Inter 1st Year Maths Straight Lines Solutions Exercise 9a

Practicing the AP Board Solutions Class 11 Maths and Chapter 9 Inter 1st Year Maths Straight Lines Solutions Exercise 9a Pdf Download will help students to clear their doubts quickly.

Intermediate 1st Year Maths Straight Lines Solutions Exercise 9a

Straight Lines Exercise 9a Solutions

Straight Lines Class 11 Exercise 9a Solutions – Straight Lines 9a Exercise Solutions

Question 1.
Draw a quadrilateral in the Cartesian plane, whose vertices are (-4, 5), (0, 7), (5, -5), and (4, -2). Also, find its area.
Solution:
Let ABCD be the given quadrilater with vertices A(4, -5) B(0, 7) C(5, -5), and D(-4, -2).
Now, let us plot the points on the Cartesian plane by joining the points AB, BC, CD, and AD, which give us the required quadrilateral.

To find the area, draw diagonal AC.
So, area (ABCD) = area(∆ABC) + area(∆ADC)
Then, area of triangle with vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is
Are of ∆ABC = \(\frac {1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
= \(\frac {1}{2}\) [-4(7 + 5) + 0(-5 – 5) + 5(5 – 7)] unit²
= \(\frac {1}{2}\) [-4(12) + 5(-2)] unit²
= \(\frac {1}{2}\) [58] unit²
= 29 unit²
Area of ∆ACD = \(\frac {1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
= \(\frac {1}{2}\) [-4(-5 + 2) + 5(-2 – 5) + (-4) (5 – (-5)] unit²
= \(\frac {1}{2}\) [-4(-3) + 5(-7) – 4(10)] unit²
= \(\frac {1}{2}\) [-63] unit²
= \(\frac {-63}{2}\) unit²
Since area cannot be negative,
Area of ∆ACD = \(\frac {-63}{2}\) unit²
Area (ABCD) = 29 + \(\frac {63}{2}\) = \(\frac {121}{2}\) unit²

Question 2.
The base of an equilateral triangle with side 2a lies along the y-axis such that the midpoint of the base is at the origin. Find the vertices of the triangle.
Solution:
Let us consider ABC, the given equilateral with side 2a.
Where, AB = BC = AC = 2a

In the above figure, assuming that the base BC lies on the x-axis such that the mid-point of BC is at the origin
i.e., BO = OC = a, where O is the origin.
The coordinates of point c are (0, a), and those of B are (0, -a).
The line joining a vertex of an equilateral ∆ with the mid-point of its opposite side is perpendicular.
So, vertex A lies on the y-axis.
By applying Pythagoras’ theorem
(AC)² = OA² + OC²
⇒ (2a)² = a² + OC²
⇒ 4a² – a² = OC²
⇒ 3a² = OC²
⇒ OC = √3a
Co-ordinates of point C = ±√3a, 0.
∴ The vertices of the given equilateral triangle are (0, a), (0, -a), (√3a, 0).

Question 3.
Find the distance between P(x₁, y₁) and Q(x₂, y₂) when:
(i) PQ is parallel to the y-axis.
(ii) PQ is parallel to the x-axis.
Solution:
Given Points P(x₁, y₁) and Q(x₂, y₂)
(i) When PQ is parallel to the y-axis, then x₁ = x₂
So, the distance between P and Q is given by \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{\left(y_2-y_1\right)^2}\)
= |y₂ – y₁|
(ii) When PQ is parallel to the x-axis, then y₁ = y₂
So, the distance between P and Q is given by \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2}\)
= |x₂ – x₁|

Question 4.
Find a point on the x-axis that is equidistant from the points (7, 6) and (3, 4).
Solution:
Let us consider (a, 0) to be the point on the x-axis that is equidistant from the points (7, 6) and (3, 4).
So, \(\sqrt{(7-a)^2+(6-0)^2}=\sqrt{(3-a)^2+(4-0)^2}\)
⇒ \(\sqrt{49+a^2-14 a+36}=\sqrt{9+a^2-6 a+16}\)
⇒ \(\sqrt{a^2-14 a+85}=\sqrt{a^2-6 a+25}\)
Now, let us square on both sides. We get
⇒ a² – 14a + 85 = a² – 6a + 25
⇒ -8a = -60
⇒ a = \(\frac {60}{8}\)
⇒ a = \(\frac {15}{2}\)
∴ The required point is (\(\frac {15}{2}\), 0)

Question 5.
Find the slope of a line that passes through the origin, and the mid-point of the line segment joining the points P (0, -4) and B(8, 0).
Solution:
The co-ordinates of the mid-point of the line segment joining the points P(0, -4) and B(8, 0) are (\(\frac{0+8}{2}\), \(\frac{-4+0}{2}\)) = (4, -2)
The slope “m” of the non-vertical line passing through the point (x₁, y₁) and (x₂, y₂) is given by
m = \(\frac{y_2-y_1}{x_2-x_1}\) Where x₁ ≠ x₂
The slope of the line passing through (0, 0) and (4, -2) is (-2, 0)/(4 – 0) = \(\frac {-1}{2}\)
∴ The required slope is \(\frac {-1}{2}\).

Question 6.
Without using the Pythagoras theorem, show that the points (4, 4), (3, 5), and (-1, -1) are the vertices of a right-angled triangle.
Solution:
The vertices of the given triangle are (4, 4), (3, 5), and (-1, -1)
The slope(m) of the line non-vertical line passing through the point (x₁, y₁) and (x₂, y₂) is given by
m = \(\frac{y_2-y_1}{x_2-x_1}\) Where x₁ ≠ x₂
So, the slope of the line AB (m₁) = \(\frac{5-4}{3-4}\) = -1
The Slope of the line BC (m₂) = \(\frac{-1-5}{-1-3}\) = \(\frac {3}{2}\)
The slope of the line CA (m₃) = \(\frac{4+1}{4+1}\) = 1
It is observed that m₁. m₃ = -1 . 1 = -1
Hence, the lines AB and CA are perpendicular to each other.
Given triangle is right-angle at A(4, 4)
And the vertices of the right-angled triangle are (4, 4), (3, 5), and (-1, -1).

Question 7.
Find the slope of the line, which makes an angle of 30° with the positive direction of the y-axis measured anticlockwise.
Solution:
We know that if a line makes an angle of 30° with the positive direction of the y-axis measured anti-clockwise,
Then the x-axis is measured anti-clockwise = 90° + 30° = 120°
∴ The slope of the given line is tan 120° = tan (180° – 60°)
= -tan 60°
= -√3

Question 8.
Without using the distance formula, show that points (-2, -1), (4, 0), (3, 3), and (-3, 2) are the vertices of a parallelogram.
Solution:
Let the given point be A(-2, -1), B(4, 0), C(3, 3), and D(-3, 2)
So, now, the slope of AB = \(\frac{0+1}{4+2}=\frac{1}{6}\)
The slope of CD = \(\frac{3-2}{3+3}=\frac{1}{6}\)
Hence, the slope of AB = the slope of CD
∴ AB || CD
Now, The slope of BC = \(\frac{3-0}{3-4}=\frac{3}{-1}\) = -3
The slope of AD = \(\frac{2+1}{-3+2}=\frac{3}{-1}\) = -1
Hence, the slope of BC = the slope of AD
∴ BC || AD

Question 9.
Find the angle between the x-axis and the line joining the points (3, -1) and (4, -2).
Solution:
The slope of the line joining the points (3, -1) and (4, -2) is given by

The angle of inclination of the line joining the points (3, -1) and (4, -2) is given by tan θ = -1.
θ = (90° + 45°) = 135°
∴ The angle between the x-axis and the line joining the points (3, -1) and (4, -2) is 135°.

Question 10.
The slope of a line is double of the slope of another line. If the tangent of the angle between them is \(\frac {1}{3}\), find the slopes of the lines.
Solution:
Let us consider “m₁” and “m” to be the slope of the two given lines such that m₁ = 2m.
We know that if θ is the angle between the lines l₁ and l₂ with slopes m₁ and m₂, then
tan θ = \(\left|\frac{\left(m_2-m_1\right)}{\left(1+m_1 m_2\right)}\right|\)
Given here that the tangent of the angle between the two lines is \(\frac {1}{3}\)
So, \(\frac{1}{3}=\left|\frac{\mathrm{m}-2 \mathrm{~m}}{1+2 \mathrm{~m} \times \mathrm{m}}\right|=\left|\frac{-\mathrm{m}}{1+2 \mathrm{~m}^2}\right|\)
\(\frac{1}{3}=\frac{\mathrm{m}}{1+2 \mathrm{~m}^2}\)
Now, Case-1:
\(\frac{1}{3}=\frac{-m}{1+2 m^2}\)
⇒ 1 + 2m² = -3m
⇒ 2m² + 1 + 3m = 0
⇒ 2m(m + 1) + 1(m + 1) = 0
⇒ (2m + 1) (m + 1) = 0
⇒ m = -1 or \(\frac {-1}{2}\)
If m = -1, then the slopes of the lines are -1 and -2
If m = \(\frac {-1}{2}\), then the slopes of the lines are \(\frac {-1}{2}\) and -1
Case-2:
\(\frac{1}{3}=\frac{-m}{1+2 m^2}\)
⇒ 2m² – 3m + 1 = 0
⇒ 2m² – 2m – m + 1 = 0
⇒ 2m(m – 1) – 1(m – 1) = 0
⇒ m = 1 or \(\frac {1}{2}\)
If m = 1, then the slopes of the lines are 1 and 2.
If m = \(\frac {1}{2}\), then the slopes of the lines are \(\frac {1}{2}\) and 1.
∴ The slopes of the lines are [-1 and -2] or [\(\frac {-1}{2}\) and -1] or [1 and 2] or [\(\frac {1}{2}\) and 1].

Question 11.
A line passes through (x₁, y₁) and (h, k). If the slope of the line is m, show that k – y₁ = m(h – x₁).
Solution:
Given the slope of the line is “m.”
The slope of the line passing through (x₁, y₁) and (h, k) is (k – y₁) (h – x₁)
So, (k – y₁) / (h – x₁) = m
⇒ (k – y₁) = m(h – x₁)
Hence proved.