Ganit Prakash 2023 Co-ordinate Geometry Distance Formula Solution Class 9 Maths. Chapter 4 – Co-ordinate Geometry Distance Formula is provided here with simple step-by-step explanations. These solutions for Co-ordinate Geometry Distance Formula are extremely popular among class 9 students. The Co-ordinate Geometry Distance Formula solution is handy for quickly completing your homework and preparing for exams.
| Book Name | : Ganit Prakash |
| Subject | : Mathematics (Maths) |
| Class | : 9 (Madhyamik/WB) |
| Publisher | : Prof. Nabanita Chatterjee |
| Chapter Name | : Co-ordinate Geometry Distance Formula (4th Chapter) |
Let us do ourselves – 4
Question 1
I measure the length of the straight line joining the following pairs of points –
(i) (18,0); (8,0)
(ii) (0,5); (0,4)
(iii) (-7,0); (-2,0)
(iv) (0,-10); (0,-3)
(v) (6,0); (-2,0)
(vi) (0,-5); (0,9)
(vii) (5,0); (0,10)
(viii) (3,0); (0,4)
(ix) (4,3); (2,1)
(x) (-2,-2); (2,2)
Solution:
Here, we use the formula to measure the length of the straight line
= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\
(i) Distance = 18 – 8 =10 units (Ans)
(ii) Distance = 15 – 4 =11 (Ans)
(iii) Distance = -2-(-7) = -2 + 7 =5 units (Ans.)
(iv) Distance = -3 – (-10) = -3 + 10 = 7 units (Ans.)
(v) Distance = 6 – (-2) = 6 + 2 = 8 unit (Ans)
(vi) Distance = 9 -(-5) = 9 + 5 = 14 units (Ans.)
\text{ (vii) }\text { Distance } =\sqrt{(5-0)^2+(0-10)^2} \\
= \sqrt{25+100} \\
= \sqrt{125} \text {units} (Ans.)
\text { (viii) Distance }=\sqrt{(3-0)^2+(0-4)^2} \\
= \sqrt{(3)^2+(-4)^2} \\
= \sqrt{9+16} \\
= \sqrt{25}=5 \text { units} (Ans.)
\text {(ix) Distance } =\sqrt{(4-2)^2+(3-1)^2} \\
= \sqrt{(2)^2+(2)^2} \\
= \sqrt{4+4} \\
= \sqrt{8} \text { units} (Ans.)
\text{(x)\text Distance }=\sqrt{(-2-2)^2+(-2-2)^2} \\
= \sqrt{(-4)^2+(-4)^2} \\
= \sqrt{16+16} \\
= \sqrt{32} \text { units} (Ans.)
Let us work – 4
Question 1
Let us calculate the distances of the following points from the origin:
(i) (7, -24),
(ii) (3, -4),
(iii) (a + b, a – b)
Solution:
\text { Distance }=\sqrt{(7-0)^2+(-24-0)^2} \\
= \sqrt{(7)^2+(-24)^2} \\
= \sqrt{49+576} \\
= \sqrt{625}=25 \text { units} (Ans.)
\text { (ii) Distance }=\sqrt{(3-0)^2+(-4-0)^2} \\
= \sqrt{(3)^2+(-4)^2} \\
= \sqrt{9+16} \\
= \sqrt{25}=5 \text { units} (Ans.)
\text { (iii) }\text { Distance } =\sqrt{(\mathrm{a}+\mathrm{b}-0)^2+(\mathrm{a}-\mathrm{b}-0)^2} \\
= \sqrt{(\mathrm{a}+\mathrm{b})^2+(\mathrm{a}-\mathrm{b})^2} \\
= \sqrt{2\left(\mathrm{a}^2+\mathrm{b}^2\right)} \text { units} (Ans.)
Question 2
Let us calculate the distances between the two points given below:
(i) (5,7) and (8,3)
(ii) (7,0) and (2,-12)
(iii) (-\frac{3}{2}, 0) and (0,-2)
(iv) (3,6) and (-2,-6)
(v) (1,-3) and (8,3)
(vi) (5,7) and (8,3)
Solution:
\text {(i) Distance }=\sqrt{(5-8)^2+(7-3)^2} \\
= \sqrt{(-3)^2+(4)^2} \\
= \sqrt{9+16} \\
= \sqrt{25}=5 \text { units} (Ans.)
\text { (ii) Distance }=\sqrt{(7-2)^2+\{0-(-12)\}^2} \\
= \sqrt{(5)^2+(0+12)^2} \\
= \sqrt{25+144} \\
= \sqrt{169}=13 \text { units} (Ans.)
\text { (iii) Distance }=\sqrt{\left(-\frac{3}{2}-0\right)^2+\{0-(-2)\}^2} \\
= \sqrt{\left(-\frac{3}{2}\right)^2+(2)^2} \\
= \sqrt{\frac{9}{4}+4} \\
= \sqrt{\frac{9+16}{4}} \\
= \sqrt{\frac{25}{4}} \\
= \sqrt{5}\text{ units} (Ans.)
\text { (iv) Distance } =\sqrt{\{3-(-2)\}^2+\{6-(-6)\}^2} \\
= \sqrt{(3+2)^2+(6+6)^2} \\
= \sqrt{(5)^2+(12)^2} \\
= \sqrt{25+144}=13 \text {units} \\
= \sqrt{169}=13 \text { units} (Ans.)
\text { (v) Distance } =\sqrt{(1-8)^2+(-3-3)^2} \\
= \sqrt{(-7)^2+(-6)^2} \\
= \sqrt{49+36} \\
= \sqrt{85} \text { units (Ans.) } \\
\text{ (vi) Distance } =\sqrt{(5-8)^2+(7-3)^2} \\
= \sqrt{(-3)^2+(4)^2} \\
= \sqrt{9+16} \\
= \sqrt{25}=5 \text { units}(Ans.)
Question 3
Let us prove that point (-2, -11) is equidistant from the two points (-3, 7) and (4, 6)
Solution:
Let the coordinates of the points A, B & C be (-2, -11),(-3, 7) and (4, 6) respectively.
\mathrm{AB} = \sqrt{\left.\{(-2)-(-3)\}^2+\{(-11)-7)\right\}^2} \\
= \sqrt{(-2+3)^2+(-11-7)^2} \\
= \sqrt{(1)^2+(-18)^2} \\
= \sqrt{1+324}=\sqrt{325} \text { units (Ans.) } \\
\mathrm{AC} =\sqrt{\{4-(-2)\}^2+(-11-6)^2} \\
= \sqrt{(4+2)^2+(-17)^2} \\
= \sqrt{(6)^2+(-17)^2}\\
= \sqrt{36+289}=\sqrt{325} \text {(units Ans.)}\\
\therefore \quad \mathrm{AB} = \mathrm{AC}\\
Hence, point A is equidistant from points B & C
Question 4
Let us show that the points (7,9),(3,-7) and (-3,3) are the vertices of a right-angled triangle by calculation.
Solution:
\text { The length of } \mathrm{AB}=\sqrt{(7-3)^{2}+\{9-(-7)\}^{2}}\\
= \sqrt{(4)^{2}+(9+7)^{2}}\\
= \sqrt{16+(16)^{2}}\\
= \sqrt{16+256}\\
= \sqrt{272} \text { units (Ans.) }\\
\therefore \quad \mathrm{AB}^{2}=272 \text { Units. }\\
\text { The length of } \mathrm{BC}=\sqrt{\{3-(-3)\}^{2}+(-7-3)^{2}}\\
= \sqrt{(3+3)^{2}+(-10)^{2}} \\
= \sqrt{(6)^{2}+(100)^{2}} \\
= \sqrt{36+100} \\
= \sqrt{136} \text { units (Ans.) } \\
\therefore \quad \mathrm{BC}^{2}=136 \text { Units }^{2}
\text { The length of } \mathrm{CA}=\sqrt{(-3-7)^{2}+(3-9)^{2}}\\
= \sqrt{(-10)^{2}+(-6)^{2}} \\
= \sqrt{100+36} \\
= \sqrt{136} \text { units (Ans.) }\\
\therefore \mathrm{CA}^{2}=136 \text { Units }^2\\
\therefore \quad \mathrm{AB}^{2}=\mathrm{BC}^{2}=\mathrm{CA}^{2}=136 \text { Units }^{2}\\
This shows that A, B & C i.e. the given points are the vertices of a right-angle triangle.
Question 5
Let us prove that in both of the following cases, the three points are the vertices of an isosceles triangle: (i) (1, 4), (4, 1) and (8, 8) (ii) (-2, -2), (2, 2) and (4, -4)
Solution:
Let A (1, 4), B (4, 1) and C (8, 8)
\mathrm{AB} =\sqrt{(1-4)^{2}+(4-1)^{2}} \\
= \sqrt{(-3)^{2}+(3)^{2}} \\
= \sqrt{9+9} \\
= \sqrt{18} \text { units }
\mathrm{BC} =\sqrt{(8-4)^{2}+(8-1)^{2}} \\
= \sqrt{(4)^{2}+(7)^{2}} \\
= \sqrt{16+49} \\
= \sqrt{65} \text { units } \\
\mathrm{CA} =\sqrt{(8-1)^{2}+(8-4)^{2}} \\
= \sqrt{(7)^{2}+(4)^{2}} \\
= \sqrt{49+16} \\
= \sqrt{65} \text { units } \\
\therefore \quad \mathrm{BC}=\mathrm{CA} ≠ \mathrm{AB} \\
This shows that A; B and C i.e. the given points are the vertices of an isosceles triangle.
(ii) Let A (-2, -2), B (2, 2) and C (4, -4)
\mathrm{AB} =\sqrt{(-2-2)^{2}+(-2-2)^{2}} \\
= \sqrt{(-4)^{2}+(-4)^{2}} \\
= \sqrt{16+16} \\
= \sqrt{32} \text { units } \\
\mathrm{BC} =\sqrt{(4-2)^{2}+(-4-2)^{2}} \\
= \sqrt{(2)^{2}+(-6)^{2}}\\
= \sqrt{4+36} \\
= \sqrt{40} \text { units } \\
\mathrm{CA} =\sqrt{\{4-(-2)\}^{2}+(-4-(-2)\}^{2}} \\
= \sqrt{(4+2)^{2}+(-4+2)^{2}} \\
= \sqrt{(6)^{2}+(-2)^{2}} \\
= \sqrt{36+4} \\
= \sqrt{40} \text { units } \\
\therefore \quad \mathrm{CA} = \mathrm{BC} ≠ \mathrm{AB}
This shows that A, B and C i.e. the given points are the vertices of an isosceles triangle.
Question 6
Let us prove that the three points A(3,3), B(8,-2) and C(-2,-2) are the vertices of a right-angled triangle. Let us calculate the length of the hypotenuse of (\triangle ABC)
Solution:
\mathrm{AB}=\sqrt{(8-3)^{2}+(-2-3)^{2}} \\
= \sqrt{(5)^{2}+(-5)^{2}} \\
= \sqrt{25+25} \\
= \sqrt{50} \text { units } \\
\therefore \quad A B^{2}=50 \text { Units } \\
\mathrm{BC}=\sqrt{\{8-(-2)\}^{2}+\{-2-(-2)\}^{2}} \\
= \sqrt{(8+2)^{2}+(-2+2)^{2}} \\
= \sqrt{(10)^{2}+0} \\
= \sqrt{100 \text { units }} \\
\therefore \quad BC=100 \text { Units } \\
\mathrm{CA}=\sqrt{(-2-3)^{2}+(-2-3)^{2}} \\
= \sqrt{(-5)^{2}+(-5)^{2}}
= \sqrt{25+25} \\
= \sqrt{50} \text { units } \\
\therefore \quad \mathrm{CA}^{2}=50 \text { Units } \\
\therefore \quad \mathrm{BC}^{2}=\mathrm{AB}^{2}+\mathrm{CA}^{2}
This shows that the three points A (3, 3), B (8, -2) and C (-2, -2) are the vertices of a right-angled triangle.
Length of the hypotenuse of\triangle ABC = BC =\sqrt{100} \text{units}\\
= 10 (Ans)
Question 7
Let us show by calculation that the points (2, 1), (0, 0), (-1, 2) and (1, 3) are the angular points of a square.
Solution:
Let A (2, 1), B (0, 0), C (-1, 2) and D (1, 3)
\mathrm{AB} =\sqrt{(2-0)^{2}+(1-0)^{2}} \\
= \sqrt{(2)^{2}+(1)^{2}} \\
= \sqrt{4+1} \\
= \sqrt{5} \text { units } \\
\mathrm{BC} =\sqrt{(-1-0)^{2}+(2-0)^{2}} \\
= \sqrt{(-1)^{2}+(2)^{2}} \\
= \sqrt{1+4} \\
= \sqrt{5} \text { units }\\
\mathrm{CD} =\sqrt{(-1-1)^{2}+(2-3)^{2}} \\
= \sqrt{(-2)^{2}+(-1)^{2}} \\
= \sqrt{4+1} \\
= \sqrt{5} \text { units }\\
\mathrm{AD} =\sqrt{(2-1)^{2}+(1-3)^{2}} \\
= \sqrt{(1)^{2}+(-2)^{2}}\\
= \sqrt{1+4} \\
= \sqrt{5} \text { units } \\
\therefore \quad \mathrm{AB} =\mathrm{BC} =\mathrm {CD} = \mathrm{AD}
This shows that the four points A, B, C and D i.e. the given points are the vertices of a square.
Question 8
Let us calculate and see that for what value of 7, the distance of the two points (2, y) and (10, -2) will be 10.
Solution:
By the question,
\sqrt{(2-10)^{2}+(-9-y)^{2}}=10 \\
or, \sqrt{(-8)^{2}+(-9-y)^{2}}=10
Squaring both sides,
or, (- 8)2 + (- 9 - y)2 = 102
or, 64 + (- 9 - y)2 = 100
or, (- 9 - y)2 = 100 - 64
or, (- 9 - y)2 = 36
or, (- 9 - y)2 = 62
or, - 9 - y = ±6
Taking positive,
or, - 9 - y = 6
or, - y = - 6 + 9
or, - y = 15y = - 15
Taking negative
or, - 9 - y = 6
or, y = 6 + 9
or, y = 3
∴ y = - 3
∴ y = - 3, - 15 (Ans.)
Question 9
Let us find the point on the x-axis which is equidistant from the two points (3, 5) and (1, 3).
Solution:
Let the points on the x-axis be (x, 0) which is equidistant from the two points (3, 4) & (1, 3)
\therefore \sqrt{(x-3)^{2}+(5-0)^{2}}=\sqrt{(x-1)^{2}+(3-0)^{2}}\\
or, (x-3)^{2}+(5)^{2}=(x-1)^{2}+(3)^{2}
or, (x)^2-2.x.3 + (3)^2 + (5)^2 = (x)^2 - 2.x.1 + (1)^2 + (3)^2
or, x^2-6x +9 +25= x^2 - 2x + 1 + 9
or, -6x+25= -2x + 1
or, -6x+2x= 1-25
or, -4x=-24
x = {\frac{-24}{-4}= 6}
or, x = 6
∴ The required point is (6.0) (Ans.)
Question 10
Let us write by calculating whether the three points 0 (0, 0), A (4, 3) and B (8, 6) are collinear.
Solution:
\mathrm{OA} =\sqrt{(4-0)^{2}+(3-0)^{2}}\\
= \sqrt{4^{2}+3^{2}}\\
= \sqrt{16+9}\\
= \sqrt{25}=5\\
\mathrm{AB} =\sqrt{(8-4)^{2}+(6-3)^{2}}\\
= \sqrt{4^{2}+3^{2}}\\
= \sqrt{16+9}\\
= \sqrt{25}=5\\
\mathrm{OB} =\sqrt{(8-0)^{2}+(6-0)^{2}} \\
= \sqrt{8^{2}+6^{2}} \\
= \sqrt{64+36} \\
= \sqrt{100=10} \text { units }\\
\therefore \quad \mathrm{OA}+\mathrm{AB}=\mathrm{OB}
So, the points O (0, 0), A (4, 3) and B (8, 6) are collinear. (Proved)
Question 11
Let us show the three points (2, 2). (-2, -2) and (-2√3, 2√3) are the vertices of an equilateral triangle.
Solution:
\mathrm {AB} =\sqrt{(-2-2)^{2}+(-2-2)^{2}} \\
= \sqrt{(-4)^{2}+(-4)^{2}}\\
= \sqrt{16+16}\\
= \sqrt{32} \text { units }\\
\mathrm{BC} =\sqrt{(-2+2 \sqrt{3})^{2}+(-2-2 \sqrt{3})^{2}}
= \sqrt{(2 \sqrt{3}-2)^{2}+(2 \sqrt{3}+2)^{2}}\\
= \sqrt{2\left\{(2 \sqrt{ } 3)^{2}+(2)^{2}\right\}}\\
= \sqrt{2(12+4)}\\
= \sqrt{2 \times 16}\\
= \sqrt{32} \text { units } \\
\mathrm{CA} =\sqrt{(2+2 \sqrt{3})^{2}+(2-2 \sqrt{3})^{2}} \\
= \sqrt{2(2)^{2}+(2\sqrt{3})^{2}}
= \sqrt{2 .(4+12)}=\sqrt{2 \times 16}\\
= \sqrt{32 \text { units }} \\
\therefore \mathrm{AB} =\mathrm{BC}=\mathrm{CA}
This shows that the points A, B and C i.e. the given points are the vertices of an equilateral triangle.
Question 12
Let us show that the points (-7,2), (19,18), (15,-6) and(-11,-22) form a parallelogram when they are joined orderly.
Solution:
Let A (-7,2), B (19,18), C (15,-6) & D (-11,-22)
\mathrm{AB} =\sqrt{(19+7)^{2}+(18-2)^{2}}\\
= \sqrt{(26)^{2}+(16)^{2}} \\
= \sqrt{676+256}=\sqrt{932 \text { units }}\\
\mathrm{BC} =\sqrt{(19-15)^{2}+(18+6)^{2}} \\
= \sqrt{(4)^{2}+(24)^{2}} \\
= \sqrt{16+476}=\sqrt{492 \text { units }} \\
\mathrm{CD} =\sqrt{(15+11)^{2}+(-6+22)^{2}} \\
= \sqrt{(26)^{2}+(6)^{2}} \\
= \sqrt{676+6}=\sqrt{932 \text { units }} \\
\mathrm{AD} =\sqrt{(-7+11)^{2}+(2+22)^{2}}
= \sqrt{(4)^{2}+(24)^{2}} \\
= \sqrt{16+426}=\sqrt{492} \text { units } \\
\therefore \quad \mathrm{AB} = \mathrm{CD} \text { and } \mathrm{BC}=\mathrm{AD}
This shows that the points A, B, C and D the given points are the vertices of a parallelogram (Proved).
Question 13
Let us show that the points (2,-2), (8,4), (5,7) and (-1,1) are the vertices of a rectangle.
Solution:
Let A (2,-2), B (8,4), C (5,7) & D (-1,1)
\mathrm{AB}=\sqrt{(2-8)^{2}+(-2-4)^{2}}\\
= \sqrt{(-6)^{2}+(-6)^{2}}\\
= \sqrt{36+36}=\sqrt{72 \text { units }}\\
\mathrm{BC}=\sqrt{(8-5)^{2}+(4-7)^{2}}\\
= \sqrt{(3)^{2}+(-3)^{2}} \\
= \sqrt{9+9}=\sqrt{18 \text { units }} \\
\mathrm{CD}=\sqrt{(-1-5)^{2}+(1-7)^{2}}\\
= \sqrt{(-6)^{2}+(-6)^{2}}\\
= \sqrt{36+36}=\sqrt{72} \text { units }\\
\mathrm{DA}=\sqrt{(-1-2)^{2}+(1+2)^{2}} \\
= \sqrt{(-3)^{2}+(3)^{2}} \\
= \sqrt{9+9}=\sqrt{18 \text { units }}\\
\therefore \quad \mathrm{AB}=\mathrm{CD} \text { and } \mathrm{BC}=\mathrm{DA}\\
This shows that the points A, B, C and D i.e. the given points are the vertices of a rectangle (Proved).
Question 14
Let us show that the points (2,5), (5,9), (9,12) and (6,8) form a rhombus when they are joined orderly.
Solution:
Let A (2,5), B (5,9), C (9,12) & D (6,8)
\mathrm{AB} =\sqrt{(5-2)^{2}+(9-5)^{2}}\\
= \sqrt{(3)^{2}+(4)^{2}}\\
= \sqrt{9+16}=\sqrt{25}\\
= 5 \text { units }\\
\mathrm{BC} =\sqrt{(9-5)^{2}+(12-9)^{2}} \\
= \sqrt{(4)^{2}+(3)^{2}} \\
= \sqrt{16+9}=\sqrt{25} \\
= 5 \text { units } \\
\mathrm{CD} =\sqrt{(9-6)^{2}+(12-8)^{2}}\\
= \sqrt{(3)^{2}+(4)^{2}}\\
= \sqrt{9+16}=\sqrt{25}\\
= 5 \text { units }\\
\mathrm{DA} =\sqrt{(6-2)^{2}+(8-5)^{2}} \\
= \sqrt{(4)^{2}+(3)^{2}} \\
= \sqrt{16+9}=\sqrt{25} \\
= 5 \text { units }\\
\therefore \mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}
This shows that the points A, B, C and D i.e. the given points are the vertices of a Rhombus (Proved).
Multiple Choice Questions:
Question 15 (i)
The distance between the two points (a+b, c-d) and (a-b, c+d) is
(a)2 \sqrt{a^{2}+c^{2}}
(b)2 \sqrt{b^{2}+d^{2}}
(c)\sqrt{a^{2}+c^{2}}
(d)\sqrt{b^{2}+d^{2}}
Solution:
(i)= \sqrt{\{(a+b)-(a-b)\}^{2}\{(c-d)-(c+d)\}^{2}} \\
= \sqrt{(a+b-a+b)^{2}+(c-d-c-d)^{2}} \\
= \sqrt{(2 a)^{2}+(-2 d)^{2}} \\
= \sqrt{4} \cdot \sqrt{b^{2}+d^{2}} \\
= 2 \sqrt{b^{2}+d^{2}} \text { units. }
Question 15 (ii)
If the distance between the two points (x,-7) and (3,-3) is 5 units, then the values of x are
(a) 0 or 6 (b) 2 or 3 (c) 5 or 1 (d) -6 or 0
Solution:
\quad \therefore Distance =5 units.\\
or, \sqrt{(x-3)^{2}+(-7+3)^{2}}=5\\
or, \sqrt{(x-3)^{2}+(-4)^{2}}=5[\mathrm{\because Squaring both sides}] \\
or, (x-3)^{2}+16=25\\
or, (x-3)^{2}=25-16=9\\
or, (x-3)^{2}=3^{2}\\
\therefore x-3= \pm 3\\
Taking positive,
x - 3 = 3
or, x = 3 + 3
or, x = 6
∴ x = 6
Taking negative
x - 3 = -3
or, x = -3 + 3
or, x = 0
∴ x = 0
∴ x = 0 or 6 (Ans.)
Question 15 (iii)
If the distance of the point (x, 4) from the origin is 5 units, then the values of x are
\mathrm{(a) \pm 4(b) \pm 5(c) \pm 3(d) none of these}
Solution:
\quad \therefore Distance =5 units.\\
or, \sqrt{(x-0)^{2}+(4-0)^{2}}=5\\
or, x^{2}+4^{2}=5 [\mathrm{\because Squaring both sides}]\\
or, x^{2}+16=25\\
or, x^{2}=25-16=9\\
or, \mathrm{x}^{2}=9 \quad \therefore x= \pm \sqrt{9}\\
\therefore x= \pm 3 (Ans.)\\
Question 15 (iv)
The triangle formed by the points (3,0), (-3,0) and (0,3)
(a) equilateral (b) isosceles (c) scalene (d) isosceles right-angled
Solution:
\mathrm{Let} \mathrm{A}(3,0), \mathrm{B}(-3,0) and \mathrm{C}(0,3)\\
\mathrm{AB} =\sqrt{(-3-3)^{2}+(0-0)^{2}}\\
= \sqrt{(-6)^{2}+(0)^{2}} \\
= \sqrt{36}=6 \text { units } \\
\mathrm{BC} =\sqrt{(-3-0)^{2}+(0-3)^{2}}\\
= \sqrt{(-3)^{2}+(-3)^{2}} \\
= \sqrt{9+9}=\sqrt{18 \text { units }} \\
\mathrm{AC} =\sqrt{(0-3)^{2}+(3-0)^{2}}=\sqrt{9+9} \\
= \sqrt{18} \text { units } \\
\therefore \mathrm{A}\mathrm{C}=\mathrm{B} \mathrm{C} ≠ \mathrm{AB}\\
This shows that A, B and C are the vertices of an isosceles triangle.
Question 15 (v)
The coordinates of the centre of the circle are (0, 0) and the coordinates of a point on the circumference are (3, 4), the length of the radius of the circle is
(a) 5 units (b) 4 units (c) 3 units (d) none of these
Solution:
= \sqrt{(3-0)^{2}+(4-0)^{2}} \\
= \sqrt{3^{2}+4^{2}} \\
= \sqrt{9+16}
= \sqrt{25}
= 5 \text { units }
Short answer type questions:
Question 16 (i)
Let us write the value of y if the distance of the point (-4, y) from origin is 5 units.
Solution:
Distance = 5 units.
or, \sqrt{(-4-0)^{2}+(y-0)^{2}}=5\\
or, (-4)^{2}+y^{2}=5^{2}[\mathrm{\because Squaring both sides}] \\
or, 16+y^{2}=25\\
or, y^{2}=25-16\\
or, y^{2}=9
or, y= \pm \sqrt{ } 9 \quad \therefore y= \pm 3 (Ans).
Question 16 (ii)
Let us write the coordinates of a point on the y-axis which is equidistant from two points (2, 3) and (-1, 2).
Solution:
Let, \mathrm{A}(2,3) and \mathrm{B}(-1,2)\\
Let a point on y-axis be C(0, y)\\
\therefore \mathrm{AC}=\mathrm{BC}\\
or, \mathrm{AC}^{2}=\mathrm{BC}^{2}\\
or, (2-0)^{2}+(3-y)^{2}=(-1-0)^{2}+(2-y)^{2}\\
or, (2)^{2}+(3)^{2}-2 \cdot 3 \cdot y+(y)^{2}=(-1)^{2}+(2)^{2}-2 \cdot 2 \cdot y+(y)^{2}\\
or, 4+9-6 y+y^{2}=1+4-4 y+y^{2}\\
or, 13-6 y=5-4 y\\
or, -6 y+4 y=5-13\\
or, -2 y=-8 \quad\\
or, y=\frac{-8}{-2}\\
\therefore \quad \mathrm{y}=4\\
The point (0, 4) on the y-axis is equidistant from two points (2, 3) and (-1, 2)
Question 16 (iii)
Let us write the coordinates of two points on x- the axis and y-axis for which an isosceles right-angled triangle is formed with the x-axis, y-axis and the straight line joining the two points.
Solution:
We can choose are y two points on the x-axis and y-anis as (2, 0) and (0, 2) & (3, 0) and (0,3) when these points are joined they formed an isosceles right-angled triangle with the co-ordinate.
Question 16 (iv)
Let us write the coordinates of two points on opposite sides of the x-axis which are equidistant from the x-axis.
Solution:
(i) (5,0) and (-5,0)
(ii) (10,0) and (-10,0)
Question 16 (v)
Let us write the Co-ordinates of two points on opposite sides of the y-axis which are equidistant from the y-axis
Solution:
(i) (0, 5) and (0, -5)
(ii) (0, 10) and (0, -10)
