Coordinate Geometry Chapter 1 Distance Formulas
⇒ Coordinate Geometry: The branch of mathematics in which different problems of geometry are solved with the help of algebra is known as coordinate geometry.
⇒ Two types of co-ordinate geometry:
- Two-dimensional or plane geometry.
- Three-dimensional or solid geometry.
⇒ XOX’ and YOY’ are two perpendicular straight lines intersects at O.
⇒ They have divided the plane into four sections. Each of these sections is called a quadrant.
⇒ The section XOY, YOX’, X’OY’, and Y’OX’ are called the 1st, 2nd, 3rd, and 4th quadrants respectively.
⇒ The fixed point O is called the origin whose coordinate is (0, 0).
⇒ The straight lines together are called the coordinate axes.
⇒ \(\overleftrightarrow{\mathrm{XOX’}}\) is called the x-axis or abscissa and \(\overleftrightarrow{\mathrm{YOY’}}\) is called the y-axis or ordinate.
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Formulas at a glance:
- Distance of a given point P(x, y) from the origin (0, 0) is OP = \(\sqrt{x^2+y^2}\)
- The distance between two given points P (x1, y1) and Q (x2, y1) is \(\overline{P Q}=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\) units.
Coordinate Geometry Chapter 1 Distance Formulas True Or False
Example 1. The distance between the points A(a, 0) and B(b, 0) along the positive direction of the X-axis is (b – a) units (b > a).
Solution: The distance between the points A(a, 0)
and B (b, 0) is \(\sqrt{(b-a)^2+(0-0)^2}\) units
= \(\sqrt{(b-a)^2}\) units
= (b – a) units
∴ The statement is True.
Example 2. The distance between two points A(-p, 0) and (-q, 0) along the negative direction of X- axis (p – q) units [p < q].
Solution: The distance between A(- p, 0) and B(-q, 0) is
\(\sqrt{\{-q-(-p)\}^2+(0-0)^2}\) units
= \(\sqrt{(-q+p)^2}\) units
= \(\sqrt{{-(q-p)}^2}\) units
= \(\sqrt{(q-p)^2}\) units =(q – p) units
∴ The statement is False.
Example 3. The points (3, 4) and (-3, -4) are equidistant from origin.
Solution: The distance of point (3, 4) from origin (0, 0) is \(\sqrt{3^2+4^2}\) units = √25 units = 5 units
The distance between points (-3, -4) and (0, 0) is \(\sqrt{(-3)^2+(-4)^2}\) units
= \(\sqrt{9+16}\) units = √25 units = 5 units
∴ The statement is True.
Coordinate Geometry Chapter 1 Distance Formulas Fill In The Blanks
Example 1. The distance between the points A (0, m) and (0, – n) is _______ units.
Solution: The required distance is \(\sqrt{(0-0)^2+\{m-(-n)\}^2}\) units = \(=\sqrt{(m+n)^2}\) units = (m + n) units.
Example 2. The distance between the points (-7, 0) and (-2, 0) is ________ units.
Solution: The required distance is \(\sqrt{\{(-7)-(-2)\}^2+(0-0)^2}\) units
= \(\sqrt{(-7+2)^2+0} \text { units }\)
= \(\sqrt{(-5)^2} \text { units }\)
= √25 units = 5 units.
Example 3. If in a square (4, 4) and (-4, 4) are two adjacent vertices, then the perimeter of the square is ______ units.
Solution: The length of AD is \(\sqrt{(-4-4)^2+(4-4)^2}\) units
= \(\sqrt{(-8)^2+(0)^2}\) units
= √64 units = 8 units
Perimeter is (4 x 8) units= 32 units.
∴ The perimeter of the square is 32 units.
Coordinate Geometry Chapter 1 Distance Formulas Short Answer Type Questions
Example 1. Find the value of y if the distance of the point (-4, y) from origin is 5 units.
Solution: The distance of the point (-4, y) from the origin (0, 0) is \(\sqrt{(-4-0)^2+(y-0)^2}\) units
= \(\sqrt{16+y^2}\) units
According to question, \(\sqrt{16+y^2}\) = 5
⇒ 16 + y2 = 25
⇒ y2 = 25 – 16
⇒ y2 = 9
⇒ y = ±√9
⇒ y = ±3
∴ The value of y is ±3.
Example 2. Find the coordinates of a point on the y-axis which is equidistant from two points (2, 3) and (-1, 2).
Solution: Let the coordinates of a point of the y-axis is (0, k)
The distance between the points (2, 3) and (0, k) is \(\sqrt{(2-0)^2+(3-k)^2}\) units
The distance between the points (1, 2) and (0, k) is \(\sqrt{(-1-0)^2+(2-k)^2}\) units
According to question, \(\sqrt{(2-0)^2+(3-k)^2=\sqrt{(-1-0)^2+(2-k)^2}}\)
⇒ 4 + (3 – k)2 ⇒ 1 + (2 – k)2 [squaring both sides]
⇒ 4 + 9 – 6k + k2 = 1 + 4 – 4k + k2
⇒ -6k + k2 + 4k – k2 = 5 – 4 – 9
⇒ -2k = -8
⇒ k = 4.
∴ The coordinate of points on the Y-axis is (0, 4).
Example 3. Write the coordinates of two points on X-axis and Y-axis for which an isosceles right-angled triangle is formed with x-axis, y-axis, and straight line. joining two points.
Solution: As an isosceles right-angled triangle is formed with the x-axis, y-axis, and a straight line joining two points A and B;
∴ OA = OB
∴ The coordinates of two points on X-axis and Y-axis are (1, 0), (0, 1); (2, 0), (0, 2); (3, 0), (0, 3), etc.
Example 4. Write the coordinates of two points on opposite sides of x-axis which are equidistant from x-axis.
Solution: The coordinates of two points on opposite sides of x-axis are (2, 3), (2, -3); (-5, 6), (-5, -6) etc.
Example 5. Write the coordinates of two points on opposite sides of y-axis which are equidistant from y-axis.
Solution: The coordinates of two points on opposite sides of y-axis which are equidistant from y-axis are (2, 5), (-2, 3); (4, 6), (-4, 8) etc.
Example 6. Show that point (-2, -11) are equidistant from points (-3, 7) and (4, 6).
Solution: The distance between the points (-2, -11) and (- 3, 7) is \(\sqrt{\{-2-(-3)\}^2+(-11-7)^2} \text { units }\)
= \(\sqrt{(-2+3)^2+(-18)^2} \text { units }\)
= \(\sqrt{1+324} \text { units }=\sqrt{325} \text { units }\)
The distance between the points (-2, -11) and (4, 6) is \(\sqrt{(-2-4)^2+(-11-6)^2}\) units
= \(\sqrt{(-6)^2+(-17)^2} \text { units }\)
= \(\sqrt{36+289} \text { units }=\sqrt{325} \text { units }\)
∴ The point (-2,-11) is equidistant from points (-3, 7) and (4, 6).
Example 7. If the points A (2, -2), B (8, 4), C (5, 7), and D(-1, 1) are the vertices of a rectangle, then show that the lengths of diagonals AC and BD are equal.
Solution: The length of diagonal AC is \(\sqrt{(2-5)^2+(-2-7)^2} \text { units }\)
= \(\sqrt{(-3)^2+(-9)^2} \text { units }\)
= \(\sqrt{9+81} \text { units }=\sqrt{90} \text { units }\)
The length of diagonal BD is \(\sqrt{\{8-(-1)\}^2+(4-1)^2}\) units
= \(\sqrt{(9)^2+(3)^2}\) units
= \(\sqrt{81+9}\) units
= \(\sqrt{90}\) units
∴ AC = BD.
Example 8. Are the points A (0, 0), B (4, 3), and C (8, 6) co-linear? Verify the statement.
Solution: If A, B, and C are co-linear then AB + BC = AC
AB = \(\sqrt{(4-0)^2+(3-0)^2} \text { units }=\sqrt{25} \text { units }=5 \text { units }\)
BC = \(\sqrt{(8-4)^2+(6-3)^2} \text { units }\)
= \(\sqrt{16+9} \text { units }=\sqrt{25} \text { units }=5 \text { units. }\)
AC = \(\sqrt{(8-0)^2+(6-0)^2} \text { units }\)
= \(\sqrt{100} \text { units }=10 \text { units }\)
∴ AB+ BC= (5 + 5) units = 10 units
∴ AB + BC = AC
∴ The given points are co-linear.
Example 9. If the distance between the points (2, y) and (10,-9) is 10 units then find the value of y.
Solution: The distance between the points (2, y) and (10, -9) is \(\sqrt{(2-10)^2+\{y-(-9)\}^2} \text { units }\)
= \(\sqrt{64+(y+9)^2} \text { units }\)
According to question, \(\sqrt{64+(y+9)^2}\) = 10
⇒ 64 + y2 + 18y + 81 = 100
⇒ y2 + 18y + 45 = 0
⇒ y + 15y + 3y + 45 = 0
⇒ y (y + 15) + 3 (y + 15) = 0
⇒ (y + 15) (y + 3) = 0
either y + 15 = 0
⇒ y = – 15
or, y + 3 = 0
⇒ y = – 3
∴ The value of is – 3.
Example 10. Find the point on X-axis which are equidistant from points (3, 5) and (1, 3).
Solution: Let the point on X-axis is (h, 0)
The distance between (3, 5) and (h, 0) is \(\sqrt{(3-h)^2+(5-0)^2}\) units
The distance between (1, 3) and (h, 0) is \(\sqrt{(1-h)^2+(3-0)^2}\) units
According to question, \(\sqrt{(3-h)^2+(5-0)^2}=\sqrt{(1-h)^2+(3-0)^2}\)
⇒ 9 – 6h + h2 + 25 = 1 – 2h + h2 + 9
⇒ -6h + 2h + h2 – h2 = 1 + 9- 9 – 25
⇒ – 4h = – 24 ⇒ h = 6
∴ The required points is (6, 0).