(i) P (5, 0)<\/strong><\/p>\n\n\n\n (ii) Q (0, -2)<\/strong><\/p>\n\n\n\n (iii) R (-4, 0)<\/strong><\/p>\n\n\n\n (iv) S (0, 5)<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (i) P (5, 0) lies on x \u2013 axis<\/p>\n\n\n\n (ii) Q (0, -2) lies on y \u2013 axis (negation half)<\/p>\n\n\n\n (iii) R (-4, 0) lies on x \u2013 axis (negative half)<\/p>\n\n\n\n (iv) S (0, 5) lies on y \u2013 axis<\/p>\n\n\n\n 1. Find the distance between the following pair of points:<\/strong><\/p>\n\n\n\n (i) (- 6, 7) and (-1, -5)<\/strong><\/p>\n\n\n\n (ii) (a + b, b + c) and (a -b, c \u2013 b)<\/strong><\/p>\n\n\n\n (iii) (a sin \u03b1, \u2013 b cos \u03b1) and (- a cos \u03b1, b sin \u03b1)<\/strong><\/p>\n\n\n\n (iv) (a, 0) and (0, b)<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (i) Let the given points be P (- 6, 7) and Q (- 1, \u2013 5)<\/p>\n\n\n\n Here,<\/p>\n\n\n\n x1<\/sub> = \u2013 6, y1<\/sub> = 7 and<\/p>\n\n\n\n x2<\/sub> = -1, y2<\/sub> = \u2013 5<\/p>\n\n\n\n (ii) Let the given points be P (a + b, b + c) and Q (a \u2013 b, c \u2013 b)<\/p>\n\n\n\n Here,<\/p>\n\n\n\n x1<\/sub> = a + b, y1 <\/sub>= b + c and<\/p>\n\n\n\n x2<\/sub> = a \u2013 b, y2<\/sub> = c \u2013 b<\/p>\n\n\n\n (iii) Let the given points be P(a sin\u03b1, \u2013 b cos \u03b1) and Q(-a cos \u03b1, b sin \u03b1) here<\/p>\n\n\n\n x1 <\/sub>= a sin \u03b1, y1<\/sub> = \u2013 b cos \u03b1 and<\/p>\n\n\n\n x2 <\/sub>\u2013 a cos \u03b1, y2<\/sub> = b sin \u03b1<\/p>\n\n\n\n (iv) Let the given points be P(a, 0) and Q (0, b)<\/p>\n\n\n\n Here,<\/p>\n\n\n\n x1<\/sub> = a, y1<\/sub> = 0, x2<\/sub> = 0, y2<\/sub> = b,<\/p>\n\n\n\n 2. Find the value of a when the distance between the points (3, a) and (4, 1) is \u221a10.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Let the given points be P (3, a) and Q(4, 1).<\/p>\n\n\n\n Here,<\/p>\n\n\n\n On squaring on both sides, we have<\/p>\n\n\n\n \u21d2 10 = 2 + a2<\/sup> \u2013 2a<\/p>\n\n\n\n \u21d2 a2<\/sup> \u2013 2a + 2 \u2013 10 = 0<\/p>\n\n\n\n \u21d2 a2<\/sup> \u2013 2a \u2013 8 = 0<\/p>\n\n\n\n By splitting the middle team,<\/p>\n\n\n\n \u21d2 a2<\/sup> \u2013 4a + 2a \u2013 8 = 0<\/p>\n\n\n\n \u21d2 a(a \u2013 4) + 2(a \u2013 4) = 0<\/p>\n\n\n\n \u21d2 (a \u2013 4) (a + 2) = 0<\/p>\n\n\n\n \u21d2 a = 4, a = \u2013 2<\/p>\n\n\n\n Thus, there are 2 possible values for a which are 4 and -2.<\/p>\n\n\n\n 3. If the points (2, 1) and (1, -2) are equidistant from the point (x, y), show that x + 3y = 0.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Let the given points be P(2, 1) and Q(1,- 2) and R(x, y)<\/p>\n\n\n\n Also, PR = QR (given)<\/p>\n\n\n\n But, PR = QR<\/p>\n\n\n\n \u21d2 x2 <\/sup>+ 5 \u2013 4x + y2<\/sup> \u20132y = x2 <\/sup>+ 5 \u2013 2x + y2 <\/sup>+ 4y<\/p>\n\n\n\n \u21d2 x2 <\/sup>+ 5 \u2013 4x + y2<\/sup> \u2013 2y = x2 <\/sup>+ 5 \u2013 2x + y2 <\/sup>+ 4y<\/p>\n\n\n\n \u21d2 \u2013 4x + 2x \u2013 2y \u2013 4y = 0<\/p>\n\n\n\n \u21d2 \u2013 2x \u2013 6y = 0<\/p>\n\n\n\n \u21d2 \u2013 2(x + 3y) = 0<\/p>\n\n\n\n \u21d2 -2(x + 3y) = 0<\/p>\n\n\n\n \u21d2 x + 3y = 0\/-2<\/p>\n\n\n\n \u21d2 x + 3y = 0<\/p>\n\n\n\n 4. Find the value of x, y if the distances of the point (x, y) from (- 3, 0) as well as from (3, 0) are 4.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Let the given points be P(x, y), Q( -3, 0) and R(3, 0)<\/p>\n\n\n\n On squaring on both sides, we get<\/p>\n\n\n\n \u21d2 16 = x2<\/sup> + 9 + 6x + y2<\/sup><\/p>\n\n\n\n \u21d2 x2<\/sup> + y2<\/sup> = 7 \u2013 6x \u2026\u2026 (1)<\/p>\n\n\n\n On squaring on both sides,<\/p>\n\n\n\n \u21d216 = x2<\/sup> + 9 \u2013 6x + y2<\/sup><\/p>\n\n\n\n \u21d2 x2<\/sup> + y2<\/sup> = 16 \u2013 9 + 6x<\/p>\n\n\n\n \u21d2 x2 <\/sup>+ y2 <\/sup>= 7 + 6x \u2026. (2)<\/p>\n\n\n\n Equating (1) and (2), we have<\/p>\n\n\n\n 7 \u2013 6x = 7 + 6x<\/p>\n\n\n\n \u21d2 7 \u2013 7 = 6x + 6x<\/p>\n\n\n\n \u21d2 0 = 12x<\/p>\n\n\n\n \u21d2 x = 12<\/p>\n\n\n\n Then, substituting the value of x = 0 in (2)<\/p>\n\n\n\n x2<\/sup> + y2<\/sup> = 7+ 6x<\/p>\n\n\n\n 0 + y2<\/sup> = 7 + 6 \u00d7 0<\/p>\n\n\n\n y2<\/sup> = 7<\/p>\n\n\n\n y = + \u221a7<\/p>\n\n\n\n As y can have two values, the points are (12, \u221a7) and (12, -\u221a7).<\/p>\n\n\n\n 5. The length of a line segment is of 10 units and the coordinates of one end-point are (2, -3). If the abscissa of the other end is 10, find the ordinate of the other end.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given,<\/p>\n\n\n\n Length of the line segment is 10 units.<\/p>\n\n\n\n Coordinates of one end-point are (2, -3) and the abscissa of the other end is 10.<\/p>\n\n\n\n So, let the ordinate of the other end be k.<\/p>\n\n\n\n Therefore, the ordinates of the other end can be 3 or -9.<\/p>\n\n\n\n 6. Show that the points A(- 4, -1), B(-2, \u2013 4), C(4, 0) and D(2, 3) are the vertices points of a rectangle.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n v<\/strong><\/p>\n\n\n\n Given: Points A(- 4, -1), B(-2, \u2013 4), C(4, 0) and D(2, 3)<\/p>\n\n\n\n Required to prove: the points are the vertices points of a rectangle.<\/p>\n\n\n\n Vertices of rectangle ABCD are: A(- 4, -1), B(-2, \u2013 4), C(4, 0) and D(2, 3)<\/p>\n\n\n\n We know that,<\/p>\n\n\n\n As the opposite sides are equal and also the diagonals are equal.<\/p>\n\n\n\n Therefore, the given points are the vertices of a rectangle.<\/p>\n\n\n\n 7. Show that the points A (1,- 2), B (3, 6), C (5, 10) and D (3, 2) are the vertices of a parallelogram.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given: Points A (1,- 2), B (3, 6), C (5, 10) and D (3, 2)<\/p>\n\n\n\n Required to prove: the points are the vertices points of a parallelogram.<\/p>\n\n\n\n Vertices of a parallelogram ABCD are: A (1, -2), B (3, 6), C (5, 10) and D (3, 2)<\/p>\n\n\n\n We know that,<\/p>\n\n\n\n Finding the diagonals,<\/p>\n\n\n\n It\u2019s seen that the opposite sides of the quadrilateral formed by the given four points are equal<\/p>\n\n\n\n i.e. (AB = CD) & (DA = BC)<\/p>\n\n\n\n Also, the diagonals BD & AC are found unequal.<\/p>\n\n\n\n Hence, the given points form a parallelogram.<\/p>\n\n\n\n 8. Prove that the points A (1, 7), B (4, 2), C (-1, -1) and D (-4, 4) are the vertices of a square.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given: Points A (1, 7), B (4, 2), C (-1, -1) and D (-4, 4)<\/p>\n\n\n\n Required to prove: the points are the vertices points of a square.<\/p>\n\n\n\n Vertices of a square ABCD are: A (1, 7), B (4, 2), C (-1, -1) and D (-4, 4)<\/p>\n\n\n\n We know that,<\/p>\n\n\n\n As the opposite sides are equal and also the diagonals are equal the given vertices are therefore the vertices of a square.<\/p>\n\n\n\n 9. Prove that the points (3, 0), (6, 4) and (- 1, 3) are vertices of a right-angled isosceles triangle.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Let the vertices of the triangle ABC be: A(3, 0), B(6, 4) and C (- 1, 3)<\/p>\n\n\n\n We know that,<\/p>\n\n\n\n It\u2019s seen that AB = AC, Thus, it\u2019s an isosceles triangle.<\/p>\n\n\n\n Verifying the Pythagoras theorem, we have<\/p>\n\n\n\n BC2<\/sup> = AB2<\/sup> + AC2<\/sup><\/p>\n\n\n\n (\u221a50)2<\/sup> = (\u221a25)2<\/sup> + (\u221a25)2<\/sup><\/p>\n\n\n\n 50 = 25 + 25<\/p>\n\n\n\n 50 = 50<\/p>\n\n\n\n As BC2<\/sup> = AB2<\/sup> + AC2<\/sup><\/p>\n\n\n\n Therefore, the given vertices are of a right-angled isosceles triangle.<\/p>\n\n\n\n 10.<\/strong> Prove that (2, -2), (-2, 1) and (5, 2) are the vertices of a right angled triangle. Find the area of the triangle and the length of the hypotenuse.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n From given,<\/p>\n\n\n\n Let consider the vertices of a triangle ABC as: A(2, -2), B(-2, 1) and C(5, 2)<\/p>\n\n\n\n We know that,<\/p>\n\n\n\n It\u2019s seen that AB = AC, thus the triangle is an isosceles triangle.<\/p>\n\n\n\n Verifying the Pythagoras theorem, we have<\/p>\n\n\n\n BC2<\/sup> = AB2<\/sup> + AC2<\/sup><\/p>\n\n\n\n (\u221a50)2<\/sup> = (\u221a25)2<\/sup> + (\u221a25)2<\/sup><\/p>\n\n\n\n 50 = 25 + 25<\/p>\n\n\n\n 50 = 50<\/p>\n\n\n\n As BC2<\/sup> = AB2<\/sup> + AC2<\/sup><\/p>\n\n\n\n Therefore, the given triangle is right angled triangle.<\/p>\n\n\n\n Now,<\/p>\n\n\n\n 11. Prove that the points (2a, 4a), (2a, 6a) and (2a + <\/strong>\u221a3a, 5a) are the vertices of an equilateral triangle.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n From given,<\/p>\n\n\n\n Let\u2019s consider the vertices of a triangle ABC as: A(2 a, 4 a), B(2 a, 6 a) and C(2a + \u221a3a, 5a)<\/p>\n\n\n\n We know that,<\/p>\n\n\n\n As all the sides are equal the triangle is an equilateral triangle.<\/p>\n\n\n\n Thus, the given vertices are of an equilateral triangle.<\/p>\n\n\n\n 12. Prove that the points (2, 3), (-4, -6) and (1, 3\/2) do not form a triangle.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n From given,<\/p>\n\n\n\n Let\u2019s consider the vertices of a triangle ABC as: A(2, 3), B(-4, -6) and C(1, 3\/2)<\/p>\n\n\n\n We know that,<\/p>\n\n\n\n Thus, the given vertices do not form a triangle as the sum of two sides of a triangle is not greater than third side.<\/p>\n\n\n\n 13. The points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a triangle ABC right triangle ABC right angled at B. Find the values of a and hence the area of triangle ABC.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given,<\/p>\n\n\n\n A right triangle ABC, right angled at B.<\/p>\n\n\n\n Points A (2, 9), B (a, 5) and C (5, 5)<\/p>\n\n\n\n So, AC is the hypotenuse<\/p>\n\n\n\n Thus, from Pythagoras theorem we have<\/p>\n\n\n\n AC2<\/sup> = AB2<\/sup> + BC2<\/sup>[(5 \u2013 2)2<\/sup> + (5 \u2013 9)2<\/sup>] = [(a \u2013 2)2<\/sup> + (5 \u2013 9)2<\/sup>] + [(5 \u2013 a)2<\/sup> + (5 \u2013 5)2<\/sup>] [32 <\/sup>+ (-4)2<\/sup>] = [(a \u2013 2)2<\/sup> + (-4)2<\/sup>] + [(5 \u2013 a)2<\/sup> + 0]<\/p>\n\n\n\n 9 + 16 = a2<\/sup> \u2013 4a + 4 + 16 + 25 \u2013 10a + a2<\/sup><\/p>\n\n\n\n 2a2<\/sup> \u2013 14a + 20 = 0<\/p>\n\n\n\n a2<\/sup> \u2013 7a + 10 = 0<\/p>\n\n\n\n (a \u2013 5)(a \u2013 2) = 0 [By factorization method]<\/p>\n\n\n\n So,<\/p>\n\n\n\n a = 5 or 2<\/p>\n\n\n\n Here, a = 5 is not possible as it coincides with point C. So, for a triangle to form the value of a = 2 is correct.<\/p>\n\n\n\n Thus, the coordinates of point B is (2, 5).<\/p>\n\n\n\n Now, the area of triangle ABC<\/p>\n\n\n\n Therefore, the area of triangle ABC is 6 sq. units<\/p>\n\n\n\n 14. Show that the quadrilateral whose vertices are (2, -1), (3, 4), (-2, 3) and (-3, -2) is a rhombus.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Let A(2, -1), B(3 ,4), C(-2, 3) and D(-3, -2)<\/p>\n\n\n\n Then we have,<\/p>\n\n\n\n Length of AB = \u221a[(3 \u2013 2)2<\/sup> + (4 \u2013 (-1))2<\/sup>] = \u221a[(1)2<\/sup> + (5)2<\/sup>] = \u221a[1 + 25] = \u221a26 units<\/p>\n\n\n\n Length of BC = \u221a[(3 \u2013 (-2))2<\/sup> + (4 \u2013 3)2<\/sup>] = \u221a[(5)2<\/sup> + (1)2<\/sup>] = \u221a[25 + 1] = \u221a26 units<\/p>\n\n\n\n Length of CD = \u221a[(-2 \u2013 (-3))
\n\n\n\nExercise 14.2 Page No: 14.15<\/h3>\n\n\n\n
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