Solutions
1. Find the distance between the following pairs of points : (i) (2, 3), (4, 1) Let $A(x_1,y_1)$ = (2,3) and $B(x_2,y_2)=(4,1)$ Distance between A and B = $√{(x_1-x_2)^2+(y_1-y_2)^2}$ = $√{(2-4)^2+(3-1)^2}$ = $√{(-2)^2+2^2}$ = $√{4+4}$ = $2√{2}$ unit(ii) (– 5, 7), (– 1, 3) Distance between A and B = $√{(x_1-x_2)^2+(y_1-y_2)^2}$ =$√{((-5)-(-1))^2+(7-3)^2}$ =$√{(-4))^2+(4)^2}$ =$√{16+16}$ =$4√{2}$ unit(iii) (a, b), (– a, – b) Distance between A and B = $√{(x_1-x_2)^2+(y_1-y_2)^2}$ =$√{(a-(-a))^2+(b-(-b))^2}$ =$√{(2a)^2+(2b)^2}$ =$2√{a^2+b^2}$2. Find the distance between the points A(0, 0) and B(36, 15). Can you now find the distance between the two towns A and B . Distance between A and B = $√{(x_1-x_2)^2+(y_1-y_2)^2}$ = $√{(0-36)^2+(0-15)^2}$ = $√{36^2+15^2}$ = $√{(3× 12)^2+(3× 5)^2}$ = $3√{(12)^2+(5)^2}$ = $3√{144+25}$ = $3√169$ = $3×13$ = $39$ unit3. Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear. 3 points A,B,C are collinear if AB+BC=AC Distance AB =$√{(x_1-x_2)^2+(y_1-y_2)^2}$ = $√{(1-2)^2+(5-3)^2}$ = $√{(-1)^2+(2)^2}$ = $√{1+4}$ = $√{5}$ Distance BC= $√{(2-(-2))^2+(3-(-11))^2}$ = $√{(4)^2+(14)^2}$ = $√{16+196}$ = $√{212}$ Distance AC = $√{(1-(-2))^2+(5-(-11))^2}$ = $√{(3)^2+(16)^2}$ = $√{9+256}$ = $√{265}$ Ans: Since AB+BC≠AC, hence the points are not collinear .4. Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle. For any 3 points to be triangle, they must not be collinear. i.e AB+BC≠AC Distance AB = $√{(x_1-x_2)^2+(y_1-y_2)^2}$ =$√{(5-6)^2+((-2)-4)^2}$ =$√{(-1)^2+(-6)^2}$ =$√{1+36}$ =$√{37}$ Distance BC = $√{(6-7)^2+(4-(-2))^2}$ = $√{(-1)^2+(6)^2}$ = $√{37}$ Distance AC = $√{(5-7)^2+((-2)-(-2))^2}$ = $√{(-2)^2+0^2}$ = $√{4}$ = $2$ Since AB+BC≠AC , hence the point are not collinear . Also since AB=BC, therfore the 3 points form a isosceles triangle.6. 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) DO IT YOURSELF AS GIVEN IN PREVIOUS EXAMPLE.7. Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9). Let the coordinate of point P on the x-axis be $(x,0)$ Since P is equidistant from A and B, i.e AP=BP $√{(2-x)^2+(-5-0)^2}$ = $√{(-2-x)^2+(9-0)^2}$ $√{(2-x)^2+25}$ = $√{(-2-x)^2+81}$ Since $(-2-x)^2$ = $(x+2)^2$ therefore $√{(2-x)^2+5^2}$ = $√{(x+2)^2+9^2}$ Squaring both the side ${(2-x)^2+5^2}$ = ${(x+2)^2+9^2}$ $(2-x)^2 - (x+2)^2$ = $81-25$ $(2-x+x+2)(2-x-x-2)$ = $56$ $4(-2x)=56$ $x= -56/8$ $x=-7$ Coordinates of the point P on x-axis is (-7,0)8. Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units. Distance = $√{(x_1-x_2)^2+(y_1-y_2)^2}$ 10 = $√{(10-2)^2+(y-(-3))^2}$ 100 = ${(8)^2+(y+3)^2}$ $(y+3)^2 = 100 - 64$ $(y+3)^2 = 36$ $y+3 = ±6$ Ans: Therefore y= -9 or 39. 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. Since Q is equidistant from P and R, Therefore PQ=RQ $√{(5-0)^2+(-3-1)^2}$ = $√{(x-0)^2+(6-1)^2}$ $√{25+16}$ = $√{x^2+25}$ $√41$ = $√{x^2+25}$ Squaring both the sides $41$ = ${x^2+25}$ $41-25$ = ${x^2}$ $16$ = ${x^2}$ $x= ±4$ If $x=4$ P(5, –3) ,R(4, 6) ,Q(0, 1) QR = $√{(4-0)^2 + (6-1)^2}$ = $√{16 + 25}$ = $√41$ PR = $√{(5-4)^2 + (-3-6)^2}$ = $√{(1)^2 + (-9)^2}$ = $√{1 + 81}$ = $√{82}$ If $x=-4$ P(5, –3) ,R(-4, 6) ,Q(0, 1) QR = $√{(-4-0)^2 + (6-1)^2}$ = $√{16 + 25}$ = $√41$ PR = $√{(5-(-4))^2 + (-3-6)^2}$ PR = $√{9^2 + (-9)^2}$ PR = $9√{2}$10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4). Since the point $(x,y) $ is equidistant from (3, 6) and (– 3, 4). $√{(x-3)^2+(y-6)^2}$ = $√{(x-(-3))^2+(y-4)^2}$ $√{(x-3)^2+(y-6)^2}$ = $√{(x+3)^2+(y-4)^2}$ ${(x-3)^2+(y-6)^2}$ = ${(x+3)^2+(y-4)^2}$ $x^2+9-6x + y^2+36-12y$ = $x^2+9+6x + y^2+16-8y$ $x^2-6x + y^2-12y+45$ = $x^2+6x + y^2-8y+25$ $6x-8y+25 +6x +12y -45 =0$ $12x+4y-20 =0$ $3x+y-5 =0$ Ans: $3x+y-5 =0$