问题1)找到以下两对点之间的距离：
(i)(2,3)，(4,1)
(ii)(-5、7)，(-1、3)
(iii)(a，b)，(-a，-b)
解决方案：
Formula used in the above question is : √(x2 – x1)2 + (y2 – y1)2 (i.e., Distance Formula)
(i) Here, x1 = 2, y1= 3, x2 = 4, y2 = 1
Now, applyling the distance formula :
= √(4-2)2 + (1-3)2
=√(2)2 + (-2)2
= √8
= 2√2 units
(ii) Here, x1= -5, y1= 7, x2 = -1, y2 = 3
Now, applying the distance formula :
= √(-1 – (-5))2 + (3 – 7)2
= √(4)2 + (-4)2
= 4√2
(iii) Here, x1 = a, y1 = b . x2 = -a, y2 = -b
Now, applying the distance formula :
= √(-a – a)2 +(-b – b)2
= √(-2a)2 + (-2b)2
= √4a2 + 4b2
= 2√a2 + b2
问题2)找到点(0，0)和(36，15)之间的距离。您现在可以找到第7.2节中讨论的下图所示的两个镇A和B之间的距离吗?
解决方案：
Formula used in the above question is : √(x2 – x1)2 + (y2 – y1)2 (i.e, Distance Formula)
Considering, Point A as (0, 0) and Point B as (36, 15) and applying the distance formula we get :
Distance between the two points : √(36 – 0)2 + (15 – 0)2
= √(36)2 + (15)2
= √1296 + 225
= √1521
= 39 units
Hence, the distance between two towns A and B is 39 units.
问题3)确定点(1、5)，(2、3)和(–2，–11)是否共线。
解决方案：
Let (1, 5), (2, 3) and (-2, -11) be points A, B and C respectively.
Collinear term means that these 3 points lie in the same line. So, to we’ ll check it.
Using distance formula we will find the distance between these points.
AB = √(2 – 1)2 + (3 – 5)2
=√(1)2 + (-2)2 =√1 + 4 =√5
BC = √(-2 – 2)2 + (-11 – 3)2
= √(-4)2 + (-14)2 = √16 + 196 = √212
CA = √(-2 – 1)2 + (-11 – 5)2
= √(-3)2 + (-16)2 = √9 + 256 =√265
As, AB + BC ≠ AC (Since, one distance is not equal to sum of other two distances, we can say that they do not lie in the same line.)
Hence, points A, B and C are not collinear.
问题4)检查(5，– 2)，(6，4)和(7，– 2)是否为等腰三角形的顶点。
解决方案：
Let (5, – 2), (6, 4) and (7, – 2) be the points A, B and C respectively.
Using distance formula :
AB = √(6 – 5)2 + (4 – (-2))2
= √(1 + 36) = √37
BC = √(7 – 6)2 + (-2 – 4)2
= √(1 + 36) = √37
AC = √(7 – 5)2 + (-2 – (-2))2
= √(4 + 0) = 2
As, AB = BC ≠ AC (Two distances equal and one distance is not equal to sum of other two)
So, we can say that they are vertices of an isosceles triangle.
问题5)如图7.8所示，在一个教室中，有四个朋友坐在A，B，C和D点。 Champa和Chameli走进课堂，观察了几分钟后，Champa问Chameli：“您不认为ABCD是正方形吗?”沙米利不同意。使用距离公式，找出其中哪个是正确的。
解决方案：
From the given fig, find the coordinates of the points
AB = √(6 – 3) + (7 – 4)
= √9+9 = √18 = 3√2
BC = √(9 – 6) + (4 – 7)
= √9+9 = √18 = 3√2
CD = √(6 – 9) + (1 – 4)
= √9 + 9 = √18 = 3√2
DA = √(6 – 3) + (1 – 4)
= √9+9 =√18 =3√2
AB = BC = CD = DA = 3√2
All sides are of equal length. Therefore, ABCD is a square and hence, Champa was correct.
问题6)通过以下几点来说明所形成的四边形的类型(如果有的话)，并给出回答的理由：
(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)
解决方案：
(i) Here, let the given points are P(-1, -2), Q(1, 0), R(-1, 2) and S(-3, 0) respectively.
PQ = √(1 – (-1))² + (0 – (-2))²
= √(1+1)²+(0+2)²
= √8 = 2 √2
QR = √(−1−1)²+(2−0)²
= √(−2)²+(2)²
= √8 = 2 √2
RS = √(−3−(−1))²+(0−2)²
= √8 = 2 √2
PS = √((−3−(−1))²+(0−(−2))²
= √8 = 2 √2
Here, we found that the length of all the sides are equal.
Diagonal PR = √(−1−(−1))²+(2−(−2))²
= √ 0+16
= 4
Diagonal QS = √(−3−1)²+(0−0)²
= √ 16 = 4
Finally, we also found that the length of diagonal are also same.
Here, PQ = QR = RS = PS = 2√2
and QS = PR = 4
This is the property of SQURE. Hence, the given figure is SQUARE.
(ii) Let the points be P(-3, 5), Q(3, 1), R(0, 3) and S(-1, -4)
PQ = √(3−(−3))²+(1−5)²
= √(3+3)²+(−4)²
= √36+16 = √52 = 2 √13
QR = √(0−3)²+(3−1)²
= √(−3)²+(2)²
= √9 + 4 = √13
RS = √(−1−0)²+(−4−3)²
= √(−1)²+(−7)²
= √1+49 = √50 = 5 √2
PS = √(−1−(−3))²+(−4−5)²
= √(−1+3)²+(−9)²
= √4+81 = √85
Here, All the lengths of sides are unequal.
So, The given points will not create any quadrilateral.
(iii) Let the points be P(4, 5), Q(7, 6), R(4, 3) and S(1, 2)
PQ = √(7−4)²+(6−5)²
= √(3)²+(1)² = √9+1 = √10
QR = √(4−7)²+(3−6)²
= √(−3)²+(−3)² = √9+9 =3 √2
RS = √(1−4)²+(2−3)²
= √(−3)²+(−1)² = √9+1 = √10
PS = √(1−4)²+(2−5)²
= √(−3)²+(−3)² = √9+9 =3 √2
We see that the opposite sides are equal. Lets find the diagonal now.
Diagonal PR = √(4−4)²+(3−5)²
= √0+4 = 2
Diagonal QS = √(1−7)²+(2−6)²
= √36+16
= √52
Here, PQ = RS = √10
and QR = PS = 3√2
We see that the diagonals are not equal.
Hence, the formed quadrilateral is a PARALLELOGRAM.
问题7)在x轴上找到与(2，– 5)和(-2，9)等距的点。
解决方案：
Let the point on the X axis be (x, 0)
Given, distance between the points (2, -5), (x, 0) = distance between points (-2, 9), (x, 0)
[ Applying Distance Formula ]
⇒ √(x – 2)²+(0 – (-5))² = √(x – (-2))²+(0 – 9)²
On squaring both the sides, we get
⇒ (x – 2)² + 5² = (x + 2)² + 9²
⇒ x² – 4x + 4 + 25 = x² + 4x + 4 + 81
⇒ -4x -4x = 85 – 29
⇒ -8x = 56
⇒ x = -7
问题8)找到点P(2，– 3)和Q(10，y)之间的距离为10个单位的y值。
解决方案：
It is given that, the distance b/w two points is 10 units.
So, we’ll find the distance and equate
PQ = √ (10 – 2)2 + (y – (-3))2
= √ (8)2 + (y +3)2
On squaring both the sides, we get :
64 +(y+3)2 = (10)2
(y+3)2 = 36
y + 3 = ±6
y + 3 = +6 or y + 3 = −6
Hence, y = 3 or -9.
问题9)如果Q(0，1)与P(5，– 3)和R(x，6)等距，则求出x的值。还要找到距离QR和PR。
解决方案：
Given, PQ = QR
We will apply distance formula and find the distance between them,
PQ = √(5 – 0)2 + (-3 – 1)2
= √ (- 5)2 + (-4)2
= √ 25 + 16 = √41
QR = √ (0 – x)2 + (1 – 6)2
= √ (-x)2 + (-5)2
= √ x2 + 25
As they both are equal so on equating them, x2 + 25 = 41
x2 =16, x = ± 4
So, putting the value of x and obtaining the value of QR and PR through distance formula
For x = +4, PR = √ (4 – 5)2 + (6 – (-3))2
= √ (-1) 2+ (9)2
= √ 82
QR = √ (0 – 4)2 + (1 – 6)2
= √ 41
For x = -4, QR = √ (0 – (-4))2 + (1 – 6)2
= √ 16 + 25 = √ 41
PR = √ (5 + 4)2 + (-3 -6)2
= √ 81 + 81 = 9 √ 2
问题10)找到x和y之间的关系，使点(x，y)与点(3，6)和(-3，4)等距。
解决方案：
Let, (x, y) be point P and (3, 6), (-3, 4) be pt A and B respectively.
It is given that their distance is equal, so we will equate the equations.
PA = PB (given)
⇒ √(x – 3)2 +(y – 6)2 = √(x-(-3))2+ (y – 4)2 [ By applying distance formula ]
On squaring both sides,
(x-3)2+(y-6)2 = (x +3)2 +(y-4)2
x2 +9-6x+y2+36-12y = x2 +9+6x+y2 +16-8y
36-16 = 6x+6x+12y-8y
20 = 12x+4y
3x+y = 5
3x+y-5 = 0