RS Aggarwal Solutions Class 10 Ex 16A

Q1) Find the co-ordinates of the point equidistant from the three given points A(5, 3), B(5, -5) and C(1, -5)

Solution

Let the required point be P(x,y). Then AP = BP = CP

That is, (AP)2 = (BP)2 = (CP)2

This means, (AP)2 = (BP)2

\(\Rightarrow\)\(\left (x-5\right )^{2}+\left (y-3\right )^{2}=\left (x-5\right )^{2}+\left (y+5\right )^{2}\)

\(\Rightarrow\)x2 – 10x +25 + y2 – 6y + 9 = x2 – 10x + 25 + y2 + 10y + 25

\(\Rightarrow\) x2 + y2 -10x -6y +34 = x2 + y2 – 10x + 10y + 50

\(\Rightarrow\) x2 + y2 -10x -6y – x2 – y2 + 10x – 10y = 50-34

\(\Rightarrow\) -16y = 16

\(\Rightarrow\) y = \(\frac{-16}{16}\) = -1

And (BP)2 = (CP)2

\(\Rightarrow\)\(\left (x-5\right )^{2}+\left (y+5\right )^{2}=\left (x-1\right )^{2}+\left (y+5\right )^{2}\)

\(\Rightarrow\)x2 – 10x +25 + y2 +10y + 25 = x2 – 2x + 1 + y2 + 10y + 25

\(\Rightarrow\) x2 + y2 -10x +10y +50 = x2 + y2 – 2x + 10y + 26

\(\Rightarrow\) x2 + y2 -10x +10y – x2 – y2 + 2x – 10y = 26-50

\(\Rightarrow\) -8x = -24

\(\Rightarrow\) y = \(\frac{-24}{-8}\) = 3

Hence the required point is (3,-1)

Q2) If the points A(4, 3) and B(x, 5)lie on a circle with the center O(2, 3). Find the value of x. [ Hint : (OA)2 = (OB)2 ]

Solution

Given, the points A(4,3) and B(x,5) lie on a circle with the center O (2,3)

Then OA = OB

Also (OA)2 = (OB)2

\(\Rightarrow\)\(\left (4-2\right )^{2}+\left (3-3\right )^{2}=\left (x-2\right )^{2}+\left (5-3\right )^{2}\)

\(\Rightarrow\)\(\left (2\right )^{2}+\left (0\right )^{2}=\left (x-2\right )^{2}+\left (2\right )^{2}\)

\(\Rightarrow\) 4 = (x-2)2 +4

\(\Rightarrow\) (x-2)2 = 0

\(\Rightarrow\) x-2 = 0

\(\Rightarrow\) x = 2

Therefore, x =2

Q3) If the point C (-2, 3) is equidistant from the points A (3, -1) and B (x, 8).Find the values of x. Also, find the distance BC.

Solution

As per the question, we have

AC = BC

\(\Rightarrow\)\(\sqrt{\left ( -2-3 \right )^{2}+\left ( 3+1 \right )^{2}}\) = \(\sqrt{\left ( -2-x \right )^{2}+\left ( 3-8 \right )^{2}}\)

\(\Rightarrow\)\(\sqrt{\left ( 5 \right )^{2}+\left ( 4 \right )^{2}}\) = \(\sqrt{\left ( x+2 \right )^{2}+\left ( -5 \right )^{2}}\)

Squaring both sides, we get

\(\Rightarrow\) 25+16 = (x+2)2 + 25

\(\Rightarrow\) (x+2)2 = 16

\(\Rightarrow\) x+2 = \(\pm 4\)

\(\Rightarrow\) x = -2\(\pm 4\) = -2 – 4 or -2 + 4 = -6 or 2

Now,

BC =\(\sqrt{\left ( -2-x \right )^{2}+\left ( 3-8 \right )^{2}}\)

= \(\Rightarrow\)\(\sqrt{\left ( -2-2 \right )^{2}+\left ( -5 \right )^{2}}\)

= \(\sqrt{16+25}=\sqrt{41}\)units

Hence x = 2 or -6 and BC = \(\sqrt{41}\)units

Q4) If the point P (2, 2) is equidistant from the points A (-2, k) and B (-2k, -3).Find the value of k. Also, find the distance AP.

Solution

As per the question, we have

AP = BP

\(\Rightarrow\)\(\sqrt{\left ( 2+2 \right )^{2}+\left ( 2-k \right )^{2}}\) = \(\sqrt{\left ( 2+2k \right )^{2}+\left ( 2+3 \right )^{2}}\)

\(\Rightarrow\)\(\sqrt{\left ( 4 \right )^{2}+\left ( 2-k \right )^{2}}\) = \(\sqrt{\left ( 2+2k \right )^{2}+\left ( 5 \right )^{2}}\)

\(\Rightarrow\) 16 + 4 + k2 – 4k = 4 + 4k2 + 8k + 25

Squaring both sides

\(\Rightarrow\) k2 + 4k + 3 = 0

\(\Rightarrow\) (k+1) (k+3) = 0

\(\Rightarrow\) k = -3 or k = -1

Now for K = -1

AP = \(\sqrt{\left ( 2+2 \right )^{2}+\left ( 2-k \right )^{2}}\)

= \(\sqrt{\left ( 4 \right )^{2}+\left ( 2+1 \right )^{2}}\)

= \(\sqrt{16+9}=\sqrt{25}\) = 5 units

For k = -3

AP = \(\sqrt{\left ( 2+2 \right )^{2}+\left ( 2-k \right )^{2}}\)

= \(\sqrt{\left ( 4 \right )^{2}+\left ( 2+3 \right )^{2}}\)

= \(\sqrt{16+25}=\sqrt{41}\)units

Here, k = -1,-3; AP = 5 units for k = -1 and AP = \(\sqrt{41}\)units for k = -3

Q5) If the point (x, y) is equidistant from the points (a + b, b – a) and (a – b, a + b), Prove that bx = ay.

As per the question, we have

\(\Rightarrow\)\(\sqrt{\left ( x-a-b \right )^{2}+\left ( y-b+a \right )^{2}}\) = \(\sqrt{\left ( x-a+b \right )^{2}+\left ( y-a-b \right )^{2}}\)

Squaring both sides

\(\Rightarrow\)\(\left (x-a-b\right )^{2}+\left (y-b+a\right )^{2}=\left (x-a+b\right )^{2}+\left (y-a-b\right )^{2}\)

\(\Rightarrow\) x2 + (a+b)2– 2x(a+b) + y2 + (a-b)2 -2y(a-b) = x2 + (a-b)2 – 2x(a-b) + y2 + (a+b)2 – 2y(a+b)

\(\Rightarrow\) – x(a+b) – y(a-b) = – x(a-b) – y(a+b)

\(\Rightarrow\) –xa – xb – ay + by = -xa + bx – ya – by

\(\Rightarrow\) by = bx

Hence , bx = ay

Q6) Using the distance formula, show that the given points are collinear

i) (1, -1), (5, 2) and (9, 5)

Solution

Let A (1,-1), B(5,2) and C(9,5) be the given points. Then

AB = \(\sqrt{\left ( 5-1 \right )^{2}+\left ( 2+1 \right )^{2}}\)

= \(\sqrt{\left (4\right )^{2}+\left ( 3 \right )^{2}}\)

= \(\sqrt{25}=5\;units\)

BC = \(\sqrt{\left ( 9-5 \right )^{2}+\left ( 5-2 \right )^{2}}\)

= \(\sqrt{\left (4\right )^{2}+\left ( 3 \right )^{2}}\)

= \(\sqrt{25}=5\;units\)

AC = \(\sqrt{\left ( 9-1 \right )^{2}+\left ( 5+1 \right )^{2}}\)

= \(\sqrt{\left (8\right )^{2}+\left ( 6 \right )^{2}}\)

= \(\sqrt{100}= 10\;units\)

\(Therefore, \) AB +BC = (5+5) units = 10 units = AC

Hence, the given points are collinear

ii) (6, 9), (0, 1) and (-6, -7)

Solution

Let A (6,9), B(0,1) and C(-6,-7) be the given points. Then

AB = \(\sqrt{\left ( 0-6 \right )^{2}+\left ( 1-9 \right )^{2}}\)

= \(\sqrt{\left (-6\right )^{2}+\left ( -8 \right )^{2}}\)

= \(\sqrt{100}= 10\;units\)

BC = \(\sqrt{\left ( -6-0 \right )^{2}+\left ( -7-1 \right )^{2}}\)

= \(\sqrt{\left (-6\right )^{2}+\left ( -8 \right )^{2}}\)

= \(\sqrt{100}= 10\;units\)

AC = \(\sqrt{\left ( -6-6 \right )^{2}+\left ( -7-9 \right )^{2}}\)

= \(\sqrt{\left (-12\right )^{2}+\left ( -16 \right )^{2}}\)

= \(\sqrt{400}= 20\;units\)

\(Therefore, \) AB +BC = (10+10) units = 20 units = AC

Hence, the given points are collinear

iii) (-1, -1), (2, 3) and (8, 11)

Solution

Let A (-1,-1), B(2,3) and C(8,11) be the given points. Then

AB = \(\sqrt{\left ( 2+1 \right )^{2}+\left ( 3+1 \right )^{2}}\)

= \(\sqrt{\left (3\right )^{2}+\left ( 4 \right )^{2}}\)

= \(\sqrt{25}= 5\;units\)

BC = \(\sqrt{\left ( 8-2 \right )^{2}+\left ( 11-3 \right )^{2}}\)

= \(\sqrt{\left (6\right )^{2}+\left ( 8 \right )^{2}}\)

= \(\sqrt{100}= 10\;units\)

AC = \(\sqrt{\left ( 8+1 \right )^{2}+\left ( 11+1 \right )^{2}}\)

= \(\sqrt{\left (9\right )^{2}+\left ( 12 \right )^{2}}\)

= \(\sqrt{225}= 15\;units\)

\(Therefore, \) AB +BC = (5+10) units = 15 units = AC

Hence, the given points are collinear

iv) (-2, 5), (0, 1) and (2, -3)

Solution

Let A (-2,5), B(0,1) and C(2,-3) be the given points. Then

AB = \(\sqrt{\left ( 0+2 \right )^{2}+\left ( 1-5 \right )^{2}}\)

= \(\sqrt{\left (2\right )^{2}+\left ( -4 \right )^{2}}\)

= \(\sqrt{20}= 2\sqrt{5}\;units\)

BC = \(\sqrt{\left ( 2-0 \right )^{2}+\left ( -3-1 \right )^{2}}\)

= \(\sqrt{\left (2\right )^{2}+\left ( -4 \right )^{2}}\)

= \(\sqrt{20}=2\sqrt{5}\;units\)

AC = \(\sqrt{\left ( 2+2 \right )^{2}+\left ( -3-5 \right )^{2}}\)

= \(\sqrt{\left (4\right )^{2}+\left ( -8 \right )^{2}}\)

= \(\sqrt{80}=4\sqrt{5}\;units\)

\(Therefore, \) AB +BC = (\(2\sqrt{5}\)+\(2\sqrt{5}\)) units = \(4\sqrt{5}\) units = AC

Hence, the given points are collinear

Q7) Show that the points A (7, 10), B (-2, 5) and C (3, -4) are the vertices of an isosceles right triangle.

Solution

The given points are A(7,10), B(-2,5) and C(3,-4)

AB = \(\sqrt{\left ( -2-7 \right )^{2}+\left ( 5-10 \right )^{2}}\)

= \(\sqrt{\left (-9\right )^{2}+\left ( -5 \right )^{2}}\)

= \(\sqrt{81+25}= \sqrt{106}\;units\)

BC = \(\sqrt{\left ( 3-\left ( -2\right ) \right )^{2}+\left ( -4-5 \right )^{2}}\)

= \(\sqrt{\left (5\right )^{2}+\left ( -9 \right )^{2}}\)

= \(\sqrt{25+81}=\sqrt{106}\;units\)

AC = \(\sqrt{\left ( 3-7 \right )^{2}+\left ( -4-10 \right )^{2}}\)

= \(\sqrt{\left (-4\right )^{2}+\left ( -14 \right )^{2}}\)

= \(\sqrt{16+196}= \sqrt{212}\;units\)

Since, AB and BC are equal, they form the vertices of an isosceles triangle

Also,(AB)2 + (BC)2 = \(\left ( \sqrt{106} \right )^{2}\) + \(\left ( \sqrt{106} \right )^{2}\) = \(\sqrt{212}\)

and (AC)2 = \(\left ( \sqrt{212} \right )^{2}\) = 212

Thus, (AB)2 + (BC)2 = (AC)2

This shows that \(\Delta\;ABC\) is right angled at B

Therefore, the given points A(7,10), B(-2,5) and C(3,-4) are the vertices of an isosceles right-angled triangle.

Q8) Show that the points A (3, 0), B (6, 4) and C (-1, 3) are the vertices of an isosceles right triangle.

Solution

Let A (3,0), B(6,4) and C(-1,3) be the given points. Then

AB = \(\sqrt{\left ( 3-6 \right )^{2}+\left ( 0-4 \right )^{2}}\)

= \(\sqrt{\left (-3\right )^{2}+\left ( -4 \right )^{2}}\)

= \(\sqrt{9+16}= \sqrt{25}\) = 5

BC = \(\sqrt{\left ( 6+1 \right )^{2}+\left ( 4-3 \right )^{2}}\)

= \(\sqrt{\left (7\right )^{2}+\left ( 1 \right )^{2}}\)

= \(\sqrt{49+1}=\sqrt{50}\) = \(5\sqrt{2}\)

AC = \(\sqrt{\left ( 3+1 \right )^{2}+\left ( 0-3 \right )^{2}}\)

= \(\sqrt{\left (4\right )^{2}+\left ( -3 \right )^{2}}\)

= \(\sqrt{16+9}\)= \(\sqrt{25}\) = 5

\(Therefore, \)AB =AC and (AB)2 + (AC)2 = (BC)2

Therefore, A (3,0), B(6,4) and C(-1,3) are the vertices of an isosceles right triangle.

Q9) If A(5, 2), B(2, -2) and C(-2, t) are the vertices of a right triangle with \(\angle B = 90^{\circ}\), then find the value of t.

Solution

\(Therefore, \) \(\angle B=90^{\circ}\)

\(Therefore, \) (AC)2 = (AB)2 + (BC)2

\(\Rightarrow\) (5+2)2 + (2-t)2 = (5-2)2 + (2+2)2 + (2+2)2 + (-2-t)2

\(\Rightarrow\) 72 + (t-2)2 = (3)2 + (4)2 + (4)2 + (t+2)2

\(\Rightarrow\) 49 + t2 – 4t + 4 = 9 +16 +16+ t2 + 4t +4

\(\Rightarrow\) 8 -4t = 4t

\(\Rightarrow\) 8t = 8

\(\Rightarrow\) t = 1

Hence, t = 1

Q10) Prove that the points A(2, 4), B(2, 6) and \(C(2+\sqrt{3},5)\) are the vertices of an equilateral triangle.

Solution

The given points are A (2,4), B(2,6) and C(\(2+\sqrt{3}\),5). Now,

AB = \(\sqrt{\left ( 2-2 \right )^{2}+\left ( 4-6 \right )^{2}}\)

= \(\sqrt{\left (0\right )^{2}+\left ( -2 \right )^{2}}\)

= \(\sqrt{0+4}=\sqrt{4}\) = 2

BC = \(\sqrt{\left (2-2-\sqrt{3} \right )^{2}+\left ( 6-5 \right )^{2}}\)

= \(\sqrt{\left (\sqrt{-3}\right )^{2}+\left ( 1 \right )^{2}}\)

= \(\sqrt{3+1}= \sqrt{4}\) = 2

AC = \(\sqrt{\left ( 2-2-\sqrt{3}\right )^{2}+\left ( 4-5 \right )^{2}}\)

= \(\sqrt{\left (\sqrt{-3}\right )^{2}+\left ( -1 \right )^{2}}\)

= \(\sqrt{3+1}=\sqrt{4}\) = 2

Hence the points are A (2,4), B(2,6) and C(\(2+\sqrt{3}\),5) are the vertices of an equilateral triangle.

Q11) Show that the points (-3, -3),(3, 3) and \((-3\sqrt{3},3\sqrt{3})\) are the vertices of an equilateral triangle.

Solution

The given points are A (-3,-3), B(3,3) and C(\(-3\sqrt{3}\),\(-3\sqrt{3}\)). Now,

AB = \(\sqrt{\left ( -3-3 \right )^{2}+\left ( -3-3 \right )^{2}}\)

= \(\sqrt{\left (-6\right )^{2}+\left ( -6 \right )^{2}}\)

= \(\sqrt{36+36}= \sqrt{72}\) = \(6\sqrt{2}\),).

BC = \(\sqrt{\left (3+3\sqrt{3} \right )^{2}+\left ( 3-3\sqrt{3}\right )^{2}}\)

= \(\sqrt{9+27+18\sqrt{3}+9+27-18\sqrt{3}}\)

= \(\sqrt{72}= 6\sqrt{2}\)

AC = \(\sqrt{\left ( -3+3\sqrt{3}\right )^{2}+\left (-3-3\sqrt{3}\right )^{2}}\)

= \(\sqrt{9+27-18\sqrt{3}+9+27+18\sqrt{3}}\)

= \(\sqrt{72}\)= \(6\sqrt{2}\)

Hence the given points are the vertices of an equilateral triangle

Q12) Show that the points A(-5, 6), B(3, 0) and C(9, 8) are the vertices of an isosceles right-angled triangle. Calculate its area.

Solution

Let the given points be A(-5,6), B(3,0) and C(9,8)

AB = \(\sqrt{\left ( 3-\left ( -5\right ) \right )^{2}+\left ( 0-6 \right )^{2}}\)

= \(\sqrt{\left (8\right )^{2}+\left ( -6 \right )^{2}}\)

= \(\sqrt{64+36}= \sqrt{100}\) = 10 units

BC = \(\sqrt{\left ( 9-3 \right )^{2}+\left ( 8-0 \right )^{2}}\)

= \(\sqrt{\left (6\right )^{2}+\left ( 8 \right )^{2}}\)

= \(\sqrt{36+64}= \sqrt{100}\) = 10 units

AC = \(\sqrt{\left ( 9-\left ( -5\right ) \right )^{2}+\left ( 8-6 \right )^{2}}\)

= \(\sqrt{\left (14\right )^{2}+\left ( 2 \right )^{2}}\)

= \(\sqrt{196+4}=\sqrt{200}\) = \(10\sqrt{2}\) units

Therefore, AB = BC = 10 units

Also, (AB)2 + (BC)2 = 102 + 102 = 200

And (AC)2 = \(\left (10\sqrt{2}\right)^{2}\) = 200

Thus (AB)2 + (BC)2 = (AC)2

This shows that\(\Delta\;ABC\) is right angled at B

Therefore, the points A(-5,6) B(3,0) and C(9,8) are the vertices of an isosceles right-angled triangle

Also, area of a triangle = \(\frac{1}{2}*base*height\)

If AB is the height and BC is the base,

Area = \(\frac{1}{2}*10*10\)

= 50 square units

Q13) Show that the points O (0, 0), \(A(3,\sqrt{3})\) and \(B(3,-\sqrt{3})\) are the vertices of an equilateral triangle. Find the area of this triangle.

Solution

The given points are O (0,0), A(\(\left(3,\sqrt{3}\right)\)) and B(\(\left(3,-\sqrt{3}\right)\)

OA = \(\sqrt{\left ( 3-0\right )^{2}+\left ( \sqrt{3}-0\right )^{2}}\)

= \(\sqrt{\left ( 3\right )^{2}+\left ( \sqrt{3}\right )^{2}}\)

= \(\sqrt{9+3}\) = \(\sqrt{12}\)= \(2\sqrt{3}\) units

AB = \(\sqrt{\left ( 3-3\right )^{2}-\left (-\sqrt{3}-\sqrt{3}\right )^{2}}\)

= \(\sqrt{0+\left(2\sqrt{3} \right)^{2}}\)

= \(\sqrt{4*3}\)

= \(\sqrt{12}\)

= \(2\sqrt{3}\) units

OB = \(\sqrt{\left ( 3-0\right )^{2}+\left ( -\sqrt{3}-0\right )^{2}}\)

= \(\sqrt{\left ( 3\right )^{2}+\left ( \sqrt{3}\right )^{2}}\)

= \(\sqrt{9+3}\) = \(\sqrt{12}\)= \(2\sqrt{3}\)

Therefore, OA = AB = OB = \(2\sqrt{3}\) units

The given points are O (0,0), A(\(\left(3,\sqrt{3}\right)\)) and B(\(\left(3,-\sqrt{3}\right)\) are the vertices of an equilateral triangle

Also, the area of the equilateral triangle OAB = \(\frac{\sqrt{3}}{4}*\left (side\right)^{2}\)

= \(\frac{\sqrt{3}}{4}*\left ( 2\sqrt{3} \right )^{2}\)

= \(\frac{\sqrt{3}}{4}*12\)

= \(3\sqrt{3}\) square units


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