COORDINATE GEOMETRY-Exercise 7.1
Maths - Exercise
Q1. Find the distance between the following pairs of points :
(i) (2, 3), (4, 1)
(ii) (– 5, 7), (– 1, 3)
(iii) (a, b), (– a, – b)
Solution:
i. A (2,3), B(4,1) \[\text{Distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\] $$\small\begin{aligned}AB&=\sqrt{\left( 4-2\right) ^{2}+\left( 1-3\right) ^{2}}\\ &=\sqrt{2^{2}+\left( -2\right) ^{2}}\\ &=\sqrt{4+4}\\ &=\sqrt{8}\\ &=2\sqrt{2}\end{aligned}$$ ii. Points X(-5,7), Y(-1, 3) $$\small\begin{aligned}XY&=\sqrt{\left( -1+5\right) ^{2}\times \left( 3-7\right) ^{2}}\\ &=\sqrt{4^{2}+\left( -4\right) ^{2}}\\ &=\sqrt{16+16}\\ &=4\sqrt{2}\end{aligned}$$ iii. A (a, b),B(-a,-b) $$\small\begin{aligned}AB&=\sqrt{\left[ a-\left( -a\right) \right] ^{2}+\left[ b-\left( -b\right) \right] ^{2}}\\ &=\sqrt{\left( 2a\right) ^{2}+\left( 2b\right) ^{2}}\\ &=\sqrt{4a^{2}+4b^{2}}\\ &=2\sqrt{a^{2}+b^{2}}\end{aligned}$$Q2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.
Solution:
Points (say) A(0,0) and B (36,15)
Distance between A and B (Say AB)
Distance between town A and town B is 39 kms
Q3. Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.
Solution:
3. Lets points be A (1,5), B (2,3), C(-2,-11) Let's find distance AB, BC, CA $$\small\begin{aligned}AB&=\sqrt{\left( 2-1\right) ^{2}+\left( 3-5\right) ^{2}}\\ &=\sqrt{1^{2}+\left( -2\right) ^{2}}\\ &=\sqrt{1^{2}+4}\\ &=\sqrt{5}\\\\ BC&=\sqrt{\left( -2-2\right) ^{2}+\left( -11-3\right) ^{2}}\\ &=\sqrt{\left( -4\right) ^{2}+\left( -14\right) ^{2}}\\ &=\sqrt{16+196}\\ &=\sqrt{212}\\\\ CA&=\sqrt{\left( -21\right) ^{2}+\left[ \left( 11-5\right) ^{2}\right] }\\ &=\sqrt{\left( -3\right) ^{2}+\left( -16\right) ^{2}}\\ &=\sqrt{9+256}\\ &=\sqrt{265}\end{aligned}$$For these points to be collinear, largest distance must equal to sum of the other two \(AB + BC = AC\) which is not true in this case
$$\sqrt{5}+\sqrt{212}\neq \sqrt{265}$$ therefore, points are not collinearQ4. Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.
Solution:
4- vertices of Δ are (say) A (5,-2) B (6,4), C (7,-2)To check-if these points forms Isosceles triangle lets calculate Distances and compare $$\small\begin{aligned}AB&=\sqrt{\left( 5-6\right) ^{2}+\left( -2-4\right) ^{2}}\\ &=\sqrt{\left( -1\right) ^{2}+\left( -6\right) ^{2}}\\ &=\sqrt{1^{2}+36}\\ &=\sqrt{37}\\\\ BC&=\sqrt{\left( 6-7\right) ^{2}+\left( 4-\left( -2\right) \right) ^{2}}\\ &=\sqrt{1^{2}+\left( 6\right) ^{2}}\\ &=\sqrt{37}\\\\ AC&=\sqrt{\left( 5-7\right) ^{2}+\left( -2-2\right) ^{2}}\\ &=\sqrt{\left( -2\right) ^{2}+\left( -4\right) ^{2}}\\ &=\sqrt{4+16}\\ &=\sqrt{20}\\\\ AB&=BC=\sqrt{37}\end{aligned}$$
\(AB=BC\), two sides of triangle is equal, therefore, given points forms an isosceles triangle.
Q5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.
Solution:
Points A (3,4), B(6,7),C(9,4) D (6,1)
Lets find distance between given points
All sides \(AB = BC = CD = AD\) are equal and diagonals (AD = BD) also equal, therefore, Champa is correct to say that ABCD is a square
Q6. Name the type of quadrilateral formed, if any, by the following
points, and give reasons for your answer:
(i) (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0)
(ii) (–3, 5), (3, 1), (0, 3), (–1, – 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Solution:
To find type of Quadrilaterals
i. (-1,-2), (1,0), (-1,2) (-3,0)Lets find distances between the given points (say)
A (-1,-2), B (1,0), C (-1,2), D (-3,0) $$\small\begin{aligned}AB&=\sqrt{\left( 1-1\right) ^{2}+\left( -2-0\right) ^{2}}\\ &=\sqrt{\left( -2\right) ^{2}+\left( -2\right) ^{2}}\\ &=\sqrt{4+4}\\ &=2\sqrt{2}\\\\ BC&=\sqrt{\left[ 1-\left( -1\right) \right] ^{2}+\left( 0-2\right) ^{2}}\\ &=\sqrt{2^{2}+2^{2}}\\ &=\sqrt{4+4}\\ &=2\sqrt{2}\\\\ CD&=\sqrt{\left[ -1-\left( -3\right) \right] ^{2}+\left( 2-0\right) ^{2}}\\ &=\sqrt{\left( 2\right) ^{2}+2^{2}}\\ &=\sqrt{4+4}\\ &=2\sqrt{2}\\\\ AD&=\sqrt{\left[ -1-\left( -3\right) \right] ^{2}+\left( 2-0\right) ^{2}}\\ &=\sqrt{\left( 2\right) ^{2}+2^{2}}\\ &=\sqrt{4+4}\\ &=2\sqrt{2}\\\\ AB=BC&=CD=AD\end{aligned}$$ Distance between given points is equal, therefore, points form a square
ii. let's (say) points
A (-3, 5), B (3,1), C (0,3), D (-1,-4)
Let's find the distance between points
iii. Lets (say) Points are
A (4,5), B(7,6), C (4, 3), D (1,2)
Lets find distances between given points
Q7. Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).
Solution:
Let the required point on the x-axis be \((x, 0)\). Since it is equidistant from the points \((2, -5)\) and \((-2, 9)\), the distance from \((x, 0)\) to \((2, -5)\) will be equal to the distance from \((x, 0)\) to \((-2, 9)\).
$$\scriptsize\begin{aligned}\sqrt{\left( x-2\right) ^{2}+\left[ 0-\left( -5\right) \right] ^{2}}&=\sqrt{\left[ x-\left( -2\right) \right] ^{2}+\left( 0-9\right) ^{2}}\end{aligned}$$ Squaring both side $$\scriptsize\begin{aligned}\Rightarrow \left( x-2\right) ^{2}+\left( 0+5\right) ^{2}&=\left( x+2\right) ^{2}+\left( 0-9\right) ^{2}\\ \Rightarrow \left( x-2\right) ^{2}-\left( x+2\right) ^{2}&=\left( 0-9\right) ^{2}-\left( 0+5\right) ^{2}\\ x^{2}+4-4x-\left( x^{2}+4x+4\right) &=81-25\\ x^{2}+4-4x-x^{2}-4x-4&=+56\\ -8x&=+56\\ 8x&=-56\\ x&=\dfrac{-56}{8}\\ &=-7\\ \left( x,y\right) &=\left( -7,0\right) \end{aligned}$$Therefore, the required point on the x-axis is \(\,(-7, 0)\). Hence, the point \((-7, 0)\) is equidistant from \((2, -5)\) and \((-2, 9)\).
Q8. Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.
Solution:
Consider the points \(P(2, -3)\) and \(Q(10, y)\). It is given that the distance between these two points is 10 units. Using the distance formula, we write
\[\small \begin{aligned} PQ &= \sqrt{(2 - 10)^2 + (-3 - y)^2} \\ 10 &= \sqrt{(-8)^2 + (-3 - y)^2} \end{aligned} \]
Squaring both sides to remove the square root, we get
\[\small \begin{aligned} 10^2 &= (-8)^2 + (-3 - y)^2 \\ 100 &= 64 + (-3 - y)^2 \\ 100 - 64 &= (-3 - y)^2 \\ 36 &= (-3 - y)^2 \end{aligned} \]
Taking square roots on both sides, we obtain
\[ \begin{aligned} -3 - y &= \pm 6 \end{aligned} \]
Solving these two cases separately,
\[\small \begin{aligned} -3 - y &= 6 \\ \Rightarrow -y &= 6 + 3 \\ \Rightarrow y &= -9 \\\\ -3 - y &= -6 \\ \Rightarrow -y &= -6 + 3 \\ \Rightarrow y &= 3 \end{aligned} \]
Therefore, the required values of \(y\) for which the distance between \(P(2, -3)\) and \(Q(10, y)\) is 10 units are \(y = 3\) and \(y = -9\).
Q9. If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR.
Solution:
Q(0, 1) is equidistant from P(5,-3) and R (x, 6) therefore
$$\scriptsize\begin{aligned}\sqrt{\left( 0-5\right) ^{2}+\left[ 1-\left( -3\right) \right] ^{2}}&=\sqrt{\left( 0-x\right) ^{2}+\left( 1-6\right) ^{2}}\\ \sqrt{5^{2}+4^{2}}&=\sqrt{x^{2}+5^{2}}\\ \sqrt{25+16}&=\sqrt{x^{2}+25}\\ 41&=x^{2}+{25}\\ x^{2}&=41-25\\ x^{2}&=16\\ x&=\pm 4\end{aligned}$$for \(x=\pm4\)
$$\small\begin{aligned} QR&=\sqrt{\left( 0-4\right) +\left( 1-6\right) +?}\\ &=\sqrt{4^{2}+5^{2}}\\ &=\sqrt{16+25}\\ &=\sqrt{41}\end{aligned}$$for \(x=-4\)
$$\small\begin{aligned}PR&=\sqrt{\left[ 5-\left( -4\right) \right] ^{2}+\left( -3-6\right) ^{2}}\\ &=\sqrt{\left( -9\right) ^{2}+\left( -9\right) ^{2}}\\ &=\sqrt{81+81}\\ &=9\times \sqrt{2}\end{aligned}$$for \(x=+4\)
$$\small\begin{aligned}PR&=\sqrt{\left( 5-4\right) ^{2}+\left( -3-6\right) ^{2} }\\ &=\sqrt{1^{2}+q^{2}}\\ &=\sqrt[2] {1+81} \\ &=\sqrt{82}\end{aligned}$$Q10.Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4).
Solution:
Let \((x, y)\) be a point which is equidistant from \((3, 6)\) and \((-3, 4)\). This means that the distance from \((x, y)\) to \((3, 6)\) is equal to the distance from \((x, y)\) to \((-3, 4)\).
$$\scriptsize\begin{aligned}\sqrt{\left( x-3\right) ^{2}+\left( y-6\right) ^{2}}&=\sqrt{\left[ x-\left( -3\right) \right] ^{2}+\left( y-4\right) ^{2}}\end{aligned}$$ Squaring both side $$\scriptsize\begin{aligned}\left( x-3\right) ^{2}+\left( y-6\right) ^{2}&=\left( x+3\right) ^{2}+\left( y-4\right) ^{2}\\ \left( x-3\right) ^{2}-\left( x+3\right) ^{2}&=\left( y-4\right) ^{2}-\left( y-6\right) ^{2}\\ x^{2}-6x+9-\left( x^{2}+6x+9\right) &=y^{2}-8y+16-\left( y^{2}-12y+36\right) \\ x^{2}-6x+9-x^{2}-6x-9&=y^{2}-8y+16-y^{2}+12y-36\\ -12x&=4y-20\\ \Rightarrow -3x&=y-5\\ \Rightarrow -3x-y&=-5\\ 3x+y&=+5\\ 3x+y-5&=0\end{aligned}$$Hence, the required relation between \(x\) and \(y\) is \(3x + y - 5 = 0\).
