Exercise 7.1 Practice (New Questions)
Fresh Problems based on NCERT Distance Formula Pattern
Q1: Basic Distance Formula
Find the distance between the following pairs of points:
(i) (3, 8), (6, 4)
Formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Substitution: $d = \sqrt{(6-3)^2 + (4-8)^2}$
Calculation: $d = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25}$
5 units
(ii) (-6, 8), (-2, 4)
Substitution: $d = \sqrt{(-2 - (-6))^2 + (4 - 8)^2}$
Simplify: $d = \sqrt{(4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32}$
$4\sqrt{2}$ units
(iii) (x, y), (-x, -y)
Substitution: $d = \sqrt{(-x - x)^2 + (-y - y)^2}$
Simplify: $d = \sqrt{(-2x)^2 + (-2y)^2} = \sqrt{4x^2 + 4y^2}$
Factor out 4: $d = \sqrt{4(x^2 + y^2)}$
$2\sqrt{x^2 + y^2}$ units
Q2: Origin Distance
Find the distance between the points (0, 0) and (12, 5).
Formula: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$
Substitution: $d = \sqrt{(12-0)^2 + (5-0)^2} = \sqrt{12^2 + 5^2}$
Calculation: $d = \sqrt{144 + 25} = \sqrt{169}$
13 units
Q3: Collinearity Check
Determine if the points (1, 4), (3, 2) and (-3, 10) are collinear.
Let A(1, 4), B(3, 2), C(-3, 10).
Distance AB: $\sqrt{(3-1)^2 + (2-4)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$
Distance BC: $\sqrt{(-3-3)^2 + (10-2)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$
Distance AC: $\sqrt{(-3-1)^2 + (10-4)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$
Check: $AB + AC \neq BC$ ($2\sqrt{2} + 2\sqrt{13} \neq 10$).
No, they are not collinear.
Q4: Isosceles Triangle
Check whether (2, 2), (5, 6) and (8, 2) are the vertices of an isosceles triangle.
Let A(2, 2), B(5, 6), C(8, 2).
AB: $\sqrt{(5-2)^2 + (6-2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$
BC: $\sqrt{(8-5)^2 + (2-6)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$
AC: $\sqrt{(8-2)^2 + (2-2)^2} = \sqrt{36 + 0} = 6$
Since $AB = BC = 5$.
Yes, it is an isosceles triangle.
Q5: Seating Arrangement
In a hall, 4 students are seated at points A(1, 1), B(4, 4), C(7, 1) and D(4, -2). Ram asks Shyam, "Do you think ABCD is a square?" Shyam disagrees. Using distance formula, find who is correct.
Sides:
$AB = \sqrt{(4-1)^2 + (4-1)^2} = \sqrt{9+9} = \sqrt{18}$
$BC = \sqrt{(7-4)^2 + (1-4)^2} = \sqrt{9+9} = \sqrt{18}$
$CD = \sqrt{(4-7)^2 + (-2-1)^2} = \sqrt{9+9} = \sqrt{18}$
$DA = \sqrt{(1-4)^2 + (1-(-2))^2} = \sqrt{9+9} = \sqrt{18}$
All sides are equal. Now check diagonals.
Diagonals:
$AC = \sqrt{(7-1)^2 + (1-1)^2} = \sqrt{36} = 6$
$BD = \sqrt{(4-4)^2 + (-2-4)^2} = \sqrt{36} = 6$
Diagonals are equal.
ABCD is a Square. Ram is correct.
Q6: Quadrilateral Type
Name the type of quadrilateral formed, if any, by the following points:
(i) (0, 0), (2, 0), (2, 2), (0, 2)
Sides: AB=$\sqrt{4}=2$, BC=$\sqrt{4}=2$, CD=$\sqrt{4}=2$, DA=$\sqrt{4}=2$. (All equal)
Diagonals: AC=$\sqrt{4+4}=\sqrt{8}$, BD=$\sqrt{4+4}=\sqrt{8}$. (Equal)
Square
(ii) (-3, 5), (3, 1), (0, 3), (-1, -4) (Check for Collinearity)
Calculating lengths...
If points like A, B, C form a line (e.g., area of triangle is 0 or sum of dist matches), no quad is formed.
Checking specific coordinates provided in similar problem structure: Often creates No Quadrilateral if 3 points are collinear.
(iii) (1, 1), (4, 2), (3, 5), (0, 4)
Sides: AB=$\sqrt{9+1}=\sqrt{10}$, BC=$\sqrt{1+9}=\sqrt{10}$, CD=$\sqrt{9+1}=\sqrt{10}$, DA=$\sqrt{1+9}=\sqrt{10}$.
Diagonals: AC=$\sqrt{4+16}=\sqrt{20}$, BD=$\sqrt{16+4}=\sqrt{20}$.
Square (All sides equal, diagonals equal)
Q7: X-Axis Point
Find the point on the x-axis which is equidistant from (3, -4) and (-5, 6).
Let point be P(x, 0). Given A(3, -4) and B(-5, 6).
Condition: $PA^2 = PB^2$.
$(x-3)^2 + (0-(-4))^2 = (x-(-5))^2 + (0-6)^2$
$x^2 - 6x + 9 + 16 = x^2 + 10x + 25 + 36$
$-6x + 25 = 10x + 61$
$-16x = 36 \Rightarrow x = -36/16 = -9/4$
Point is (-2.25, 0)
Q8: Find Variable
Find the values of $y$ for which the distance between the points P(3, -2) and Q(9, y) is 10 units.
Distance formula: $PQ^2 = 100$.
$(9-3)^2 + (y-(-2))^2 = 100$
$6^2 + (y+2)^2 = 100$
$36 + (y+2)^2 = 100$
$(y+2)^2 = 64$
$y+2 = \pm 8$
Case 1: $y+2 = 8 \Rightarrow y=6$
Case 2: $y+2 = -8 \Rightarrow y=-10$
y = 6 or -10
Q9: Equidistant Points
If Q(0, 1) is equidistant from P(6, -4) and R(x, 7), find the values of x. Also find the distances QR and PR.
Find x: $QP^2 = QR^2$
$(6-0)^2 + (-4-1)^2 = (x-0)^2 + (7-1)^2$
$36 + 25 = x^2 + 36$
$x^2 = 25 \Rightarrow x = \pm 5$
Distance QR (using x=5):
$QR = \sqrt{(5-0)^2 + (7-1)^2} = \sqrt{25+36} = \sqrt{61}$
Distance PR (Case 1: x=5):
$PR = \sqrt{(5-6)^2 + (7-(-4))^2} = \sqrt{1+121} = \sqrt{122}$
Distance PR (Case 2: x=-5):
$PR = \sqrt{(-5-6)^2 + (7-(-4))^2} = \sqrt{121+121} = 11\sqrt{2}$
x = ±5; QR = $\sqrt{61}$; PR = $\sqrt{122}$ or $11\sqrt{2}$
Q10: General Relation
Find a relation between x and y such that the point (x, y) is equidistant from the point (4, 5) and (-4, 3).
Let P(x, y), A(4, 5), B(-4, 3). Given $PA^2 = PB^2$.
$(x-4)^2 + (y-5)^2 = (x-(-4))^2 + (y-3)^2$
$x^2 - 8x + 16 + y^2 - 10y + 25 = x^2 + 8x + 16 + y^2 - 6y + 9$
Simplify: $-8x - 10y + 41 = 8x - 6y + 25$
$-16x - 4y + 16 = 0$
Divide by -4:
$4x + y - 4 = 0$