Tuesday, June 9th, 2020
There are all sorts of ways that we can find the measurements of lines and angles. We can use rulers to measure lines and protractors to measure angles. In coordinate geometry, we can use graphs and coordinates to find measurements and other useful information about geometric figures.
A coordinate graph is a rectangular grid with two number lines called axes. The x-axis is the horizontal number line and the y-axis is the vertical number line. The axes intersect at the origin which is the point (0,0).
Now, Let us have a look at some formulas for coordinate geometry. We will use the below picture as a reference for the formulas.
or y = mx + c
where a, b and c are the lengths of the BC, AC and AB respectively.
tan θ = (m1−m2) / (1+m1m2)
P = (Ax1+By1+C) / (A2+B2)
→ General equation of a line Ax + By = C
→ Slope intercept form y = mx + c (c is y intercept)
→ Point-slope form y – y1 = m (x – x1 ) (m is the slope of the line)
→ Intercept form x / a + y / b = 1 (where a and b are x and y intercepts respectively)
→ Two point form: (y−y1) / (y2−y1) = (x−x1) / (x2−x1)
Quadrant I => X is Positive Y is Positive
Quadrant II => X is Negative Y is Positive
Quadrant III => X is Negative Y is Negative
Quadrant IV => X is Positive Y is Negative
Example 1: What is the distance between the points A (3,8) and B(-2,-7) ?
Solution: The distance between 2 points (x1, y1) and (x2, y2) is given as
Sqrt ((x2-x1)2 + (y2-y1)2)
Hence, required distance = sqrt((-2-3)2 + (-7-8)2)
Example 2: The points of intersection of three lines 2X + 3Y – 5=0 and 5X – 7Y + 2=0 and 9X – 5Y – 4 = 0
a.) Form a triangle
b.) Are on lines perpendicular to each other
c.) Are on lines parallel to each other
d.) Are coincident
Solution: To solve the question above, we should remember the properties of the lines for being parallel, perpendicular or intersecting
The three lines can be expressed in the y=mx + c format as:
Y = (5/3) – (2X/3), Y = (5X/7) + (2/7) , Y = (9X/5) – (4/5)
Therefore, the slopes of the three lines are -2/3, 5/7, 9/5 and their Y intercepts are 5/3, 2/7 and 4/5 respectively.
We see above that the product of slopes of none of the lines is -1. Thus, lines are not perpendicular to each other.
Also, slopes of the no two lines is same. Thus, lines are not parallel to each other.
Solving the first two equations we get X=1 and Y = 1. If we substitute (1,1) in the third equation Y=(9X/5 – 4/5), we find that it also satisfies the equation. This suggests that the three lines intersect at a common point and hence coincident.
Example 3: The area of the triangle whose vertices are (a + 1, a + 1), (a, a) and (a+2, a) is
Solution: Let a = 0, Thus the three vertices of the triangle becomes (1, 1) (0, 0) and (2, 0)
If we look at the below figure,
Area = ½ * base * height = ½ * 2 * 1 = 1
Imp: The main point to note here is that area will be independent of a.
Example 4: Consider a triangle drawn on the X – Y plane with its three vertices of (41,0) , (0,41) and (0,0), each vertex being represented by its (X,Y) coordinates. The number of points with integer coordinates inside the triangle (excluding all the points on the boundary) is:
Solution: Equation of the line will be of the form => x + y = 41.
Now, we know that if the x,y coordinates of a point are integer, the sum will also be an integer
X + y = k (k, a variable)
As per the question we need to exclude all the values lying on the boundary of triangle, k can take all values from 1 to 40 only. K = 0 is rejected as at k =0 will give the point at A which is also not allowed.
With K = 40, x + y = 40; this will be satisfied by points (1, 39), (2, 38), (3, 37)…………………………………….. (38, 2), (39, 1). That is a total of 39 points
Similarly x + y = 38, will be satisfied by 37 points.
X + Y = 37, will be satisfied by 36 points
X + Y = 3 will be satisfied by 2 points
X + Y = 2 will be satisfied by 1 point
X + Y = 1 will not be satisfied by any point
So, the total number of all such points is
39 + 38 + 37 + 36 + ……………………. + 3 + 2 + 1 = n(n+1)/2 points = (39*40) / 2 = 780 points
Example 5: Two lines P and Q intersect at point (3, 2) in the x-y plane. The slope of line P is 45 degrees and line Q is parallel to the X axis. What is the area (in sq. units) of the triangle formed by P, Q and a line perpendicular to P passing through point (5, 4) ?
Solution: Let us look at the image below:
As slope of line P is 45 degree. Therefore, ∠ABC = 45 degree
In triangle ABC, length of AB = SQRT [(5-3)2 + (4-2)2] = 2√2 units
Therefore, length of line AC = 2√2 units (Since ABC is an isosceles triangle. Thus AB = AC )
Thus, required area = ½ * 2√2 * 2√2 = 4 sq. units
Example 6: The line √3 Y = x is the radius of the circle. It meets the circle o=centered at origin O at point M (√3, 1). If PQ is the tangent to the circle at M as shown, find the length of the PQ.
a.) (5/2)√3 units
b.) 3 √3 units
c.) 2√3 units
d.) 8/√3 units
Solution: PQ is perpendicular to line Y = X / √3 (Since, radius of a circle is perpendicular to the tangent of the circle)
Therefore, slope of PQ = -1 / (1/ √3) = – √3 (Since, product of slopes of line perpendicular to each other is -1)
Therefore, Let equation of the line PQ be y = – √3x + c
Now at the point M, when x = √3, y = 1
Putting the above values of x and y in the above equation, we get c = 4
The equation of the line becomes, Y = – √3x + 4
Thus, by using the above equation, we get:
Coordinates of point P = (0, 4) and coordinates of point Q = (4/√3, 0) (Putting x = 0 in above equation, we find value of P and putting Y = 0 in above equation, we find value of Q)
Hence PQ = sqrt [(4/√3) 2 + 42 ] = 8/√3 units
Summary: For solving problems in coordinate geometry, we need to know concepts of lines, circles, square, rectangle etc. If we know the basic formulas of coordinate geometry, and have a good idea of basic geometry, it will be easy to solve problems. Most of the questions in CAT exam is of the pattern as given above.
All questions from CAT Exam Quantitative Aptitude – Geometry
Quantitative Aptitude – Geometry – Triangles – Q1: Let P be an interior point of a right-angled isosceles triangle ABC with hypotenuse AB.
Quantitative Aptitude – Geometry – Triangles – Q2: Let ABC be a right-angled triangle with BC as the hypotenuse. Lengths of AB and AC are 15 km and 20 km, respectively.
Quantitative Aptitude – Geometry – Triangles – Q3: From a triangle ABC with sides of lengths 40 ft, 25 ft and 35 ft, a triangular portion GBC is cut off where G is the centroid of ABC.
Quantitative Aptitude – Geometry – Circles – Q1: ABCD is a quadrilateral inscribed in a circle with centre O. If ∠COD = 120 degrees and ∠BAC = 30 degrees
Quantitative Aptitude – Geometry – Circles – Q2: Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC.
Quantitative Aptitude – Geometry – Coordinate – Q1: The points (2, 5) and (6, 3) are two end points of a diagonal of a rectangle. If the other diagonal has the equation y = 3x + c, then c is
Quantitative Aptitude – Geometry – Coordinate – Q2: The shortest distance of the point (½, 1) from the curve y = |x -1| + |x + 1| is
Quantitative Aptitude – Geometry – Mensuration – Q1: The base of a vertical pillar with uniform cross section is a trapezium whose parallel sides are of lengths 10 cm and 20 cm
Quantitative Aptitude – Geometry – Mensuration – Q2: A ball of diameter 4 cm is kept on top of a hollow cylinder standing vertically.
Quantitative Aptitude – Geometry – Mensuration – Q3: A solid metallic cube is melted to form five solid cubes whose volumes are in the ratio 1 : 1 : 8: 27: 27.
Quantitative Aptitude – Geometry – Polygon – Q1: Let ABCDEF be a regular hexagon with each side of length 1 cm. The area (in sq cm) of a square with AC as one side is
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