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Chapter 7: Coordinate Geometry

    Introduction

    Distance Between Two Points

    Distance Between Two Points on a Line Parallel To The Coordinate Axis

    Distance Between Any Two Points In the X-Y Plane

    Section Formula

    TriSectional Points of a Line

    Centroid Triangle

    Area of the Triangle

    Straight Lines

    Summary

    Enhanced Curriculum Support

    Easy Level Worksheet

    Moderate Level Worksheet

    Hard Level Worksheet

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Chapter 7: Coordinate Geometry > Distance Between Any Two Points In the X-Y Plane

Distance Between Any Two Points In the X-Y Plane

Let us now find the distance between any two points P(x1,y1) and Q(x2,y2).

Then, OR = x1, OS = x2.

So, RS = x2−x1 =

Also, SQ = y2, ST = = y1

So, = y2−y1

Now, applying the Pythagoras theorem in triangle PTQ, we get:

PQ2 = PT2 + QT2 = x2−x12+ y2−y12

Therefore, PQ = x2−x12+y2−y12

Note that since distance is always , we take only the square root.

So, the distance between the points P(x1,y1) and Q(x2,y2) is:

PQ = x2−x12+y2−y12

which is called the distance formula.

Remarks:

(1) In particular, the distance of a point P(x, y) from the origin O(0, 0) is given by OP = x2+y2

(2) We can also write, PQ = x1−x22+y1−y22 as the points are interchangeable.

1. Do the points (3, 2), (–2, –3) and (2, 3) form a triangle? If so, name the type of triangle formed.

Solution : Let us apply the distance formula to find the distances PQ, QR and PR, where P(3, 2), Q(–2, –3) and R(2, 3) are the given points. We have:

Check for triangle

  • Let's find the length of the sides using distance formula.
  • We have: PQ = 3+22+2+32 = 2+2= = (upto two decimal places)
  • QR = −2−22+−3−32 = −42+−62 = = (upto two decimal places)
  • PR = 3−22+2−32 = 12+−12 = = (upto two decimal places)
  • Check if the sum of any two of these sides is greater than the third side.
  • Therefore, the points P, Q and R form a triangle.
  • Also, PQ2+PR2=QR2 , by the converse of Pythagoras theorem, we have angle P = 90°.
  • Therefore, PQR is a right triangle.

2. Show that the points (1, 7), (4, 2), (–1, –1) and (– 4, 4) are the vertices of a square.

Solution : Let A(1, 7), B(4, 2), C(–1, –1) and D(– 4, 4) be the given points. One way of showing that ABCD is a square is to use the property that all its sides should be equal and both its digonals should also be equal. Now,

AB = 1−42+7−22 = 9+25=

BC = 4+12+2+12 = =

CD = −1+42+−1−42 = =

DA = 1+42+7−42 = =

AC = 1+12+7+12 = =

BD = 4+42+2−42 = =

Since, AB = BC = CD = DA and AC = BD, all the four sides of the quadrilateral ABCD are and its diagonals AC and BD are also .

Thereore, ABCD is a square.

Alternative Solution : We find the four sides and one diagonal, say, AC as above. Here AD2+DC2 = + = = AC2.

Therefore, by the converse of Pythagoras theorem, angle D = °.

A quadrilateral with all four sides equal and one angle ° is a square. So, ABCD is a square.

3. Above figure shows the arrangement of desks in a classroom. Ashima, Bharti and Camella are seated at A(3, 1), B(6, 4) and C(8, 6) respectively. Do you think they are seated in a line? Give reasons for your answer.

Check for line

  • Let's find the distance between the students using distance formula.
  • We have: AB = 6−32+4−12 = 9+9= =
  • BC = 8−62+6−42 = 4+4= =
  • AC = 8−32+6−12 = 25+25= =
  • We see that: AB + BC = 32+22 =
  • This is further equal to
  • Thus, the points A, B and C are collinear. Therefore, they are seated in a line.

4. Find a relation between x and y such that the point (x , y) is equidistant from the points (7, 1) and (3, 5).

Solution : Let P(x, y) be equidistant from the points A(7, ) and B(, 5).

We are given that AP = BP. So, AP2 =

Therefore, x−72+y−12=x−32+y−52

On expanding, we get:

x2 + + + y2−2y+1 = x2 – + 9+y2−10y +

x – y = which is the required relation.

Remark: Note that the graph of the equation x – y = 2 is a line. From earlier studies, we know that a point which is equidistant from A and B lies on the perpendicular bisector of .

Therefore, the graph of x – y = 2 is the perpendicular bisector of AB (see below).

5. Find a point on the y-axis which is equidistant from the points A(6, 5) and B(– 4, 3).

Solution : We know that a point on the y-axis is of the form .

So, let the point P(0, y) be equidistant from A and B. Then:

6−02+5−y2=−4−02+3−y2

On expanding,

+ 25 + y2−10y = + 9 + y2 – 6y

4y = 36

y =

So, the required point is (0,9).

Let us check our solution : AP = 6−02+5−92 = + =

BP = −4−02+3−92 = + =

Note : Using the remark above, we see that (0, 9) is the intersection of the y-axis and the perpendicular bisector of AB.