# Co-ordinate Geometry

Coordinate geometry is used to represent algebraic relations on graphs.
We shall be dealing with two-dimensional problems, where there are two variables to be handled.  The variables are normally denoted by the ordered pair (x, y).
The horizontal axis is the X -axis and the vertical axis is the Y-axis. If the coordinates of a point on the XY plane is (x, y), it implies that it is at a perpendicular distance of x from the Y-axis and at a perpendicular distance y from the X-axis. The point of intersection of the X and Y-axis is called the origin and the coordinates of this point is (0, 0). Signs of the coordinate ‘x’ and ‘y’ depends on the quadrant in which the point lies

I quadrant :        x coordinate is positive; y coordinate is positive
II quadrant :      x coordinate is negative; y coordinate is positive
III quadrant :     x coordinate is negative: y coordinate is negative
IV quadrant :     x coordinate is positive; y coordinate is negative

Some fundamental formulae:
1. Distance between the points: Distance between two points $\left( {{x_{1,}}{y_1}} \right)$ and $\left( {{x_{2,}}{y_2}} \right)$ is $= \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$

Slope:
Angle made by the line with the positive direction of x - axis is called the inclination of the line.
If $\theta$ is the inclination, then ‘tan $\theta$’ denotes the slope of the line.
Slope of the line joining the points $\left( {{x_{1,}}{y_1}} \right)$ and $\left( {{x_{2,}}{y_2}} \right)$ is $\displaystyle\frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$; . The slope is also indicated by m.

Slope intercept form:
All straight lines can be written as y = mx + c,  where m is the slope of the straight line, c is the Y intercept or the Y coordinate of the point at which the straight line cuts the Y-axis.

Point slope form:
The equation of a straight line passing through $\left( {{x_{1,}}{y_1}} \right)$ and having a slope m is $y - {y_1} = m\left( {x - {x_1}} \right)$

Two point form:
The equation of a straight line passing through two points $\left( {{x_{1,}}{y_1}} \right)$ and $\left( {{x_{2,}}{y_2}} \right)$ is $y - {y_1} = \displaystyle\frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}(x - {x_1})$

Intercept form:
The intercept form of a line is $\displaystyle\frac{x}{a}\; + \;\frac{y}{b}\; = \;1$
Where ‘a’ is the intercept on x-axis and ‘b’ is the intercept on y-axis.
The point of intersection of any two lines of the form y = ax + b and y = cx + d is same as the solution arrived at when these two equations are solved.

Perpendicular Distance:
The length of perpendicular from a given point $\left( {{x_{1,}}{y_1}} \right)$ to a given line ax + by + c = 0 is $\left| {\displaystyle\frac{{a{x_1} + b{y_1} + c}}{{\sqrt {({a^2} + {b^2})} }}} \right| = p$, where p is the length of perpendicular. In particular, the length of perpendicular from origin

Distance Between two parallel lines:
Distance between two parallel straight lines $ax + by + {c_1}$ = 0 and $ax + by + {c_2}$ = 0 is given by $\displaystyle\frac{{\left| {{c_1}\; - \;{c_2}} \right|}}{{\sqrt {{a^2}\; + \;{b^2}} }}$

Acute Angle between two lines:
If ${a_1}x + {b_1}y + {c_1}$ = 0  and ${a_2}x + {b_2}y + {c_2}$ = 0 are equations of two lines, then acute angle (q) between them is given by $Cos\;\theta \; = \displaystyle\frac{{\left| {\;{a_1}\;{a_2}\; + \;{b_1}\;{b_2}\;} \right|}}{{\sqrt {\left( {{a^2}_1\; + {b^2}_1} \right)\;\left( {{a^2}_2\; + {b^2}_2} \right)} }}$

If ${m_1}$ and ${m_2}$ are slopes of two straight lines, then acute angle ($\theta$) between them is given by $\tan \theta = \dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}$

Area of the Triangle:
The area of a triangle whose vertices are $\left( {{x_{1,}}{y_1}} \right)$, $\left( {{x_{2,}}{y_2}} \right)$ and $\left( {{x_{3,}}{y_3}} \right)$ = $\displaystyle\frac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]$

Solved Examples:

1.  What is the distance between points A (4, 3) and B (7, 7)
Solution:
$\left( {{x_{1,}}{y_1}} \right)$ = (4, 3);  $\left( {{x_{2,}}{y_2}} \right)$ = (7, 7)
Distance between points = $\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$
=$\sqrt {{{\left( {7 - 4} \right)}^2} + {{\left( {7 - 3} \right)}^2}}$ = $\sqrt {{3^2}\: + \:{4^2}}$ = $\sqrt {25}$ = 5 units

2. If the distance between the points (3, k) and (4, 1) is , then the value of k is
a. 4 or 2
b. –4
c. –4 or 2
d. 4 or –2
Sol: $\sqrt {{{\left( {4 - 3} \right)}^2} + {{\left( {1 - k} \right)}^2}} = \sqrt {10}$
$\Rightarrow \,\,1 + 1 + {k^2} - 2k = 10$
$\Rightarrow \,\,{k^2} - 2k - 8 = 0$
$\Rightarrow \,\,\left( {k - 4} \right)\left( {k + 2} \right) = 0$
$\Rightarrow$ k = 4 or k = –2

3. The point on the y-axis which is equidistant from A(–5, –2) and B(3, 2) is
a. (–4, 0)
b. (0, 2)
c. (0, –2)
d. (0, –4)
Sol: Let P(0, y) be the coordinates of the point on y-axis.
Then PA = PB $\Rightarrow \,\,\sqrt {{{\left( { - 5 - 0} \right)}^2}$ + ${{\left( { - 2 - y} \right)}^2}}$  = $\sqrt {{{\left( {3 - 0} \right)}^2} + {{\left( {2 - y} \right)}^2}}$
$\Rightarrow$ 25 + 4 + ${y^2}$ + 4y = 9 + 4 + ${y^2}$ – 4y
$\Rightarrow$ 8y = –16
$\Rightarrow$ y = -2

4. The vertices of a triangle are (–2, 0), (2, 3) and (1, –3), then the type of the triangle is
a. Right angle
b. Equilateral
c. Isosceles
d. Scalene
Sol: A(–2, 0), B(2, 3) and C(1, –3)
$AB = \sqrt {{{\left( {2 + 2} \right)}^2} + {{\left( {3 - 0} \right)}^2}} = 5$
$BC = \sqrt {{{\left( {1 - 2} \right)}^2} + {{\left( { - 3 - 3} \right)}^2}} = \sqrt {37}$
$CA = \sqrt {{{\left( { - 2 - 1} \right)}^2} + {{\left( {0 + 3} \right)}^2}} = 3\sqrt 2$
$\,\,AB \ne BC \ne CA$  $\Rightarrow \Delta$ ABC is a scalene triangle

5. The points which are not collinear are
(I) (0, –1), (8, 3) and (6, 7)
(II) (4, 3), (5, 1) and (1, 9)
(III) (2, 5), (–1, 2) and (4, 7)
(IV) (–3, 2), (1, –2) and (9, –10)
a. I
b. II
c. III
d. IV
Sol: From option (a)  A(0, –1), B(8, 3) and C(6, 7)
$AB = \sqrt {{{\left( {8 - 0} \right)}^2} + {{\left( {3 + 1} \right)}^2}} = 4\sqrt 5$
$BC = \sqrt {{{\left( {6 - 8} \right)}^2} + {{\left( {7 - 3} \right)}^2}} = 2\sqrt 5$
$AC = \sqrt {{{\left( {6 - 0} \right)}^2} + {{\left( {7 + 1} \right)}^2}} = 10$
$AB + BC \ne AC$
$\Rightarrow$  Points are not collinear.

6. The coordinates of the point which divides the line segment joining the points (–7, 4) and  (–6, –5) internally in the ratio 7 : 2 is
a. $\left( {\displaystyle\frac{{56}}{9},\,\,3} \right)$
b. $\left( { - \displaystyle\frac{{56}}{9},\,\, - 3} \right)$
c. $\left( { - \displaystyle\frac{{61}}{9},\,\,2} \right)$
d. $\left( {\displaystyle\frac{{61}}{9},\,\, - 2} \right)$
Sol: Let (x, y) be the coordinates of the required point.
$x = \displaystyle\frac{{7\left( { - 6} \right) + 2\left( { - 7} \right)}}{{7 + 2}} = \displaystyle\frac{{ - 42 - 14}}{9} = - \displaystyle\frac{{56}}{9}$
$y = \displaystyle\frac{{7\left( { - 5} \right) + 2\left( 4 \right)}}{{7 + 2}} = \displaystyle\frac{{ - 35 + 8}}{9} = - 3$
$\,\,\left( {x,\,\,y} \right) \equiv \left( { - \displaystyle\frac{{56}}{9}, - 3} \right)$

7.  If the coordinates of the centroid of a triangle are (1, 3) and two of its vertices are (–7, 6) and  (8, 5), then the third vertex of the triangle is
a. $\left( {\displaystyle\frac{2}{3},\,\,\frac{{14}}{3}} \right)$
b. $\left( { - \displaystyle\frac{2}{3},\, - \,\frac{{14}}{3}} \right)$
c. (2, –2)
d. (–2, 2)
Sol: Let (x, y) be the coordinates of the third vertex of the triangle. Then $\displaystyle\frac{{ - 7 + 8 + x}}{3}$ = 1 and $\displaystyle\frac{{6 + 5 + y}}{3} = 3$ $\Rightarrow$ x + 1 = 3 and y + 11 = 9 $\Rightarrow$ x = 2 and y = –2 $\Rightarrow$ (x, y) = (2, –2)

8. The ratio in which the line segment joining (3, 4) and (–2, –1) is divided by the x-axis is
a. 3 : 2
b. 1 : 4
c. 4 : 3
d. None of these
Sol: Let $\lambda$ : 1 be the required ratio. Then
$\displaystyle\frac{{\lambda \left( { - 1} \right) + 1\left( 4 \right)}}{{\lambda + 1}} = 0$ [Ordinate of the point on the x-axis will be zero.]
$\Rightarrow \lambda$ = 4
The ratio is 4 : 1.

9. If a line passes through the mid-point of the line joining (–3, –4) and (–5, 6) and has a slope of $\displaystyle\frac{3}{4}$ , then its equation is
a. 4x – 3y + 16 = 0
b. 3x – 4y + 16 = 0
c. 4x + 3y – 16 = 0
d. 3x + 4y – 16 = 0
Sol. Mid-points of (–3, –4) and (–5, 6) is $\left( {\displaystyle\frac{{ - 3 - 5}}{2},\,\;\displaystyle\frac{{ - 4\; + \;6}}{2}} \right)$
$\left[ {Mid{\rm{ - }}point \equiv \left( {\displaystyle\frac{{{x_1} + \;{x_2}}}{2},\;\displaystyle\frac{{{y_1}\; + \;{y_2}}}{2}} \right)} \right]$ i.e., (–4, 1)
The required equation of the line is $y\; - \;1\; = \;\displaystyle\frac{3}{4}\;\left( {x\; + \;4} \right)$ [Using $y - {y_1} = m(x - {x_1})$]
$\Rightarrow$  3x – 4y + 16 = 0

10. The equation of the line through (2, –4) and parallel to the line joining the points (2, 3) and (–4, 5) is
a. 3x – y = 0
b. x + 3y = 0
c. x + 3y + 10 = 0
d. 3x – y = 14
Sol: Slope of the line joining (2, 3) and (–4, 5) is $\displaystyle\frac{{{y_2}\; - \;{y_1}}}{{{x_2}\; - \;{x_1}}}\; = \displaystyle\frac{{5\; - \;3}}{{ - 4\; - 2}}\; = \; - \displaystyle\frac{1}{3}$
Slope of any line parallel to it $= \; - \displaystyle\frac{1}{3}$
The required equation of the line is y + 4 = $- \displaystyle\frac{1}{3}\;\left( {x\; - \;2} \right)$ [Using $y - {y_1} = m(x - {x_1})$]
$\Rightarrow$ x + 3y + 10 = 0

11. If the vertices of a triangle are (1, 3), (–2, 4) and (3, –5), then the equation of the altitude from (1, 3) to  the opposite side is
a. 9x + 5y – 24 = 0
b. 9x – 5y + 24 = 0
c. 5x – 9y + 22 = 0
d. 5x + 9y – 22 = 0
Sol: Let the vertices be A(1, 3), B(–2, 4) and C(3, – 5).
Then the slope of BC =$\displaystyle\frac{{y{\;_2}\; - \;{y_1}}}{{{x_2}\; - \;{x_1}}}\; = \;\displaystyle\frac{{ - 5\; - \;4}}{{3\; + \;2}}\; = \; - \displaystyle\frac{9}{5}$
Slope of altitude through A =$\displaystyle\frac{5}{9}$
The equation of the altitude passes through A (1, 3) is $y\; - \;3\; = \;\displaystyle\frac{5}{9}\;\left( {x\; - \;1} \right)$
$\Rightarrow \;5x\; - \;9y\; + \;22\; = \;0$

12. The equation of the line which is concurrent with the lines x – y – 2 = 0 and 3x + 4y + 15 = 0 and is perpendicular to the line joining the points (2, 3) and (1, 1) is
a. x + 2y + 1 = 0
b. x – 2y + 7 = 0
c. x + 2y + 5 = 0
d. None of these
Sol: Given lines are x – y = 2      ... (i) 3x + 4y = –15   ... (ii) Solving (i) and (ii), we get  x = – 1 and y = – 3
Slope of the line joining the points (2, 3) and (1, 1) is = $\displaystyle\frac{{1\; - \;3}}{{1\; - \;2}}\; = \;2$
Slope of its perpendicular = $\displaystyle\frac{{ - 1}}{2}$
The required equation of the line is y + 3 = -$\displaystyle\frac{1}{2}\;\left( {x\; + \;1} \right)$
$\Rightarrow$ x + 2y + 7 = 0

13. The point on the x-axis which is equidistant from B(5, 9) and C(– 4, 6) is
a. (–3, 0)
b. (3, 0)
c. (–,4 0)
d. (0, –4)
Sol: Let P(x, 0) be the coordinates of the point on x-axis. Then PB = PC
$\Rightarrow \sqrt {{{\left( {5 - x} \right)}^2} + {{\left( {9 - 0} \right)}^2}} = \sqrt {{{\left( { - 4 - x} \right)}^2} + {{\left( {6 - 0} \right)}^2}}$
$\Rightarrow$ 25 – 10x + ${x^2}$ + 81 = 16 + 8x + ${x^2}$ + 36
$\Rightarrow$ – 18x = – 54 $\Rightarrow$  x = 3
The coordinates of the point is (3, 0).

14. The ratio in which the line 3x + y – 9 = 0 divides the segment joining the points (1, 3) and (2, 7) is
a. 3 : 4
b. 4 : 3
c. 2 : 3
d. 3 : 2
Sol: Let the line 3x + y – 9 = 0 divide the segment joining the points (1, 3) and (2, 7) at point P in the ratio k : 1.
The point P is $\left( {\displaystyle\frac{{2k + 1}}{{k + 1}},\displaystyle\frac{{7k + 3}}{{k + 1}}} \right)$
P lies on 3x + y – 9 = 0
$3\left( {\displaystyle\frac{{2k + 1}}{{k + 1}}} \right) + \displaystyle\frac{{7k + 3}}{{k + 1}} - 9 = 0$
$\Rightarrow$6k + 3 + 7k + 3 – 9k – 9 = 0 4k – 3 = 0 $\Rightarrow$
$k = \displaystyle\frac{3}{4}$
The ratio is or 3 : 4

15. If the line through A(3, y) and B(2, 7) is parallel to the line through the points P(–1, 4) and Q(0, 6),  then the value of y is
a. – 7
b. 7
c. – 9
d. 9
Sol. AB and PQ are parallel.
$\Rightarrow$  Slope of AB = Slope of PQ
$\Rightarrow \displaystyle\frac{{7 - y}}{{2 - 3}} = \displaystyle\frac{{6 - 4}}{{0 + 1}}$
$\Rightarrow$7 – y = – 2 $\Rightarrow$y = 9

16. The equation of the perpendicular bisector of the line joining (5, 6) and (2, –2) is
a. 6x + 16y = 53
b. 3x – 8y = 5
c. 16x – 6y + 44 = 0
d. None of these
Sol. Let the points be A(5, 6) and B(2, –2)
Mid-points of AB $\equiv$ $\left( {\displaystyle\frac{{5 + 2}}{2},\displaystyle\frac{{6 - 2}}{2}} \right)$
i.e$\left( {\displaystyle\frac{7}{2},2} \right)$
Slope of AB =$\displaystyle\frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \displaystyle\frac{{ - 2 - 6}}{{2 - 5}} = \displaystyle\frac{8}{3}$
The required equation of the line is y – 2 = $- \displaystyle\frac{3}{8}\left( {x - \displaystyle\frac{7}{2}} \right)$  [Using $y - {y_1} = m(x - {x_1})$]
$\Rightarrow$ 6x + 16y = 53

17. If a line passes through (3, 4) and (8, 5), then a point common to this line and 2x + y + 1 = 0 is
a. (– 2, 3)
b. (– 4, 6)
c. (3, – 4)
d. (– 2, 6)
Sol. The equation of the line passing through (3, 4) and (8, 5) is
$y - 4 = \displaystyle\frac{{5 - 4}}{{8 - 3}}\left( {x - 3} \right)$
$\Rightarrow y - 4 = \displaystyle\frac{1}{5}\left( {x - 3} \right)$
$\Rightarrow x - 5y = 17\;$ .........(1) $2x + y = - 1\;$ ....(2)
Given that these two linke have a point in common.
Solving (i) and (ii), we get x = – 2 and y = 3
Common point is (– 2, 3).

18. The equation of the line parallel to x-axis and passing through the point of intersection of the lines x + y – 5 = 0 and 2x + 3y = 13 is
a. y = 2
b. y = 3
c. x = 2
d. x = 3
Sol. Given lines are  x + y = 5          ... (i) 2x + 3y = 13    ... (ii)
Solving (i) and (ii), we get x = 2, y = 3 The required line is parallel to x-axis. Its slope = 0
The required equation of the line is  y – 3 = 0 (x – 2) $\Rightarrow$  y = 3

19. The vertices of a quadrilateral ABCD are A(1, 0), B(5, 3), C(3, 7) and D(–1, 4), then the type
a. parallelogram
b. rectangle
c. rhombus
d. square
Sol. $AB = \sqrt {{{\left( {5 - 1} \right)}^2} + {{\left( {3 - 0} \right)}^2}} = 5$
$BC = \sqrt {{{\left( {3 - 5} \right)}^2} + {{\left( {7 - 3} \right)}^2}} = 2\sqrt 5$
$CD = \sqrt {{{\left( { - 1 - 3} \right)}^2} + {{\left( {4 - 7} \right)}^2}} = 5$
$AD = \sqrt {{{\left( { - 1 - 1} \right)}^2} + {{\left( {4 - 0} \right)}^2}} = 2\sqrt 5$
$AC = \sqrt {{{\left( {3 - 1} \right)}^2} + {{\left( {7 - 0} \right)}^2}} = \sqrt {53}$
$BD = \sqrt {{{\left( { - 1 - 5} \right)}^2} + {{\left( {4 - 3} \right)}^2}} = \sqrt {37}$
AB = CD, BC = AD and AC $\ne$ BD.
$\Rightarrow$ABCD is a parallelogram

20. In what ratio does the point $R \equiv \left( {\displaystyle\frac{3}{5},\frac{{11}}{5}} \right)$  divide the line segment joining the points P (3, 5) and Q (–3, –2)?
a. 2 : 3
b. 3 : 2
c. 2 : 5
d. 3 : 5
Sol. Let the point R divide PQ in the ratio .
Then $R \equiv \left( {\displaystyle\frac{{ - 3\lambda + 3}}{{\lambda + 1}},\frac{{ - 2\lambda + 5}}{{\lambda + 1}}} \right)$  [Using section formula]
But, given $R \equiv \left( {\displaystyle\frac{3}{5},\frac{{11}}{5}} \right)$
$\displaystyle\frac{{ - 3\lambda + 3}}{{\lambda + 1}} = \frac{3}{5}$ and $\displaystyle\frac{{ - 2\lambda + 5}}{{\lambda + 1}} = \frac{{11}}{5}$
$\Rightarrow 18\lambda = 12$ and $21\lambda = 14$
$\Rightarrow \lambda = \displaystyle\frac{2}{3}$ and $\lambda = \displaystyle\frac{2}{3}$
Therefore, the required ratio is 2 : 3.
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