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Coordinate geometry helps to locate the points on the plane, also called coordinates. The distance of a point from the y-axis is called the abscissa (x-coordinate), and the distance from the x-axis is called ordinate (y-coordinate).
Distance Formula: The coordinates of point P are (x1, y1), and Q are (x2, y2).
\(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
At origin it becomes,
\(\sqrt{x^2 + y^2}\)
Section Formula: If the point P(x, y) divides the line segment joining A (x1, y1) and B(x2, y2) internally in the ratio m1 : m2 then, the coordinates of P are
\((\frac{m_1x_2 + m_2 x_1}{m_1 + m_2},\frac{m_1y_2 + m_2 y_1}{m_1 + m_2})\)
Mid-point: The mid-point of any line segment divides the line segment in the ratio of 1 : 1.
A(x1, y1) and B(x2, y2).
\((\frac{x_1 + x_2}{2},\frac{y_1 +y_2}{2})\)
Area of a Triangle: Area of the triangle formed by the point (x1, y1), (x2, y2) and (x3, y3) is
The video below explains this:
Coordinate Geometry Detailed Video Explanation:
Sample Questions
Ques. The area of a triangle with vertices (a, b + c), (b, c + a) and (c, a + b) is
(A) (a + b + c)2
(B) 0
(C) a + b + c
(D) ab
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Ans: (b) 0
Explanation:
In triangle ABC,
A = (x1, y1) = (a, b + c)
B = (x2, y2) = (b, c + a)
C = (x3, y3) = (c, a + b)
Area of triangle ABC = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
= (1/2)[a(c + a – a – b) + b(a + b – b – c) + c(b + c – c – a)]
= (1/2)[a(c – b) + b(a – c) + c(b – a)]
= (1/2)(0) = 0
Ques: The point which lies on the perpendicular bisector of the line segment joining the points A(–2, –5) and B(2, 5) is
(a) (0, 0)
(b) (0, 2)
(c) (2, 0)
(d) (–2, 0)
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Ans: (a) (0, 0)
Explanation:
The perpendicular bisector of a line segment divides the line segment into two equal parts. By mid-point theorem,
[(-2 + 2)/2, (-5 + 5)/2]
= (0, 0)
Ques. If the points (a, 0), (0, b) and (1, 1) are collinear, then which of the following is true?
(a) (1/a) + (1/b)= 2
(b) (1/a) + (1/b)= 1
(c) (1/a) + (1/b)= 0
(d) (1/a) + (1/b)= 4
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Ans: (b) (1/a) + (1/b) = 1
Explanation:
The area of a triangle whose vertices are (x1, y1), (x2, y2), (x3, y3)
= 1/2 [x1(y2 – y3) + x2(y3 – y1)
The area of the triangle = zero, if points are collinear
= 1/2a(b – 1) + 0(1 – 0) + 1(0 – b) = 0
= (1/a) + (1/b) = 1
Ques. The points on X-axis at a distance of 10 units from (11, –8) are
(a) (5, 2) (17, 0)
(b) (5, 0) (17, 0)
(c) (6, 0) (17, 0)
(d) (5, 0) (16, 0)
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Ans: (b) (5, 0) (17, 0)
Explanation:
Suppose point A = (x,0)
By using the distance formula, (x−11)2+82=102
Hence, x2−22x+121+64=100
x2 −22x+85=0,
By mid-term split, sum of two numbers is -22 and product = 85, we get,
x2−17x−5x+85=0
(x−17) (x−5)=0
x=17, 5
Thus, the points are (5,0) and (17,0).
Ques. Mid-point of the line-segment joining the points (– 5, 4) and (9, – 8) is:
(a) (-2,2)
(b) (7,-6)
(c) (2,-2)
(d) (-7,6)
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Ans: (c) (2,-2)
Explanation:
The midpoint divides it in equal ratio, so it is 1:1.
[(x1+ x2)/2, (y1+y2)/2]
The mid-point of the line-segment joining the points (– 5, 4) and (9, – 8) = (2, -2)
Ques. A line intersects the y-axis and x-axis at the points P and Q, respectively. If (2, -5) is the midpoint of PQ, then the coordinates of P and Q are, respectively
(a) (0, -5) and (2, 0)
(b) (0, 10) and (-4, 0)
(c) (0, 4) and (-10, 0)
(d) (0, -10) and (4, 0)
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Ans: (d) (0, -10) and (4, 0)
Explanation:
Let (0, y) and (x, 0) be the vertices of points P and Q.
As (2, -5) is the midpoint of PQ, then [(x1+ x2)/2, (y1+y2)/2]
[(0 + x)/2, (y + 0)/2] = (2, -5)
(x/2, y/2) = (2, -5)
If we equate the corresponding coordinates,
x/2 = 2, y/2 = -5
x = 4, y = -10
Hence, the coordinates of P and Q are (0, -10) and (4, 0).
Ques. The points (-4, 0), (4, 0), (0, 3) are the vertices of a
(a) Right triangle
(b) Isosceles triangle
(c) Equilateral triangle
(d) Scalene triangle
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Ans: (b) Isosceles triangle
Explanation.
Let A = (-4, 0), B = (4, 0), C = (0, 3)
By distance formula,
= \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
We found AB = 8
BC = 5
Ac = 5
So, BC = AC and hence it is an isosceles triangle, where two sides of a triangle.
Ques. The point which divides the lines segment joining the points (7, -6) and (3, 4) in ratio 1 : 2 internally lies in the
(a) I quadrant
(b) II quadrant
(c) III quadrant
(d) IV quadrant
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Ans (d) IV quadrant
Explanation:
By section formula, \((\frac{m_1x_2 + m_2 x_1}{m_1 + m_2},\frac{m_1y_2 + m_2 y_1}{m_1 + m_2})\)
We get, (17/3, -8/3), where m1, : m2 = 1:2.
A = (7, -6)
B= (3,4)
Ques: The distance between the points (– 1, – 5) and (– 6, 7) is
(a) 144 units
(b) 13 units
(c) 12 units
(d) 169 units
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Answer: (d) 13 units
Explanation:
By using distance formula, \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
Square root of (-6+1)2 +(7+5)2
Hence, Square root of 25+144 = 13 units.
Ques: The midpoint of the line segment joining the points A (-2, 8) and B (-6, -4) is
(a) (-4, -6)
(b) (2, 6)
(c) (-4, 2)
(d) (4, 2)
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Ans: (c) (-4, 2)
Explanation:
By midpoint formula, \((\frac{m_1x_2 + m_2 x_1}{m_1 + m_2},\frac{m_1y_2 + m_2 y_1}{m_1 + m_2})\)
1/2 [-2+(-6), 1/2 (-4+8)], so we get (-4, 2).
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