y respectively
q is given by
x¢ A O x O A x
x/a + y/b = 1
y
Fig. 12.10
B
y-intercept

Illustration
A
Question 9: Find the equation of a straight line passing
O x-intercept x
through (2, –3) and having a slope of 1 unit.
Answer: Here slope = 1
And point given is (2, –3). Fig. 12.11
So, we will use point-slope formula for finding the
If a straight line cuts x-axis at A and the y-axis at B
equation of straight line. This formula is given by:
then OA and OB are known as the intercepts of the line
( y – y1) = m (x – x1) on x-axis and y-axis respectively.
So, equation of the line will be y – (– 3) = 1 (x – 2) 11. Perpendicularity and Parallelism
fi y+3= x–2 Condition for two lines to be parallel: Two lines are said
fi y–x+5=0 to be parallel if their slopes are equal.
 Coordinate Geometry IV.109

For this to happen, ratio of coefficient of x and y in both 12. Length of perpendicular or Distance of a point
the lines should be equal. from a line The length of perpendicular from a given
In a general form, this can be stated as: line parallel to point (x1, y1) to a line ax + by + c = 0 is
ax + by + c = 0 is ax + by + k = 0 | ax1 + by1 + c |
or dx + ey + k = 0 if a/d = b/e where k is a constant.
a 2 + b2
Corollary:
Illustration
(a) Distance between two parallel lines.
Question 10: Which of the lines represented by the fol- If two lines are parallel, the distance between them
lowing equations are parallel to each other? will always be the same.
1. x + 2y = 5 2. 2x – 4y = 6 When two straight lines are parallel whose equa-
3. x – 2y = 4 4. 2x + 6y = 8 tions are ax + by + c = 0 and ax + by + c1 = 0, then
| c - c1 |
(a) 1 and 2 (b) 2 and 4 (c) 2 and 3 (d) 1 and 4 the distance between them is given by .
a 2 + b2
Answer: Go through the options and check which of the
two lines given will satisfy the criteria for two lines to be (b) The length of the perpendicular from the origin to
parallel. It will be obvious that option c is correct, that is, |c|
the line ax + by + c = 0 is given by
the line 2x – 4y = 6 is parallel to the line x – 2y = 4. a + b2
2

Question 11: Find the equation of a straight line parallel

to the straight line 3x + 4y = 7 and passing through the
Illustration
point (3, –3).
Answer: Equation of the line parallel to 3x + 4y = 7 will Question 14: Two sides of a square lie on the lines x +
be of the form 3x + 4y = k. y = 2 and x + y = –2. Find the area of the square formed
This line passes through (3, –3), so this point will satisfy in this way.
the equation of straight line 3x + 4y = k. So, 3.3 + 4. (–3) Answer: Obviously, the difference between the parallel
= k fi k = –3. lines will be the side of the square.
Hence, equation of the required straight line will be To convert it into the form of finding the distance of a
3x + 4y + 3 = 0. point from a line, we will have to find out a point at which
Condition for two lines to be perpendicular: Two lines any one of these two lines cut the axes and then we will
are said to be perpendicular if product of the slopes of the draw a perpendicular from that point to the other line, and
lines is equal to –1. this distance will be the side of the square.
For this to happen, the product of the coefficients of x + To find the point at which the equation of the line x + y
the product of the coefficients of y should be equal to zero. = 2 cut the axes, we will put once x = 0 and then again y
= 0.
Illustration When x = 0, y = 2, so the coordinates of the point where
it cuts y-axis is (0, 2).
Question 12: Which of the following two lines are per- Now the point is (0, 2), and the equation of line on which
pendicular? perpendicular is to be drawn is x + y = –2.
1. x + 2y = 5 2. 2x – 4y = 6 |1.0 + 1.2 + 2 | 4
3. 2x + 3y = 4 4. 2x – y = 4 So, distance = = .
2
1 +1 2 2
(a) 1 and 2 (b) 2 and 4 (c) 2 and 3 (d) 1 and 4
Check the equations to get option 4 as the correct 16
\ Area = =8
answer. 2
Question 13: Find the equation of a straight line perpen- Alternatively: Draw the points on the paper and you will
dicular to the straight line 3x + 4y = 7 and passing through get the length of diagonal as 4 units; so, length of side will
the point (3, –3).
be 2 2 and, therefore, the area will be 8 sq units.
Answer: Equation of the line perpendicular to 3x + 4y =
7 will be of the form 4x – 3y = K. Alternatively: you can also use the formula for the distance
This line passes through (3, –3), so this point will satisfy between two parallel lines as
the equation of straight line 4x – 3y = K. So, 4.3 – 3.–3 | 2 + 2| 4
fi K = 21. =
2
1 +1 2 2
Hence, equation of required straight line will be 4x – 3y = 21.
IV.110 How to Prepare for Quantitative Aptitude for CAT

13. Change of axes If origin (0, 0) is shifted to (h, k) Answer: Let (X, Y) be the coordinates of the point in
then the coordinates of the point (x, y) referred to the old the new axis.
axes and (X, Y) referred to the new axes can be related
Then, 1=X+5 \ X = –4
with the relation x = X + h and y = Y + k.
2=Y+2 \ Y =0
y
Y So, the new coordinates of the point will be (– 4, 0).
14. Point of intersection of two lines Point of inter-
(h, k)
X section of two lines can be obtained by solving the equa-
tions as simultaneous equations.
O x
An Important Result
Fig. 12.12 If all the three vertices of a triangle have integral
coordinates, then that triangle cannot be an Equilateral
Illustration triangle.

Question 15: If origin (0, 0) is shifted to (5, 2), what

will be the coordinates of the point in the new axis which
was represented by (1, 2) in the old axis?

Space for Rough Work

 Coordinate Geometry IV.111

Level of Difficulty (I)

1. Find the distance between the points (3, 4) and 11. Find the coordinates of the points that trisect the line
(8, –6). segment joining (1, –2) and (–3, 4).
(a) 5 (b) 5 5 -1 -5 ˆ
(a) ÊÁ , 0ˆ˜ (b) ÊÁ ,2
Ë 3 ¯ Ë 3 ˜¯
(c) 2 5 (d) 4 5
2. Find the distance between the points (5, 2) and (0, 0). (c) Both (a) and (b) (d) None of these
12. Find the coordinates of the point that divides the line
(a) 27 (b) 21 segment joining the points (6, 3) and (– 4, 5) in the
(c) 29 (d) 31 ratio 3 : 2 internally.
3. Find the value of p if the distance between the points -21ˆ 21
(8, p) and (4, 3) is 5. (a) ÊÁ 0, ˜ (b) ÊÁ 0, ˆ˜
Ë 5 ¯ Ë 5¯
(a) 6 (b) 0
11 14 - 11 - 14 ˆ
(c) Both (a) and (b) (d) None of these (c) ÊÁ , ˆ˜ (d) ÊÁ , ˜
Ë 2 3¯ Ë 2 3 ¯
4. Find the value of c if the distance between the point
13. In Question 12 question, find the coordinates of the
(c, 4) and the origin is 5 units.
point if it divides the points externally.
(a) 3 (b) –3
(a) (24, –9) (b) (3, –5)
(c) Both a and b (d) none of these
(c) (–24, 9) (d) (5, –3)
5. Find the mid-point of the line segment made by
14. In what ratio is the line segment joining (–1, 3) and
joining the points (3, 2) and (6, 4).
(4, –7) divided at the point (2, –3)?
9 -3 (a) 3 : 2 (b) 2 : 3
(a) ÊÁ , 3ˆ˜ (b) ÊÁ , - 1ˆ˜
Ë2 ¯ Ë 2 ¯ (c) 3 : 5 (d) 5 : 3
9 3 Ê 3ˆ 15. In question 14, find the nature of division?
(c) ÊÁ , - ˆ˜ (d) Á ˜ (a) Internal (b) External
Ë2 2¯ Ë - 1¯
(c) Cannot be said
6. If the origin is the mid-point of the line segment
16. In what ratio is the line segment made by the points
joined by the points (2, 3) and (x, y), find the value
(7, 3) and (– 4, 5) divided by the y-axis?
of (x, y).
(a) 2 : 3 (b) 4 : 7
(a) (2, 3) (b) (–2, 3)
(c) 3 : 5 (d) 7 : 4
(c) (­–2, –3) (d) (2, –3)
17. What is the nature of the division in the above
7. Find the points that divide the line segment joining question?
(2, 5) and (–1, 2) in the ratio 2 : 1 internally.
(a) External (b) Internal
(a) (1, 2) (b) (–3, 2)
(c) Cannot be said
(c) (3, 1) (d) (0, 3) 18. If the coordinates of the mid-point of the line segment
8. In what ratio does the x-axis divide the line segment joining the points (2, 1) and (1, –3) is (x, y) then the
joining the points (2, –3) and (5, 6)? relation between x and y can be best described by
(a) 2 : 1 (b) 1 : 2 (a) 3x + 2y = 5 (b) 6x + y = 8
(c) 3 : 4 (d) 2 : 3 (c) 5x – 2y = 4 (d) 2x – 5y = 4
9. How many squares are possible if two of the vertices 19. Points (6, 8), (3, 7), (–2, –2) and (1, –1) are joined
of a quadrilateral are (1, 0) and (2, 0)? to form a quadrilateral. What will be this structure?
(a) 1 (b) 2 (a) Rhombus (b) Parallelogram
(c) 3 (d) 4 (c) Square (d) Rectangle
10. If the point R (1, –2) divides externally the line seg- 20. Points (4, –1), (6, 0), (7, 2) and (5, 1) are joined
ment joining P (2, 5) and Q in the ratio 3 : 4, what to be a vertex of a quadrilateral. What will be the
will be the coordinates of Q? structure?
(a) (–3, 6) (b) (2, –4) (a) Rhombus (b) Parallelogram
(c) (7/3, 28/3) (d) (1, 2) (c) Square (d) Rectangle
IV.112 How to Prepare for Quantitative Aptitude for CAT

21. What will be the centroid of a triangle whose vertices 29. Which of the following will be the equation of a
are (2, 4), (6, 4) and (2, 0)? straight line that is parallel to the y-axis at a distance
11 units from it?
7 5ˆ
(a) ÊÁ , ˜ (b) (3, 5) (a) x = + 11, x = –11 (b) y = 11, y = –11
Ë2 2¯
(c) y = 0 (d) None of these
10 8
(c) ÊÁ , ˆ˜ (d) (1, 4) 30. Which of the following will be the equation of a
Ë 3 3¯ straight line parallel to the y-axis at a distance of 9
22. The distance between the lines 4x + 3y = 11 and 8x units to the left?
+ 6y = 15 is (a) x = – 9 (b) x = 9
7 (c) y = 9 (d) y = – 9
(a) 4 (b) 31. What can be said about the equation of the straight
10
line x = 7?
5
(c) (d) 26 (a) It is the equation of a straight line at a distance of
7 7 units towards the right of the y-axis.
23. If the mid-point of the line joining (3, 4) and (p, 7) (b) It is the equation of a straight line at a distance of
is (x, y) and 2x + 2y + 1 = 0, then what will be the 7 units towards the left of the y-axis.
value of p? (c) It is the equation of a straight line at a distance of
-17 7 units below the x-axis.
(a) 15 (b)
2 (d) It is the equation of a straight line at a distance of
17 7 units above the x-axis.
(c) –15 (d) 32. What can be said about the equation of the straight
2
line y = –8?
24. Find the third vertex of the triangle whose two ver- (a) It is the equation of a straight line at a distance of
tices are (–3, 1) and (0, –2) and the centroid is the 8 units below the x-axis.
origin.
(b) It is the equation of a straight line at a distance of
- 4 14 ˆ
(a) (2, 3) (b) ÊÁ , 8 units above the x-axis.
Ë 3 3 ˜¯
(c) It is the equation of a straight line at a distance of
(c) (3, 1) (d) (6, 4) 8 units towards the right of the y-axis.
25. Find the area of the triangle whose vertices are (d) It is the equation of a straight line at a distance of
(1, 3), (–7, 6) and (5, –1). 8 units towards the left of the y-axis.
(a) 20 (b) 10 33. Which of the following straight lines passes through
(c) 18 (d) 24 the origin?
26. Find the area of the triangle whose vertices are (a) x + y = 4 (b) x2 + y2 = – 6
(a, b + c), (a, b – c) and (– a, c).
(c) x + y = 5 (d) x = 4y
(a) 2ac (b) 2bc
(c) b (a + c) (d) c (a – b) 34. What will be the point of intersection of the equation
of lines 2x + 5y = 6 and 3x + 4y = 7?
27. The number of lines that are parallel to 2x + 6y + 7
= 0 and have an intercept of length 10 between the 11 4ˆ - 11 ˆ
(a) ÊÁ , ˜ (b) ÊÁ , 4˜
coordinate axes is Ë7 7¯ Ë 7 ¯
(a) 0 (b) 1
- 2ˆ - 2ˆ
(c) 2 (d) infinite (c) ÊÁ 3, (d) ÊÁ 4,
Ë 7 ˜¯ Ë ˜
5 ¯
28. Which of the following three points represent a
straight line? 35. If P (6, 7), Q ( 2, 3) and R (4, –2) be the vertices
of a triangle, then which of the following is not a
-1
(a) ÊÁ , 3ˆ˜ , (–5, 6) and (– 8, 8) point contained in this triangle?
Ë 2 ¯ (a) (4, 3) (b) (3, 3)
-1 (c) (4, 2) (d) (6, 1)
(b) ÊÁ , 3ˆ˜ , (5, 6) and (– 8, 8)
Ë 2 ¯ 36. What will be the reflection of the point (4, 5) in the
1 second quadrant?
(c) ÊÁ , 3ˆ˜ , (–5, 6) and (– 8, 8) (a) (– 4, – 5) (b) (– 4, 5)
Ë2 ¯
(c) (4, – 5) (d) None of these
-1 5
(d) ÊÁ , 3ˆ˜ , ÊÁ ˆ˜ and (8, 8) 37. What will be the reflection of the point (4, 5) in the
Ë 2 ¯ Ë 6¯
third quadrant?
 Coordinate Geometry IV.113

(a) (– 4, – 5) (b) (– 4, 5) 40. What will be the length of the perpendicular drawn
(c) (4, – 5) (d) None of these from the point (4, 5) upon the straight line 3x + 4y
38. What will be the reflection of the point (4, 5) in the = 10?
fourth quadrant? 12 32
(a) (b)
(a) (– 4, – 5) (b) ( – 4, 5) 5 5
(c) (4, –5) (d) None of these 22 42
(c) (d)
39. If the origin gets shifted to (2, 2), then what will be 5 5
the new coordinates of the point (4, –2)?
(a) (– 2, 4) (b) (2, 4)
(c) (4, 2) (d) (2, – 4)

Space for Rough Work

IV.114 How to Prepare for Quantitative Aptitude for CAT

Level of Difficulty (II)

1. Find the area of the quadrilateral the coordinates 10. Find the distance between the two parallel straight
of whose angular points taken in order are (1, 1), lines y = mx + c and y = mx + d? [Assume c > d ]
(3, 4), (5, –2) and (4, –7).
Ê (c - d ) ˆ Ê (d - c) ˆ
(a) 20.5 (b) 41 (a) Á (b) Á
1 ˜ 1 ˜
(c) 82 (d) 61.5 Ë (1 + m 2 ) 2 ¯ Ë (1 + m 2 ) 2 ¯
2. Find the area of the quadrilateral the coordinates Ê ˆ Ê -d ˆ
of whose angular points taken in order are (–1, 6), d
(c) Á (d) Á
2 12 ˜ 1 ˜
(–3, –9), (5, –8) and (3, 9). Ë (1 + m ) ¯ Ë (1 + m) 2 ¯
(a) 48 (b) 96 11. What will be the equation of the straight line that
(c) 192 (d) 72 passes through the intersection of the straight lines
3. Two vertices of a triangle are (5, –1) and (–2, 3). 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 and is perpen-
If the orthocenter of the triangle is the origin, what dicular to the straight line 3x – 4y = 5?
will be the coordinates of the third point?
32 84
(a) (4, 7) (b) (– 4, 7) (a) 8x + 6y = (b) 4x + 3y =
7 17
(c) (– 4, –7) (d) (4, –7)
4. Find the equation of the straight line passing through 62 58
(c) 4x + 3y = (d) 8x + 6y =
the origin and the point of intersection of the lines 17 17
x/a + y/b = 1 and x/b + y/a = 1. 12. In question 11, find the equation of the straight line
(a) y = x (b) y = – x if it is parallel to the straight line 3x + 4y = 5?
(c) y = 2x (d) y = –2x 58 58
(a) 12x + 16y = (b) 3x + 4y =
5. One side of a rectangle lies along the line 4x + 7y 17 17
+ 5 = 0. Two of its vertices are (–3, 1) and (1, 1).
Which of the following may be an equation which 58
(c) 6x + 8y = (d) None of these
represents any of the other three straight lines? 17
(a) 7x – 4y = 3 (b) 7x – 4y + 3 = 0 13. The orthocenter of the triangle formed by the points
(c) y + 1 = 0 (d) 4x + 7y = 3 (0, 0), (8, 0) and (4, 6) is
8
6. The points (p–1, p + 2), (p, p + 1), (p + 1, p) are (a) ÊÁ 4, ˆ˜ (b) (3, 4)
collinear for Ë 3¯
5
(a) p = 0 (b) p = 1 (c) (4, 3) (d) ÊÁ 3, ˆ˜
(c) p = –1/2 (d) any value of p Ë 2¯
7. The straight line joining (1, 2) and (2, –2) is per- 14. The area of a triangle is 5 square units, two of its
pendicular to the line joining (8, 2) and (4, p). What vertices are (2, 1) and (3, –2). The third vertex lies
will be the value of p? on y = x + 3. what will be the third vertex?
(a) –1 (b) 1 5 13 7 13
(a) ÊÁ , ˆ˜ (b) ÊÁ , ˆ˜
(c) 3 (d) None of these Ë3 3 ¯ Ë2 2¯
8. What will be the length of the perpendicular drawn (c) (3, 4) (d) (1, 2)
from the point (–3, –4) to the straight line 12 (x + 6) 15. The equations of two equal sides AB and AC of an
= 5 (y – 2)? isosceles triangle ABC are x + y = 5 and 7x – y = 3
4 1 respectively. What will be the equation of the side
(a) 5 ÊÁ ˆ˜ (b) 5 ÊÁ ˆ˜ BC if area of triangle ABC is 5 square units.
Ë 13 ¯ Ë 13 ¯
2 1 (a) x + 3y – 1 = 0 (b) x – 3y + 1 = 0
(c) 3 ÊÁ ˆ˜ (d) 3 ÊÁ ˆ˜ (c) 2x – y = 5 (d) x + 2y = 5
Ë 13 ¯ Ë 13 ¯
9. The area of the triangle with vertices at (a, b + c), 16. Three vertices of a rhombus, taken in order are (2,
(b, c + a) and (c, a + b) is –1), (3, 4) and (–2, 3). Find the fourth vertex.
(a) 0 (b) a + b + c (a) (3, 2) (b) (–3, –2)
(c) a2 + b2 + c2 (d) 1 (c) (–3, 2) (d) (3, –2)
 Coordinate Geometry IV.115

17. Four vertices of a parallelogram taken in order are 19. What will be the value of p if the equation of straight
(–3, –1), (a, b), (3, 3) and (4, 3). What will be the line 2x + 5y = 4 gets changed to 2x + 5y = p after
ratio of a to b? shifting the origin at (3, 3)?
(a) 4 : 1 (b) 1 : 2 (a) 16 (b) –17
(c) 12 (d) 10
(c) 1 : 3 (d) 3 : 1
20. A line passing through the points (a, 2a) and (–2,
18. What will be the new equation of straight line 3x + 3) is perpendicular to the line 4x + 3y + 5 = 0. Find
4y = 6 if the origin gets shifted to (3, – 4)? the value of a?
(a) 3x + 4y = 5 (b) 4x – 3y = 4 (a) –14/3 (b) 18/5
(c) 3x + 4y + 1 = 0 (d) 3x + 4y – 13 = 0 (c) 14/3 (d) –18/5

Space for Rough Work

IV.116 How to Prepare for Quantitative Aptitude for CAT

Level of Difficulty (III)

1. The area of a triangle is 5 square units. Two of its 9. If the points (a, 0), (0, b) and (1, 1) are collinear
vertices are (2, 1) and (3, –2). The third vertex lies then which of the following is true?
on y = x + 3. What will be the third vertex? 1 1 1 1
(a) (4, –7) (b) (4, 7) (a) + =2 (b) – =1
a b a b
(c) (– 4, –7) (d) (– 4, 7)
1 1 1 1
2. One side of a rectangle lies along the line 4x + 7y (c) – =2 (d) + =1
+ 5 = 0. Two of its vertices are (–3, 1) and (1, 1). a b a b
Which of the following is not an equation of the 10. If P and Q are two points on the line 3x + 4y = –15,
other three straight lines? such that OP = OQ = 9 units, the area of the triangle
POQ will be
(a) 14x – 8y = 6 (b) 7x – 4y = –25
(c) 4x + 7y = 11 (d) 14x – 8y = 20 (a) 18 2 sq units (b) 3 2 sq units
3. The area of triangle formed by the points (p, 2–2p), (c) 6 2 sq units (d) 15 2 sq units
(1–p, 2p) and (–4–p, 6–2p) is 70 units. How many 11. If the coordinates of the points A, B, C and D are
integral values of p are possible? (6, 3), (–3, –5), (4, –2) and (a, 3a) respectively and
(a) 2 (b) 3 if the ratio of the area of triangles ABC and DBC is
(c) 4 (d) None of these 2 : 1, then the value of a is
4. What are the points on the axis of x whose perpen-
dicular distance from the straight line x/p + y/q = 1 (a) - 9 (b) 9
is p? 2 2
pÈ - 23 23
(a) q + ( p2 + q2 ) ˘ , 0 (c) (d)
qÎ ˚
36 18
pÈ 12. The equations of two equal sides AB and AC of an
(b) q - ( p2 + q2 ) ˘ , 0
qÎ ˚ isosceles triangle ABC are x + y = 5 and 7x – y = 3
(c) Both (a) and (b) respectively. What will be the length of the intercept
(d) None of these cut by the side BC on the y-axis?
5. If the medians PT and RS of a triangle with vertices (a) 9 (b) 8
P (0, b), Q (0, 0) and R (a, 0) are perpendicular 5
to each other, which of the following satisfies the
(c) 1.5 (d) No unique solution
relationship between a and b?
13. A line is represented by the equation 4x + 5y = 6 in
(a) 4b2 = a2 (b) 2b2 = a2
the coordinate system with the origin (0, 0). You are
(c) a = –2b (d) a2 + b2 = 0 required to find the equation of the straight line per-
6. The point of intersection of the lines x/a + y/b = 1 pendicular to this line that passes through the point
and x/b + y/b = 1 lies on the line (1, –2) [which is in the coordinate system where
(a) x + y = 1 (b) x + y = 0 origin is at (–2, –2)].
(c) x – y = 1 (d) x – y = 0 (a) 5x – 4y = 11 (b) 5x – 4y = 13
7. PQR is an isosceles triangle. If the coordinates of the (c) 5x – 4y = –3 (d) 5x – 4y = 7
base are Q (1, 3) and R (–2, 7), then the coordinates 14. P (3, 1), Q (6, 5) and R (x, y) are three points such
of the vertex P can be that the angle PRQ is a right angle and the area of
7 D PRQ is 7. the number of such points R that are
(a) ÊÁ 4, ˆ˜ (b) (2, 5)
Ë 2¯ possible is
5 1 (a) 1 (b) 2
(c) ÊÁ , 6ˆ˜ (d) ÊÁ , 2ˆ˜ (c) 3 (d) 4
Ë6 ¯ Ë3 ¯
15. Two sides of a square lie on the lines x + y = 1 and
8. The extremities of a diagonal of a parallelogram are x + y + 2 = 0. What is its area?
the points (3, – 4) and (– 6, 5). If the third vertex is
the point (–2, 1), the coordinate of the fourth vertex is (a) 11 (b) 9
(a) (1, 0) (b) (–1, 0) 2 2
(c) (–1, 1) (d) (1, –1) (c) 5 (d) 4
 Coordinate Geometry IV.117

16. Find the value of k if the straight line 2x + 3y + 4 + k (a) 1 (b) 2

(6x – y + 12) = 0 is perpendicular to the line 7x + 5y (c) 3 (d) None of these
= 4. 19. What will be the area of the rhombus ax ± by ± c
- 29
(a) - 33 (b) = 0?
37 37 3c 2 4c 2
(a) (b)
19 ab ab
(c) (d) None of these
37
2c 2 c2
17. If p is the length of the perpendicular from the origin (c) (d)
x y ab ab
to the line + = 1, then which of the following
a b 20. The coordinates of the mid-points of the sides of a
is true? triangle are (4, 2), (3, 3) and (2, 2). What will be
1 1 1 1 1 1 the coordinates of the centroid of the triangle?
(a) = 2 – 2 (b) = 2 – 2
p2 p2 7 - 7ˆ
b a a b (a) ÊÁ 3, ˆ˜ (b) ÊÁ - 3, ˜
1 1 1 Ë 3¯ Ë 3 ¯
(c) 2
= 2 + 2 (d) None of these
p a b - 7ˆ 7
(c) ÊÁ 3, (d) ÊÁ - 3, ˆ˜
18. How many points on x + y = 4 are there that lie at Ë 3 ˜¯ Ë 3¯
a unit distance from the line 4x + 3y = 10?

Space for Rough Work

IV.118 How to Prepare for Quantitative Aptitude for CAT

formed by joining circumcenter and the orthocenter

answer key in the ratio 2 : 1.
Let the vertices of the triangle be O(0, 0), A(8, 0)
Level of Difficulty (I) and B(4, 6).
1. (b) 2. (c) 3. (c) 4. (c) The equation of an altitude through O and perpen-
5. (a) 6. (c) 7. (d) 8. (b) dicular to AB is y = 2/3x and similarly the equation
9. (c) 10. (c) 11. (c) 12. (b) of an altitude through A and perpendicular to OB is
13. (c) 14. (a) 15. (a) 16. (d) 2x + 3y = 16. Now find the point of intersection of
17. (b) 18. (b) 19. (b) 20. (a) these two straight lines.
21. (c) 22. (b) 23. (c) 24. (c)
14. Use the options.
25. (b) 26. (a) 27. (c) 28. (a)
29. (a) 30. (a) 31. (a) 32. (a) Alternative: Draw the points in the cartesian
33. (d) 34. (a) 35. (d) 36. (b) co-ordinate system and then use the simple geometry
37. (a) 38. (c) 39. (d) 40. (c) formula to calculate the point using the options.
15. Draw the points and then check with the options.
Level of Difficulty (II)
Alternative: Find out the point of intersection with the
1. (a) 2. (b) 3. (c) 4. (a)
help of options and then use the formula for area of ?.
5. (a) 6. (d) 7. (b) 8. (b)
17. Sum of x and y co-ordinates of opposite vertices in
9. (a) 10. (a) 11. (c) 12. (d)
a parallelogram are same.
13. (a) 14. (b) 15. (d) 16. (b)
17. (a) 18. (c) 19. (b) 20. (b) 19. If the origin gets changed to (h, k) from (0, 0) then
Old x co-ordinate = New x co-ordinate + h
Level of Difficulty (III)
Old y co-ordinate = New y co-ordinate + k
1. (c) 2. (d) 3. (d) 4. (c)
5. (b) 6. (d) 7. (c) 8. (b) 20. Equation of any straight line perpendicular to the
9. (d) 10. (a) 11. (c) 12. (b) line 4x + 3y + 5 = 0 will be of the form of 3x – 4y
13. (a) 14. (b) 15. (b) 16. (b) = k, where k is any constant.
17. (c) 18. (b) 19. (c) 20. (a) Now form the equation of the straight line with the
given two points and then equate.
Hints Level of Difficulty (III)
Level of Difficulty (II) 1. First check the options to see that which of the points
1. Use the area of a triangle formula for the two parts lie on the equation of straight line y = x + 3.
of the quadrilateral separately and then add them.
And then again check the options, if needed, to
4. Find the point of intersection of the lines by solving
the simultaneous equations and then use the two- confirm the second constraint regarding area of tri-
point formula of a straight line. angle.
Alternative: After finding out the point of intersec- 2. Use the options.
tion, use options to check. 3. Use the formula of area of a quadrilateral which will
6. For 3 points to be collinear, lead to a quadratic equation. Now solve the quadratic
(i) Either the slope of any two of the 3 points should equation to see the number of integral solutions it
be equal to the slope of any other two points. OR can have.
(ii) The area of the triangle formed by the three points 4. Use the formula of distance of a point from the
should be equal to zero. straight line using the options.
Solve using options.
7. Use the options to find the length using the distance
7. Form the equation of the straight lines and then use
formula.
the options.
9. Make the slope of any two points equal to the slope
10. Point of intersection of y = mx + c with x-axis is
of any other two points.
(–c/m, 0).
Now use the formula for the distance of a point to Slope = Difference of Y coordinates/Difference of X
a straight line. coordinates.
11. Find the point of intersection of the lines and then 14. Draw the points on cartesian coordinate system.
put the coordinate of this point into the equation 15. Length of the square can be find out using the meth-
4x + 3y = K, which is perpendicular to the equation od of finding out the distance between two parallel
of straight line 3x – 4y = 5, to find out K. lines.
13. Orthocenter is the point of intersection of altitudes 17. Use the formula (perpendicular distance of a point
of a triangle and centroid divides the straight line from a straight line.)