Class 9 Advanced Maths Chapter 4 Coordinate Geometry Ex 4.3 Solutions

Students often refer to NCERT Class 9 Advanced Maths Solutions Chapter 4 Coordinate Geometry Ex 4.3 to verify their answers.

Advanced Maths Class 9 Exercise 4.3 Solutions

Class 9 Advanced Maths Ex 4.3 Solutions

Question 1.
A straight line with a slope of \(\frac{1}{2}\) passes through the points A(x, 2), B(3, 4) and C(7, y). Find the coordinates of points A and C and calculate the value of x + y.
Solution:
Since the line has slope \(\frac{1}{2}\), any two points on the line must satisfy:
m = \(\frac{y_2-y_1}{x_2-x_1}=\frac{1}{2}\)
Using points A(x, 2) and B(3,4):
\(\frac{4-2}{3-x}=\frac{1}{2}\)
\(\frac{2}{3-x}=\frac{1}{2}\)
4 = 3 – x
x = -1
So, A=(-1,2)
Using points B(3, 4) and C(7, y):
\(\frac{y-4}{7-3}=\frac{1}{2}\)
y – 4 = 2
y = 6
Then, C(7, 6)
So, x + y = -1 + 6 = 5

Question 2.
Points P(2, 3), Q(5, 7) and R (13, k) are three consecutive vertices of a rectangle. What is the value of k?
Solution:
Since P(2, 3), Q(5, 7) and R(13, k) are three consecutive vertices of a rectangle, the sides PQ and QR must be perpendicular.
So, their slopes satisfy:
m_PQ.m_QR = -1
Slope of PQ, m_PQ = \(\frac{7-3}{5-2}=\frac{4}{3}\)
Slope of QR, m_QR = \(\frac{k-7}{13-5}=\frac{k-7}{8}\)
Since the lines are perpendicular:
Then, \(\frac{k-7}{8} \times \frac{4}{3}=-1\)
\(\frac{k-7}{6}=-1\)
k – 7 = -6
k = 1

Question 3.
A line Z passes through the origin (0, 0). It intersects the line segment joining the points A(2, 8) and B(6, 2). What is the possible range of values for the slope’m’ of line l?
Solution:
The line segment joins: A(2, 8) and B(6, 2).
A line l passes through the origin (0, 0) and intersects this segment. If a line through the origin intersects the segment AB, then its slope must lie between the slopes of the lines joining the origin to A and B.
Slope of OA:
m_OA = \(\frac{8-0}{2-0}\) = 4
Slope of OB:
m_OB = \(\frac{2-0}{6-0}=\frac{2}{6}=\frac{1}{3}\)
Therefore, the slope m of line l must lie between these two values.
\(\frac{1}{3}\) ≤ m ≤ 4
So, the possible range of values is:
m ∈ [latex]\frac{1}{3}[/latex]

Question 4.
Line l₁ passes through the points A( 1, k) and B(k, 7). Line l₂ passes through the points C(-2, 4) and D( 1,10). If line l₁ is parallel to line l₂ find the value of k and evaluate the expression k² + 5.
Solution:
For two lines to be parallel, their slopes must be equal.
Line l₂ passes through C(-2, 4) and D(1, 10).
m₂ = \(\frac{10-4}{1-(-2)}\)
= \(\frac{6}{3)}\) = 2
Linel₁ passes through A(1, k) and B(k, 7).
m₁ = \(\frac{7-k}{k-1}\)
Since the lines are parallel:
\(\frac{7-k}{k-1}\) = 2
7 – k = 2k – 2
k = 3
Then, k² + 5 = 3² + 5
= 9 + 5 = 14

Question 5.
A straight line p has a slope 2. Another straight line q passes through the points M(3a, 4) and N(a, 5). The line p and line q intersect at a right angle. Calculate the value of a and find the coordinates of points M and N.
Solution:
Line p has slope:
m_p = 2
Since, line q is perpendicular to line p, m_p = \(-\frac{1}{2}\)
Line q passes through: M(3a, 4), N(a, 5)
So its slope is: m_q = \(\frac{5-4}{a-3 a}=\frac{1}{-2 a}\)
Given:
\(\frac{1}{-2 a}=-\frac{1}{2}\)
Multiply both sides by -2a: a = 1
For a = 1:
M = (3a, 4) = (3, 4)
N = (a, 5) = (1, 5)
Thus, M = (3, 4), N = (1, 5)